How to make a matlab code decrease the values step by step and plot a graph?

Hello sir, kindly share your suggestions to solve this problem, input is speed = 150 m/s, convert to km/hr =150*3.6 = 540 for minutes =540/60 = 9 range = 60

manually I done It

t0 = sqrt((5*9)*(5*9)+(60.^2)) = 75

t1 = sqrt((4*9)*(4*9)+(60.^2)) =69.9

t2 = sqrt((3*9)*(3*9)+(60.^2)) =65.7

t3 = sqrt((2*9)*(2*9)+(60.^2)) =62.6

t4 = sqrt((1*9)*(1*9)+(60.^2)) =60.6

t5 = sqrt((60.^2)) =60

t6 = sqrt((1*9)*(1*9)+(60.^2)) =60.6

t7 = sqrt((2*9)*(2*9)+(60.^2)) =62.6

t8 = sqrt((3*9)*(3*9)+(60.^2)) =65.7

t9 = sqrt((5*9)*(5*9)+(60.^2)) =69.9

t10 = sqrt((4*9)*(4*9)+(60.^2)) = 75

My code:

a = input(speed);

spd = a*3.6;

spd_1 = spd/60;

b =input(range);

for bb = 0:10

 c = (sqrt(spd_1*bb).^2+(b.^2))

end

kindly share the details ,how to make a loop for t0 to t10 values and plot a graph

Answer 1

[Birdman]

You do not need a for loop in this case. Use vectorized code:

a=150;b=60;
spd=a*3.6;
spd_1=spd/60;
bb=[5:-1:1 0 1:5];
c=sqrt((spd_1.*bb).^2+b.^2)

Answer 2

[KSSV]

t0 = sqrt((5*9)*(5*9)+(60.^2)) 

t1 = sqrt((4*9)*(4*9)+(60.^2)) 

t2 = sqrt((3*9)*(3*9)+(60.^2)) 

t3 = sqrt((2*9)*(2*9)+(60.^2)) 

t4 = sqrt((1*9)*(1*9)+(60.^2)) 

N = 5 ;
T = zeros(N,1) ;
for i = 1:N
    T(i) = sqrt((i*9)*(i*9)+(60.^2)) ;
end

[sonu]

thank you!!!! with slight modification i got it