Seee the code below:
% [Car B] is traveling a distance d ahead of [Car A]. Both cars are traveling
% at the same initial velocity [50 km/h to 100 km/h] until the autonomous vehicle
% [Car B] suddenly applies the brakes. This is due to a pedestrian walking onto the
% street, causing the car to decelerate at 9 m/s^2. It takes the autonomous vehicle
% [Car A] 0.75 s to react. When [Car A] applies it's breaks, it decelerates at
% 12 m/s^2.
% Determine the minimum distance d between the cars to avoid a collision.
% velocity is v
% initial velocity is v0
% initial position is s0
% position of object after time is s
% Declare the needed symbolic variables at the start
syms v0 time d t tb
% First calculate the velocity of [Car B] as a function of time with constant
% acceleration
% initial velocity will run from 50 km/h to 100 km/h
for v0=50:1:100
% vb=v0b-ab*tb
vb=v0-9*t;
end
% Calculate the displacement of the [Car B] as a function of time with
% constant acceleration
for v0=50:1:100
% sb=s0b+v0b*tb+0.5*ab*t^2
s0b=v0*t;
sb=s0b+v0*tb+0.5*(-9)*t^2;
end
% Next calculate the velocity of [Car A] as a function of time with constant
% acceleration
for v0=50:1:100
% va=v0+aa*ta
va=v0-12*(t-0.75);
end
% Calculate the displacement of the [Car A] as a function of time with
% constant acceleration
for v0=50:1:100
s0a=v0*t;
% sa=s0a+v0*ta+0.5*aa*ta^2
sa=s0a*(0.75)+v0*(t-0.75)+0.5*(-12)*(t-0.75)^2;
end
% Next calculate the time taken for the moment of closest approach by
% equating the velocity of [Car B] to [Car A]
for v0=50:1:100
eqn1 = v0-9*t == v0-12*(t-0.75);
time=solve(eqn1,t);
end
% Calculate the minimum distance between the cars to avoid collision by
% equating the displacement of [Car A] and [Car B]
for v0=50:1:100
eqn2 = d+v0*time-0.5*(9)*time^2==(v0*0.75)+(v0*time)-(v0*0.75)+0.5*(-12)*(time-0.75)^2
distance=solve(eqn2,d)
end
Maybe consider to save the results of the loop in an symbolic array to work with them later.
