{"group":{"id":1,"name":"Community","lockable":false,"created_at":"2012-01-18T18:02:15.000Z","updated_at":"2025-12-14T01:33:56.000Z","description":"Problems submitted by members of the MATLAB Central community.","is_default":true,"created_by":161519,"badge_id":null,"featured":false,"trending":false,"solution_count_in_trending_period":0,"trending_last_calculated":"2025-12-14T00:00:00.000Z","image_id":null,"published":true,"community_created":false,"status_id":2,"is_default_group_for_player":false,"deleted_by":null,"deleted_at":null,"restored_by":null,"restored_at":null,"description_opc":null,"description_html":null,"published_at":null},"problems":[{"id":61183,"title":"Estimate brake line pressure required for a given force.","description":"Hydraulic braking systems amplify pedal input to generate braking force. Given braking force and piston area, compute the hydraulic pressure required inside the brake lines.","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(33, 33, 33); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none; white-space: normal; \"\u003e\u003cdiv style=\"block-size: 42px; display: block; min-width: 0px; padding-block-start: 0px; padding-inline-start: 2px; padding-left: 2px; padding-top: 0px; perspective-origin: 407px 21px; transform-origin: 407px 21px; vertical-align: baseline; \"\u003e\u003cdiv style=\"font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eHydraulic braking systems amplify pedal input to generate braking force. Given braking force and piston area, compute the hydraulic pressure required inside the brake lines.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function P = brakePressure(F,A)\r\nP = 0;\r\nend\r\n","test_suite":"%%\r\nF = 4000; A = 0.004;\r\nP_correct = 1e6;\r\nassert(abs(brakePressure(F,A)-P_correct) \u003c 1)\r\n\r\n%%\r\nF = 3000; A = 0.003;\r\nP_correct = 1e6;\r\nassert(abs(brakePressure(F,A)-P_correct) \u003c 1)\r\n\r\n%%\r\nF = 0; A = 0.005;\r\nP_correct = 0;\r\nassert(isequal(brakePressure(F,A),P_correct))\r\n","published":true,"deleted":false,"likes_count":1,"comments_count":0,"created_by":2305225,"edited_by":2305225,"edited_at":"2026-02-02T06:25:47.000Z","deleted_by":null,"deleted_at":null,"solvers_count":40,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2026-02-02T06:25:43.000Z","updated_at":"2026-04-04T03:06:32.000Z","published_at":"2026-02-02T06:25:47.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eHydraulic braking systems amplify pedal input to generate braking force. Given braking force and piston area, compute the hydraulic pressure required inside the brake lines.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":61188,"title":"Compute optimal regenerative and friction brake torque blending.","description":"Electric and hybrid vehicles combine regenerative braking with traditional friction braking to maximize energy recovery while ensuring safety.\r\nGiven total braking torque demand and maximum regenerative torque capability, compute how braking effort should be split between regenerative and friction braking.","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(33, 33, 33); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none; white-space: normal; \"\u003e\u003cdiv style=\"block-size: 93px; display: block; min-width: 0px; padding-block-start: 0px; padding-inline-start: 2px; padding-left: 2px; padding-top: 0px; perspective-origin: 407px 46.5px; transform-origin: 407px 46.5px; vertical-align: baseline; \"\u003e\u003cdiv style=\"block-size: 42px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eElectric and hybrid vehicles combine regenerative braking with traditional friction braking to maximize energy recovery while ensuring safety.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003cdiv style=\"block-size: 42px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eGiven total braking torque demand and maximum regenerative torque capability, compute how braking effort should be split between regenerative and friction braking.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function [Tregen, Tfric] = brakeBlending(T_total, Tregen_max)\r\nTregen = 0; Tfric = 0;\r\nend\r\n","test_suite":"%%\r\nT_total = 300; Tregen_max = 200;\r\n[Tregen,Tfric] = brakeBlending(T_total,Tregen_max);\r\nassert(isequal([Tregen Tfric],[200 100]))\r\n\r\n%%\r\nT_total = 150; Tregen_max = 200;\r\n[Tregen,Tfric] = brakeBlending(T_total,Tregen_max);\r\nassert(isequal([Tregen Tfric],[150 0]))\r\n\r\n%%\r\nT_total = 0; Tregen_max = 200;\r\n[Tregen,Tfric] = brakeBlending(T_total,Tregen_max);\r\nassert(isequal([Tregen Tfric],[0 0]))\r\n\r\n","published":true,"deleted":false,"likes_count":1,"comments_count":0,"created_by":2305225,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":33,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2026-02-02T07:28:49.000Z","updated_at":"2026-04-04T03:31:36.000Z","published_at":"2026-02-02T07:28:49.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eElectric and hybrid vehicles combine regenerative braking with traditional friction braking to maximize energy recovery while ensuring safety.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eGiven total braking torque demand and maximum regenerative torque capability, compute how braking effort should be split between regenerative and friction braking.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":61182,"title":"Compute the required brake torque at wheel to stop the car","description":"Brake torque defines how effectively braking force translates into wheel deceleration. Given braking force and wheel radius, determine the resulting torque applied at the wheel hub.","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(33, 33, 33); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none; white-space: normal; \"\u003e\u003cdiv style=\"block-size: 42px; display: block; min-width: 0px; padding-block-start: 0px; padding-inline-start: 2px; padding-left: 2px; padding-top: 0px; perspective-origin: 407px 21px; transform-origin: 407px 21px; vertical-align: baseline; \"\u003e\u003cdiv style=\"font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eBrake torque defines how effectively braking force translates into wheel deceleration. Given braking force and wheel radius, determine the resulting torque applied at the wheel hub.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function T = brakeTorque(F,r)\r\nT = 0;\r\nend\r\n","test_suite":"%%\r\nF = 3000; r = 0.3;\r\nT_correct = 900;\r\nassert(isequal(brakeTorque(F,r),T_correct))\r\n\r\n%%\r\nF = 2000; r = 0.25;\r\nT_correct = 500;\r\nassert(isequal(brakeTorque(F,r),T_correct))\r\n\r\n%%\r\nF = 0; r = 0.3;\r\nT_correct = 0;\r\nassert(isequal(brakeTorque(F,r),T_correct))\r\n\r\n","published":true,"deleted":false,"likes_count":0,"comments_count":0,"created_by":2305225,"edited_by":2305225,"edited_at":"2026-02-02T06:23:14.000Z","deleted_by":null,"deleted_at":null,"solvers_count":46,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2026-02-02T06:23:10.000Z","updated_at":"2026-04-04T03:34:26.000Z","published_at":"2026-02-02T06:23:14.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eBrake torque defines how effectively braking force translates into wheel deceleration. Given braking force and wheel radius, determine the resulting torque applied at the wheel hub.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":61181,"title":"Compute braking force using vehicle mass and acceleration.","description":"Compute braking force required to stop a vehicle of mass 'm' and with acceleration 'a' \r\nRemember:  F = m × a.","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(33, 33, 33); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none; white-space: normal; \"\u003e\u003cdiv style=\"block-size: 51px; display: block; min-width: 0px; padding-block-start: 0px; padding-inline-start: 2px; padding-left: 2px; padding-top: 0px; perspective-origin: 407px 25.5px; transform-origin: 407px 25.5px; vertical-align: baseline; \"\u003e\u003cdiv style=\"block-size: 21px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 10.5px; text-align: left; transform-origin: 383px 10.5px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eCompute braking force required to stop a vehicle of mass 'm' and with acceleration 'a' \u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003cdiv style=\"block-size: 21px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 10.5px; text-align: left; transform-origin: 383px 10.5px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eRemember:  F = m × a.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function F = brakingForce(m,a)\r\nF = 0;\r\nend\r\n","test_suite":"%%\r\nassert(isequal(brakingForce(1000,5),5000))\r\n\r\n%%\r\nassert(isequal(brakingForce(0,10),0))\r\n\r\n%%\r\nassert(isequal(brakingForce(1500,3),4500))\r\n","published":true,"deleted":false,"likes_count":0,"comments_count":0,"created_by":2305225,"edited_by":2305225,"edited_at":"2026-02-02T05:31:40.000Z","deleted_by":null,"deleted_at":null,"solvers_count":44,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2026-02-02T05:09:13.000Z","updated_at":"2026-04-04T03:25:45.000Z","published_at":"2026-02-02T05:25:08.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eCompute braking force required to stop a vehicle of mass 'm' and with acceleration 'a' \u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eRemember:  F = m × a.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":61187,"title":"Estimate brake disc temperature rise during braking.","description":"During braking, kinetic energy is converted into thermal energy, causing brake discs to heat up. Excessive temperature rise can lead to brake fade and reduced braking effectiveness.\r\nGiven braking energy absorbed by the disc, disc mass, and material heat capacity, estimate the resulting temperature increase.","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(33, 33, 33); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none; white-space: normal; \"\u003e\u003cdiv style=\"block-size: 93px; display: block; min-width: 0px; padding-block-start: 0px; padding-inline-start: 2px; padding-left: 2px; padding-top: 0px; perspective-origin: 407px 46.5px; transform-origin: 407px 46.5px; vertical-align: baseline; \"\u003e\u003cdiv style=\"block-size: 42px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eDuring braking, kinetic energy is converted into thermal energy, causing brake discs to heat up. Excessive temperature rise can lead to brake fade and reduced braking effectiveness.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003cdiv style=\"block-size: 42px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eGiven braking energy absorbed by the disc, disc mass, and material heat capacity, estimate the resulting temperature increase.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function dT = brakeDiscTempRise(E, m, Cp)\r\ndT = 0;\r\nend\r\n","test_suite":"%%\r\nE = 150000; m = 7; Cp = 500;\r\ndT_correct = 42.8571;\r\nassert(abs(brakeDiscTempRise(E,m,Cp)-dT_correct) \u003c 1e-3)\r\n\r\n%%\r\nE = 100000; m = 5; Cp = 450;\r\ndT_correct = 44.4444;\r\nassert(abs(brakeDiscTempRise(E,m,Cp)-dT_correct) \u003c 1e-3)\r\n\r\n%%\r\nE = 0; m = 6; Cp = 500;\r\ndT_correct = 0;\r\nassert(isequal(brakeDiscTempRise(E,m,Cp),dT_correct))\r\n\r\n","published":true,"deleted":false,"likes_count":0,"comments_count":0,"created_by":2305225,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":38,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2026-02-02T07:26:30.000Z","updated_at":"2026-04-04T03:21:34.000Z","published_at":"2026-02-02T07:26:30.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eDuring braking, kinetic energy is converted into thermal energy, causing brake discs to heat up. Excessive temperature rise can lead to brake fade and reduced braking effectiveness.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eGiven braking energy absorbed by the disc, disc mass, and material heat capacity, estimate the resulting temperature increase.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":61184,"title":"Estimate dynamic load transfer to front axle during braking.","description":"During braking, load shifts from the rear axle to the front axle. Given mass, deceleration, center of gravity height, and wheelbase, compute this dynamic load transfer ,a key factor in vehicle stability and brake force distribution.\r\n\r\nUse: g = 9.81;","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(33, 33, 33); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none; white-space: normal; \"\u003e\u003cdiv style=\"block-size: 102px; display: block; min-width: 0px; padding-block-start: 0px; padding-inline-start: 2px; padding-left: 2px; padding-top: 0px; perspective-origin: 407px 51px; transform-origin: 407px 51px; vertical-align: baseline; \"\u003e\u003cdiv style=\"block-size: 42px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eDuring braking, load shifts from the rear axle to the front axle. Given mass, deceleration, center of gravity height, and wheelbase, compute this dynamic load transfer ,a key factor in vehicle stability and brake force distribution.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003cdiv style=\"block-size: 21px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 10.5px; text-align: left; transform-origin: 383px 10.5px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003e\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003cdiv style=\"block-size: 21px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 10.5px; text-align: left; transform-origin: 383px 10.5px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eUse: g = 9.81;\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function dW = loadTransfer(m,a,h,L)\r\ndW = 0;\r\nend\r\n","test_suite":"%%\r\nm = 1200; a = 5; h = 0.5; L = 2.5;\r\ndW_correct = 1200;\r\nassert(abs(loadTransfer(m,a,h,L)-dW_correct) \u003c 1)\r\n\r\n%%\r\nm = 1000; a = 4; h = 0.45; L = 2.6;\r\ndW_correct = 692.3;\r\nassert(abs(loadTransfer(m,a,h,L)-dW_correct) \u003c 2)\r\n\r\n%%\r\nm = 0; a = 5; h = 0.5; L = 2.5;\r\ndW_correct = 0;\r\nassert(isequal(loadTransfer(m,a,h,L),dW_correct))\r\n\r\n","published":true,"deleted":false,"likes_count":1,"comments_count":0,"created_by":2305225,"edited_by":2305225,"edited_at":"2026-02-02T06:29:05.000Z","deleted_by":null,"deleted_at":null,"solvers_count":35,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2026-02-02T06:29:00.000Z","updated_at":"2026-04-04T03:08:43.000Z","published_at":"2026-02-02T06:29:05.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eDuring braking, load shifts from the rear axle to the front axle. Given mass, deceleration, center of gravity height, and wheelbase, compute this dynamic load transfer ,a key factor in vehicle stability and brake force distribution.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eUse: g = 9.81;\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":61180,"title":"Compute vehicle stopping distance using initial speed and constant deceleration.","description":"Given vehicle speed v (m/s) and constant deceleration a (m/s²), compute stopping distance\r\nRemember: d = v² / (2a)","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(33, 33, 33); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none; white-space: normal; \"\u003e\u003cdiv style=\"block-size: 51px; display: block; min-width: 0px; padding-block-start: 0px; padding-inline-start: 2px; padding-left: 2px; padding-top: 0px; perspective-origin: 407px 25.5px; transform-origin: 407px 25.5px; vertical-align: baseline; \"\u003e\u003cdiv style=\"block-size: 21px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 10.5px; text-align: left; transform-origin: 383px 10.5px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eGiven vehicle speed \u003c/span\u003e\u003c/span\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"font-family: Menlo, Monaco, Consolas, \u0026quot;Courier New\u0026quot;, monospace; \"\u003ev\u003c/span\u003e\u003c/span\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003e (m/s) and constant deceleration \u003c/span\u003e\u003c/span\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"font-family: Menlo, Monaco, Consolas, \u0026quot;Courier New\u0026quot;, monospace; \"\u003ea\u003c/span\u003e\u003c/span\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003e (m/s²), compute stopping distance\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003cdiv style=\"block-size: 21px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 10.5px; text-align: left; transform-origin: 383px 10.5px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eRemember: d = v² / (2a)\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function d = stoppingDistance(v,a)\r\nd = 0;\r\nend","test_suite":"%%\r\nv = 20; a = 5;\r\nd_correct = 40;\r\nassert(abs(stoppingDistance(v,a)-d_correct) \u003c 1e-6)\r\n\r\n%%\r\nv = 30; a = 10;\r\nd_correct = 45;\r\nassert(abs(stoppingDistance(v,a)-d_correct) \u003c 1e-6)\r\n\r\n%%\r\nv = 0; a = 5;\r\nd_correct = 0;\r\nassert(abs(stoppingDistance(v,a)-d_correct) \u003c 1e-6)\r\n\r\n","published":true,"deleted":false,"likes_count":1,"comments_count":0,"created_by":2305225,"edited_by":2305225,"edited_at":"2026-02-02T05:29:31.000Z","deleted_by":null,"deleted_at":null,"solvers_count":50,"test_suite_updated_at":"2026-02-02T05:29:31.000Z","rescore_all_solutions":false,"group_id":1,"created_at":"2026-02-02T04:56:51.000Z","updated_at":"2026-04-04T03:22:28.000Z","published_at":"2026-02-02T05:05:46.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eGiven vehicle speed \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:rFonts w:cs=\\\"monospace\\\"/\u003e\u003c/w:rPr\u003e\u003cw:t\u003ev\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e (m/s) and constant deceleration \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:rFonts w:cs=\\\"monospace\\\"/\u003e\u003c/w:rPr\u003e\u003cw:t\u003ea\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e (m/s²), compute stopping distance\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eRemember: d = v² / (2a)\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":61189,"title":"Simulate full-stop emergency braking scenario.","description":"Emergency braking events demand rapid deceleration to bring the vehicle safely to rest. Given initial vehicle speed and constant deceleration, simulate the braking process and determine the time required to reach zero velocity.\r\nYour solution should correctly handle typical braking cases and extreme boundary conditions.","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(33, 33, 33); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none; white-space: normal; \"\u003e\u003cdiv style=\"block-size: 72px; display: block; min-width: 0px; padding-block-start: 0px; padding-inline-start: 2px; padding-left: 2px; padding-top: 0px; perspective-origin: 407px 36px; transform-origin: 407px 36px; vertical-align: baseline; \"\u003e\u003cdiv style=\"block-size: 42px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eEmergency braking events demand rapid deceleration to bring the vehicle safely to rest. Given initial vehicle speed and constant deceleration, simulate the braking process and determine the time required to reach zero velocity.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003cdiv style=\"block-size: 21px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 10.5px; text-align: left; transform-origin: 383px 10.5px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eYour solution should correctly handle typical braking cases and extreme boundary conditions.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function t_stop = emergencyBrakingTime(v0,a)\r\nt_stop = 0;\r\nend\r\n","test_suite":"%%\r\nv0 = 25; a = 5;\r\nt_correct = 5;\r\nassert(abs(emergencyBrakingTime(v0,a)-t_correct) \u003c 1e-6)\r\n\r\n%%\r\nv0 = 30; a = 6;\r\nt_correct = 5;\r\nassert(abs(emergencyBrakingTime(v0,a)-t_correct) \u003c 1e-6)\r\n\r\n%%\r\nv0 = 0; a = 5;\r\nt_correct = 0;\r\nassert(isequal(emergencyBrakingTime(v0,a),t_correct))\r\n","published":true,"deleted":false,"likes_count":1,"comments_count":0,"created_by":2305225,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":33,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2026-02-02T07:31:11.000Z","updated_at":"2026-04-04T03:23:49.000Z","published_at":"2026-02-02T07:31:11.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eEmergency braking events demand rapid deceleration to bring the vehicle safely to rest. Given initial vehicle speed and constant deceleration, simulate the braking process and determine the time required to reach zero velocity.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eYour solution should correctly handle typical braking cases and extreme boundary conditions.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":61186,"title":"Compute optimal front–rear brake force distribution.","description":"Modern braking systems dynamically distribute braking forces between front and rear axles to maintain stability, reduce stopping distance, and prevent wheel lock.\r\nGiven total braking demand and axle load distribution, compute the optimal front and rear brake force allocation that preserves balance and traction.","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(33, 33, 33); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none; white-space: normal; \"\u003e\u003cdiv style=\"block-size: 93px; display: block; min-width: 0px; padding-block-start: 0px; padding-inline-start: 2px; padding-left: 2px; padding-top: 0px; perspective-origin: 407px 46.5px; transform-origin: 407px 46.5px; vertical-align: baseline; \"\u003e\u003cdiv style=\"block-size: 42px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eModern braking systems dynamically distribute braking forces between front and rear axles to maintain stability, reduce stopping distance, and prevent wheel lock.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003cdiv style=\"block-size: 42px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eGiven total braking demand and axle load distribution, compute the optimal front and rear brake force allocation that preserves balance and traction.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function [Ff, Fr] = brakeDistribution(F_total, front_ratio)\r\nFf = 0; Fr = 0;\r\nend\r\n","test_suite":"%%\r\nF_total = 6000; front_ratio = 0.6;\r\n[Ff,Fr] = brakeDistribution(F_total,front_ratio);\r\nassert(isequal([Ff Fr],[3600 2400]))\r\n\r\n%%\r\nF_total = 8000; front_ratio = 0.7;\r\n[Ff,Fr] = brakeDistribution(F_total,front_ratio);\r\nassert(isequal([Ff Fr],[5600 2400]))\r\n\r\n%%\r\nF_total = 0; front_ratio = 0.6;\r\n[Ff,Fr] = brakeDistribution(F_total,front_ratio);\r\nassert(isequal([Ff Fr],[0 0]))\r\n","published":true,"deleted":false,"likes_count":0,"comments_count":0,"created_by":2305225,"edited_by":2305225,"edited_at":"2026-02-02T06:39:17.000Z","deleted_by":null,"deleted_at":null,"solvers_count":31,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2026-02-02T06:39:14.000Z","updated_at":"2026-04-04T03:39:45.000Z","published_at":"2026-02-02T06:39:17.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eModern braking systems dynamically distribute braking forces between front and rear axles to maintain stability, reduce stopping distance, and prevent wheel lock.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eGiven total braking demand and axle load distribution, compute the optimal front and rear brake force allocation that preserves balance and traction.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":61185,"title":"Compute wheel slip ratio during braking.","description":"During braking, a difference develops between the vehicle’s forward speed and the rotational speed of its wheels. This difference is captured by a quantity known as wheel slip ratio, which plays a critical role in traction control, ABS algorithms, and vehicle stability systems.\r\nGiven the vehicle longitudinal speed and wheel circumferential speed, determine the corresponding slip ratio. Your implementation should handle normal driving conditions, heavy braking scenarios, and boundary cases reliably.","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(33, 33, 33); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none; white-space: normal; \"\u003e\u003cdiv style=\"block-size: 114px; display: block; min-width: 0px; padding-block-start: 0px; padding-inline-start: 2px; padding-left: 2px; padding-top: 0px; perspective-origin: 407px 57px; transform-origin: 407px 57px; vertical-align: baseline; \"\u003e\u003cdiv style=\"block-size: 63px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 31.5px; text-align: left; transform-origin: 383px 31.5px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eDuring braking, a difference develops between the vehicle’s forward speed and the rotational speed of its wheels. This difference is captured by a quantity known as \u003c/span\u003e\u003c/span\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"font-style: italic; \"\u003ewheel slip ratio\u003c/span\u003e\u003c/span\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003e, which plays a critical role in traction control, ABS algorithms, and vehicle stability systems.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003cdiv style=\"block-size: 42px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eGiven the vehicle longitudinal speed and wheel circumferential speed, determine the corresponding slip ratio. Your implementation should handle normal driving conditions, heavy braking scenarios, and boundary cases reliably.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function s = wheelSlipRatio(v_vehicle, v_wheel)\r\ns = 0;\r\nend\r\n","test_suite":"%%\r\nv_vehicle = 20; v_wheel = 18;\r\ns_correct = 0.1;\r\nassert(abs(wheelSlipRatio(v_vehicle,v_wheel) - s_correct) \u003c 1e-6)\r\n\r\n%%\r\nv_vehicle = 25; v_wheel = 0;\r\ns_correct = 1;\r\nassert(abs(wheelSlipRatio(v_vehicle,v_wheel) - s_correct) \u003c 1e-6)\r\n\r\n%%\r\nv_vehicle = 0; v_wheel = 0;\r\ns_correct = 0;\r\nassert(abs(wheelSlipRatio(v_vehicle,v_wheel) - s_correct) \u003c 1e-6)\r\n\r\n","published":true,"deleted":false,"likes_count":0,"comments_count":0,"created_by":2305225,"edited_by":2305225,"edited_at":"2026-02-02T06:36:51.000Z","deleted_by":null,"deleted_at":null,"solvers_count":32,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2026-02-02T06:36:05.000Z","updated_at":"2026-04-04T03:41:48.000Z","published_at":"2026-02-02T06:36:51.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eDuring braking, a difference develops between the vehicle’s forward speed and the rotational speed of its wheels. This difference is captured by a quantity known as \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:i/\u003e\u003c/w:rPr\u003e\u003cw:t\u003ewheel slip ratio\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e, which plays a critical role in traction control, ABS algorithms, and vehicle stability systems.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eGiven the vehicle longitudinal speed and wheel circumferential speed, determine the corresponding slip ratio. Your implementation should handle normal driving conditions, heavy braking scenarios, and boundary cases reliably.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"}],"problem_search":{"errors":[],"problems":[{"id":61183,"title":"Estimate brake line pressure required for a given force.","description":"Hydraulic braking systems amplify pedal input to generate braking force. Given braking force and piston area, compute the hydraulic pressure required inside the brake lines.","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(33, 33, 33); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none; white-space: normal; \"\u003e\u003cdiv style=\"block-size: 42px; display: block; min-width: 0px; padding-block-start: 0px; padding-inline-start: 2px; padding-left: 2px; padding-top: 0px; perspective-origin: 407px 21px; transform-origin: 407px 21px; vertical-align: baseline; \"\u003e\u003cdiv style=\"font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eHydraulic braking systems amplify pedal input to generate braking force. Given braking force and piston area, compute the hydraulic pressure required inside the brake lines.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function P = brakePressure(F,A)\r\nP = 0;\r\nend\r\n","test_suite":"%%\r\nF = 4000; A = 0.004;\r\nP_correct = 1e6;\r\nassert(abs(brakePressure(F,A)-P_correct) \u003c 1)\r\n\r\n%%\r\nF = 3000; A = 0.003;\r\nP_correct = 1e6;\r\nassert(abs(brakePressure(F,A)-P_correct) \u003c 1)\r\n\r\n%%\r\nF = 0; A = 0.005;\r\nP_correct = 0;\r\nassert(isequal(brakePressure(F,A),P_correct))\r\n","published":true,"deleted":false,"likes_count":1,"comments_count":0,"created_by":2305225,"edited_by":2305225,"edited_at":"2026-02-02T06:25:47.000Z","deleted_by":null,"deleted_at":null,"solvers_count":40,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2026-02-02T06:25:43.000Z","updated_at":"2026-04-04T03:06:32.000Z","published_at":"2026-02-02T06:25:47.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eHydraulic braking systems amplify pedal input to generate braking force. Given braking force and piston area, compute the hydraulic pressure required inside the brake lines.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":61188,"title":"Compute optimal regenerative and friction brake torque blending.","description":"Electric and hybrid vehicles combine regenerative braking with traditional friction braking to maximize energy recovery while ensuring safety.\r\nGiven total braking torque demand and maximum regenerative torque capability, compute how braking effort should be split between regenerative and friction braking.","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(33, 33, 33); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none; white-space: normal; \"\u003e\u003cdiv style=\"block-size: 93px; display: block; min-width: 0px; padding-block-start: 0px; padding-inline-start: 2px; padding-left: 2px; padding-top: 0px; perspective-origin: 407px 46.5px; transform-origin: 407px 46.5px; vertical-align: baseline; \"\u003e\u003cdiv style=\"block-size: 42px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eElectric and hybrid vehicles combine regenerative braking with traditional friction braking to maximize energy recovery while ensuring safety.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003cdiv style=\"block-size: 42px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eGiven total braking torque demand and maximum regenerative torque capability, compute how braking effort should be split between regenerative and friction braking.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function [Tregen, Tfric] = brakeBlending(T_total, Tregen_max)\r\nTregen = 0; Tfric = 0;\r\nend\r\n","test_suite":"%%\r\nT_total = 300; Tregen_max = 200;\r\n[Tregen,Tfric] = brakeBlending(T_total,Tregen_max);\r\nassert(isequal([Tregen Tfric],[200 100]))\r\n\r\n%%\r\nT_total = 150; Tregen_max = 200;\r\n[Tregen,Tfric] = brakeBlending(T_total,Tregen_max);\r\nassert(isequal([Tregen Tfric],[150 0]))\r\n\r\n%%\r\nT_total = 0; Tregen_max = 200;\r\n[Tregen,Tfric] = brakeBlending(T_total,Tregen_max);\r\nassert(isequal([Tregen Tfric],[0 0]))\r\n\r\n","published":true,"deleted":false,"likes_count":1,"comments_count":0,"created_by":2305225,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":33,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2026-02-02T07:28:49.000Z","updated_at":"2026-04-04T03:31:36.000Z","published_at":"2026-02-02T07:28:49.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eElectric and hybrid vehicles combine regenerative braking with traditional friction braking to maximize energy recovery while ensuring safety.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eGiven total braking torque demand and maximum regenerative torque capability, compute how braking effort should be split between regenerative and friction braking.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":61182,"title":"Compute the required brake torque at wheel to stop the car","description":"Brake torque defines how effectively braking force translates into wheel deceleration. Given braking force and wheel radius, determine the resulting torque applied at the wheel hub.","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(33, 33, 33); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none; white-space: normal; \"\u003e\u003cdiv style=\"block-size: 42px; display: block; min-width: 0px; padding-block-start: 0px; padding-inline-start: 2px; padding-left: 2px; padding-top: 0px; perspective-origin: 407px 21px; transform-origin: 407px 21px; vertical-align: baseline; \"\u003e\u003cdiv style=\"font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eBrake torque defines how effectively braking force translates into wheel deceleration. Given braking force and wheel radius, determine the resulting torque applied at the wheel hub.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function T = brakeTorque(F,r)\r\nT = 0;\r\nend\r\n","test_suite":"%%\r\nF = 3000; r = 0.3;\r\nT_correct = 900;\r\nassert(isequal(brakeTorque(F,r),T_correct))\r\n\r\n%%\r\nF = 2000; r = 0.25;\r\nT_correct = 500;\r\nassert(isequal(brakeTorque(F,r),T_correct))\r\n\r\n%%\r\nF = 0; r = 0.3;\r\nT_correct = 0;\r\nassert(isequal(brakeTorque(F,r),T_correct))\r\n\r\n","published":true,"deleted":false,"likes_count":0,"comments_count":0,"created_by":2305225,"edited_by":2305225,"edited_at":"2026-02-02T06:23:14.000Z","deleted_by":null,"deleted_at":null,"solvers_count":46,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2026-02-02T06:23:10.000Z","updated_at":"2026-04-04T03:34:26.000Z","published_at":"2026-02-02T06:23:14.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eBrake torque defines how effectively braking force translates into wheel deceleration. Given braking force and wheel radius, determine the resulting torque applied at the wheel hub.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":61181,"title":"Compute braking force using vehicle mass and acceleration.","description":"Compute braking force required to stop a vehicle of mass 'm' and with acceleration 'a' \r\nRemember:  F = m × a.","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(33, 33, 33); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none; white-space: normal; \"\u003e\u003cdiv style=\"block-size: 51px; display: block; min-width: 0px; padding-block-start: 0px; padding-inline-start: 2px; padding-left: 2px; padding-top: 0px; perspective-origin: 407px 25.5px; transform-origin: 407px 25.5px; vertical-align: baseline; \"\u003e\u003cdiv style=\"block-size: 21px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 10.5px; text-align: left; transform-origin: 383px 10.5px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eCompute braking force required to stop a vehicle of mass 'm' and with acceleration 'a' \u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003cdiv style=\"block-size: 21px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 10.5px; text-align: left; transform-origin: 383px 10.5px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eRemember:  F = m × a.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function F = brakingForce(m,a)\r\nF = 0;\r\nend\r\n","test_suite":"%%\r\nassert(isequal(brakingForce(1000,5),5000))\r\n\r\n%%\r\nassert(isequal(brakingForce(0,10),0))\r\n\r\n%%\r\nassert(isequal(brakingForce(1500,3),4500))\r\n","published":true,"deleted":false,"likes_count":0,"comments_count":0,"created_by":2305225,"edited_by":2305225,"edited_at":"2026-02-02T05:31:40.000Z","deleted_by":null,"deleted_at":null,"solvers_count":44,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2026-02-02T05:09:13.000Z","updated_at":"2026-04-04T03:25:45.000Z","published_at":"2026-02-02T05:25:08.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eCompute braking force required to stop a vehicle of mass 'm' and with acceleration 'a' \u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eRemember:  F = m × a.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":61187,"title":"Estimate brake disc temperature rise during braking.","description":"During braking, kinetic energy is converted into thermal energy, causing brake discs to heat up. Excessive temperature rise can lead to brake fade and reduced braking effectiveness.\r\nGiven braking energy absorbed by the disc, disc mass, and material heat capacity, estimate the resulting temperature increase.","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(33, 33, 33); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none; white-space: normal; \"\u003e\u003cdiv style=\"block-size: 93px; display: block; min-width: 0px; padding-block-start: 0px; padding-inline-start: 2px; padding-left: 2px; padding-top: 0px; perspective-origin: 407px 46.5px; transform-origin: 407px 46.5px; vertical-align: baseline; \"\u003e\u003cdiv style=\"block-size: 42px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eDuring braking, kinetic energy is converted into thermal energy, causing brake discs to heat up. Excessive temperature rise can lead to brake fade and reduced braking effectiveness.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003cdiv style=\"block-size: 42px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eGiven braking energy absorbed by the disc, disc mass, and material heat capacity, estimate the resulting temperature increase.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function dT = brakeDiscTempRise(E, m, Cp)\r\ndT = 0;\r\nend\r\n","test_suite":"%%\r\nE = 150000; m = 7; Cp = 500;\r\ndT_correct = 42.8571;\r\nassert(abs(brakeDiscTempRise(E,m,Cp)-dT_correct) \u003c 1e-3)\r\n\r\n%%\r\nE = 100000; m = 5; Cp = 450;\r\ndT_correct = 44.4444;\r\nassert(abs(brakeDiscTempRise(E,m,Cp)-dT_correct) \u003c 1e-3)\r\n\r\n%%\r\nE = 0; m = 6; Cp = 500;\r\ndT_correct = 0;\r\nassert(isequal(brakeDiscTempRise(E,m,Cp),dT_correct))\r\n\r\n","published":true,"deleted":false,"likes_count":0,"comments_count":0,"created_by":2305225,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":38,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2026-02-02T07:26:30.000Z","updated_at":"2026-04-04T03:21:34.000Z","published_at":"2026-02-02T07:26:30.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eDuring braking, kinetic energy is converted into thermal energy, causing brake discs to heat up. Excessive temperature rise can lead to brake fade and reduced braking effectiveness.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eGiven braking energy absorbed by the disc, disc mass, and material heat capacity, estimate the resulting temperature increase.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":61184,"title":"Estimate dynamic load transfer to front axle during braking.","description":"During braking, load shifts from the rear axle to the front axle. Given mass, deceleration, center of gravity height, and wheelbase, compute this dynamic load transfer ,a key factor in vehicle stability and brake force distribution.\r\n\r\nUse: g = 9.81;","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(33, 33, 33); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none; white-space: normal; \"\u003e\u003cdiv style=\"block-size: 102px; display: block; min-width: 0px; padding-block-start: 0px; padding-inline-start: 2px; padding-left: 2px; padding-top: 0px; perspective-origin: 407px 51px; transform-origin: 407px 51px; vertical-align: baseline; \"\u003e\u003cdiv style=\"block-size: 42px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eDuring braking, load shifts from the rear axle to the front axle. Given mass, deceleration, center of gravity height, and wheelbase, compute this dynamic load transfer ,a key factor in vehicle stability and brake force distribution.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003cdiv style=\"block-size: 21px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 10.5px; text-align: left; transform-origin: 383px 10.5px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003e\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003cdiv style=\"block-size: 21px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 10.5px; text-align: left; transform-origin: 383px 10.5px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eUse: g = 9.81;\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function dW = loadTransfer(m,a,h,L)\r\ndW = 0;\r\nend\r\n","test_suite":"%%\r\nm = 1200; a = 5; h = 0.5; L = 2.5;\r\ndW_correct = 1200;\r\nassert(abs(loadTransfer(m,a,h,L)-dW_correct) \u003c 1)\r\n\r\n%%\r\nm = 1000; a = 4; h = 0.45; L = 2.6;\r\ndW_correct = 692.3;\r\nassert(abs(loadTransfer(m,a,h,L)-dW_correct) \u003c 2)\r\n\r\n%%\r\nm = 0; a = 5; h = 0.5; L = 2.5;\r\ndW_correct = 0;\r\nassert(isequal(loadTransfer(m,a,h,L),dW_correct))\r\n\r\n","published":true,"deleted":false,"likes_count":1,"comments_count":0,"created_by":2305225,"edited_by":2305225,"edited_at":"2026-02-02T06:29:05.000Z","deleted_by":null,"deleted_at":null,"solvers_count":35,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2026-02-02T06:29:00.000Z","updated_at":"2026-04-04T03:08:43.000Z","published_at":"2026-02-02T06:29:05.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eDuring braking, load shifts from the rear axle to the front axle. Given mass, deceleration, center of gravity height, and wheelbase, compute this dynamic load transfer ,a key factor in vehicle stability and brake force distribution.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eUse: g = 9.81;\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":61180,"title":"Compute vehicle stopping distance using initial speed and constant deceleration.","description":"Given vehicle speed v (m/s) and constant deceleration a (m/s²), compute stopping distance\r\nRemember: d = v² / (2a)","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(33, 33, 33); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none; white-space: normal; \"\u003e\u003cdiv style=\"block-size: 51px; display: block; min-width: 0px; padding-block-start: 0px; padding-inline-start: 2px; padding-left: 2px; padding-top: 0px; perspective-origin: 407px 25.5px; transform-origin: 407px 25.5px; vertical-align: baseline; \"\u003e\u003cdiv style=\"block-size: 21px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 10.5px; text-align: left; transform-origin: 383px 10.5px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eGiven vehicle speed \u003c/span\u003e\u003c/span\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"font-family: Menlo, Monaco, Consolas, \u0026quot;Courier New\u0026quot;, monospace; \"\u003ev\u003c/span\u003e\u003c/span\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003e (m/s) and constant deceleration \u003c/span\u003e\u003c/span\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"font-family: Menlo, Monaco, Consolas, \u0026quot;Courier New\u0026quot;, monospace; \"\u003ea\u003c/span\u003e\u003c/span\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003e (m/s²), compute stopping distance\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003cdiv style=\"block-size: 21px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 10.5px; text-align: left; transform-origin: 383px 10.5px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eRemember: d = v² / (2a)\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function d = stoppingDistance(v,a)\r\nd = 0;\r\nend","test_suite":"%%\r\nv = 20; a = 5;\r\nd_correct = 40;\r\nassert(abs(stoppingDistance(v,a)-d_correct) \u003c 1e-6)\r\n\r\n%%\r\nv = 30; a = 10;\r\nd_correct = 45;\r\nassert(abs(stoppingDistance(v,a)-d_correct) \u003c 1e-6)\r\n\r\n%%\r\nv = 0; a = 5;\r\nd_correct = 0;\r\nassert(abs(stoppingDistance(v,a)-d_correct) \u003c 1e-6)\r\n\r\n","published":true,"deleted":false,"likes_count":1,"comments_count":0,"created_by":2305225,"edited_by":2305225,"edited_at":"2026-02-02T05:29:31.000Z","deleted_by":null,"deleted_at":null,"solvers_count":50,"test_suite_updated_at":"2026-02-02T05:29:31.000Z","rescore_all_solutions":false,"group_id":1,"created_at":"2026-02-02T04:56:51.000Z","updated_at":"2026-04-04T03:22:28.000Z","published_at":"2026-02-02T05:05:46.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eGiven vehicle speed \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:rFonts w:cs=\\\"monospace\\\"/\u003e\u003c/w:rPr\u003e\u003cw:t\u003ev\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e (m/s) and constant deceleration \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:rFonts w:cs=\\\"monospace\\\"/\u003e\u003c/w:rPr\u003e\u003cw:t\u003ea\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e (m/s²), compute stopping distance\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eRemember: d = v² / (2a)\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":61189,"title":"Simulate full-stop emergency braking scenario.","description":"Emergency braking events demand rapid deceleration to bring the vehicle safely to rest. Given initial vehicle speed and constant deceleration, simulate the braking process and determine the time required to reach zero velocity.\r\nYour solution should correctly handle typical braking cases and extreme boundary conditions.","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(33, 33, 33); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none; white-space: normal; \"\u003e\u003cdiv style=\"block-size: 72px; display: block; min-width: 0px; padding-block-start: 0px; padding-inline-start: 2px; padding-left: 2px; padding-top: 0px; perspective-origin: 407px 36px; transform-origin: 407px 36px; vertical-align: baseline; \"\u003e\u003cdiv style=\"block-size: 42px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eEmergency braking events demand rapid deceleration to bring the vehicle safely to rest. Given initial vehicle speed and constant deceleration, simulate the braking process and determine the time required to reach zero velocity.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003cdiv style=\"block-size: 21px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 10.5px; text-align: left; transform-origin: 383px 10.5px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eYour solution should correctly handle typical braking cases and extreme boundary conditions.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function t_stop = emergencyBrakingTime(v0,a)\r\nt_stop = 0;\r\nend\r\n","test_suite":"%%\r\nv0 = 25; a = 5;\r\nt_correct = 5;\r\nassert(abs(emergencyBrakingTime(v0,a)-t_correct) \u003c 1e-6)\r\n\r\n%%\r\nv0 = 30; a = 6;\r\nt_correct = 5;\r\nassert(abs(emergencyBrakingTime(v0,a)-t_correct) \u003c 1e-6)\r\n\r\n%%\r\nv0 = 0; a = 5;\r\nt_correct = 0;\r\nassert(isequal(emergencyBrakingTime(v0,a),t_correct))\r\n","published":true,"deleted":false,"likes_count":1,"comments_count":0,"created_by":2305225,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":33,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2026-02-02T07:31:11.000Z","updated_at":"2026-04-04T03:23:49.000Z","published_at":"2026-02-02T07:31:11.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eEmergency braking events demand rapid deceleration to bring the vehicle safely to rest. Given initial vehicle speed and constant deceleration, simulate the braking process and determine the time required to reach zero velocity.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eYour solution should correctly handle typical braking cases and extreme boundary conditions.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":61186,"title":"Compute optimal front–rear brake force distribution.","description":"Modern braking systems dynamically distribute braking forces between front and rear axles to maintain stability, reduce stopping distance, and prevent wheel lock.\r\nGiven total braking demand and axle load distribution, compute the optimal front and rear brake force allocation that preserves balance and traction.","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(33, 33, 33); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none; white-space: normal; \"\u003e\u003cdiv style=\"block-size: 93px; display: block; min-width: 0px; padding-block-start: 0px; padding-inline-start: 2px; padding-left: 2px; padding-top: 0px; perspective-origin: 407px 46.5px; transform-origin: 407px 46.5px; vertical-align: baseline; \"\u003e\u003cdiv style=\"block-size: 42px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eModern braking systems dynamically distribute braking forces between front and rear axles to maintain stability, reduce stopping distance, and prevent wheel lock.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003cdiv style=\"block-size: 42px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eGiven total braking demand and axle load distribution, compute the optimal front and rear brake force allocation that preserves balance and traction.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function [Ff, Fr] = brakeDistribution(F_total, front_ratio)\r\nFf = 0; Fr = 0;\r\nend\r\n","test_suite":"%%\r\nF_total = 6000; front_ratio = 0.6;\r\n[Ff,Fr] = brakeDistribution(F_total,front_ratio);\r\nassert(isequal([Ff Fr],[3600 2400]))\r\n\r\n%%\r\nF_total = 8000; front_ratio = 0.7;\r\n[Ff,Fr] = brakeDistribution(F_total,front_ratio);\r\nassert(isequal([Ff Fr],[5600 2400]))\r\n\r\n%%\r\nF_total = 0; front_ratio = 0.6;\r\n[Ff,Fr] = brakeDistribution(F_total,front_ratio);\r\nassert(isequal([Ff Fr],[0 0]))\r\n","published":true,"deleted":false,"likes_count":0,"comments_count":0,"created_by":2305225,"edited_by":2305225,"edited_at":"2026-02-02T06:39:17.000Z","deleted_by":null,"deleted_at":null,"solvers_count":31,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2026-02-02T06:39:14.000Z","updated_at":"2026-04-04T03:39:45.000Z","published_at":"2026-02-02T06:39:17.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eModern braking systems dynamically distribute braking forces between front and rear axles to maintain stability, reduce stopping distance, and prevent wheel lock.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eGiven total braking demand and axle load distribution, compute the optimal front and rear brake force allocation that preserves balance and traction.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\",\"relationship\":null}],\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"target\":\"/matlab/document.xml\",\"relationshipId\":\"rId1\"}]}"},{"id":61185,"title":"Compute wheel slip ratio during braking.","description":"During braking, a difference develops between the vehicle’s forward speed and the rotational speed of its wheels. This difference is captured by a quantity known as wheel slip ratio, which plays a critical role in traction control, ABS algorithms, and vehicle stability systems.\r\nGiven the vehicle longitudinal speed and wheel circumferential speed, determine the corresponding slip ratio. Your implementation should handle normal driving conditions, heavy braking scenarios, and boundary cases reliably.","description_html":"\u003cdiv style = \"text-align: start; line-height: 20.44px; min-height: 0px; white-space: normal; color: rgb(33, 33, 33); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: none; white-space: normal; \"\u003e\u003cdiv style=\"block-size: 114px; display: block; min-width: 0px; padding-block-start: 0px; padding-inline-start: 2px; padding-left: 2px; padding-top: 0px; perspective-origin: 407px 57px; transform-origin: 407px 57px; vertical-align: baseline; \"\u003e\u003cdiv style=\"block-size: 63px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 31.5px; text-align: left; transform-origin: 383px 31.5px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eDuring braking, a difference develops between the vehicle’s forward speed and the rotational speed of its wheels. This difference is captured by a quantity known as \u003c/span\u003e\u003c/span\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"font-style: italic; \"\u003ewheel slip ratio\u003c/span\u003e\u003c/span\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003e, which plays a critical role in traction control, ABS algorithms, and vehicle stability systems.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003cdiv style=\"block-size: 42px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; padding-inline-start: 0px; padding-left: 0px; perspective-origin: 383px 21px; text-align: left; transform-origin: 383px 21px; white-space-collapse: preserve; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; \"\u003e\u003cspan style=\"block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 0px 0px; transform-origin: 0px 0px; unicode-bidi: normal; \"\u003e\u003cspan style=\"\"\u003eGiven the vehicle longitudinal speed and wheel circumferential speed, determine the corresponding slip ratio. Your implementation should handle normal driving conditions, heavy braking scenarios, and boundary cases reliably.\u003c/span\u003e\u003c/span\u003e\u003c/div\u003e\u003c/div\u003e\u003c/div\u003e","function_template":"function s = wheelSlipRatio(v_vehicle, v_wheel)\r\ns = 0;\r\nend\r\n","test_suite":"%%\r\nv_vehicle = 20; v_wheel = 18;\r\ns_correct = 0.1;\r\nassert(abs(wheelSlipRatio(v_vehicle,v_wheel) - s_correct) \u003c 1e-6)\r\n\r\n%%\r\nv_vehicle = 25; v_wheel = 0;\r\ns_correct = 1;\r\nassert(abs(wheelSlipRatio(v_vehicle,v_wheel) - s_correct) \u003c 1e-6)\r\n\r\n%%\r\nv_vehicle = 0; v_wheel = 0;\r\ns_correct = 0;\r\nassert(abs(wheelSlipRatio(v_vehicle,v_wheel) - s_correct) \u003c 1e-6)\r\n\r\n","published":true,"deleted":false,"likes_count":0,"comments_count":0,"created_by":2305225,"edited_by":2305225,"edited_at":"2026-02-02T06:36:51.000Z","deleted_by":null,"deleted_at":null,"solvers_count":32,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2026-02-02T06:36:05.000Z","updated_at":"2026-04-04T03:41:48.000Z","published_at":"2026-02-02T06:36:51.000Z","restored_at":null,"restored_by":null,"spam":null,"simulink":false,"admin_reviewed":false,"description_opc":"{\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eDuring braking, a difference develops between the vehicle’s forward speed and the rotational speed of its wheels. 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