Two hotrodders start from rest in a drag race. The drag race distance is D. The

Question

Two hotrodders start from rest in a drag race. The drag race distance is D. The first car has a constant acceleration A and reaches his maximum speed in a distance 1 / 2. D The second car has a constant acceleration a but reaches his maximum speed in a distance 1 / 2 D. How long does it take the first car to finish the race? How long does it take the second car to finish the race? What is the algebraic relationship between the acceleration of the first and second car such that the race ends in a tie? Plot position vs time and velocity vs time and acceleration vs time for both cars throughout the race if it ends in a tie.

Explanation / Answer

Accelerating:
1/2 D = AT12/2 => T1 = v(D/A) Top speed is V = A T1 = A v(D/A) = v(D A) So time for this part of race is rest of track/speed(t = s/v): T2 = 1/2 D / V = D/2V = D/(2v(DA))=v(D/A)/2 Overall time: T = 3/2 v(D/A)
b) second car will need t = t1+t2
Accelerating:
3/4 D = a t12/2 => t1 = v(3D/2a)
Top speed is v = a t1 = a v(3D/2a) = v(3Da/2)
So time is rest of track/speed: t1 = 1/4 D / v = D/4v = D/4v(3Da/2) = v(2D/3a)/4
t = t1+t2 = v(3D/2a) * (1 + 2/3 * 1/4) = v(3D/2a) * (1 + 1/6) = v(3D/2a) * 7/6
c) T = t
T2 = t2
(3/2)2 D/A = 3D/2a (7/6)2
9/4A = 3/2a * 49/36    /* 36
9*9/A = 49*3/2a         /* 1/3
27/A = 49/2a            /* 2
54a = 49A (i did this quickly, you should double check it) d) on each graph, on time axis the most important moments are t=0, t=t1, t=T1 and t=T(time to finish the race)
I do not have time for plotting graphs, and custom diagram tool isnt working properly    v(3D/2a) * 7/6
c) T = t
T2 = t2
(3/2)2 D/A = 3D/2a (7/6)2
9/4A = 3/2a * 49/36    /* 36
9*9/A = 49*3/2a         /* 1/3
27/A = 49/2a            /* 2
54a = 49A (i did this quickly, you should double check it) d) on each graph, on time axis the most important moments are t=0, t=t1, t=T1 and t=T(time to finish the race)
I do not have time for plotting graphs, and custom diagram tool isnt working properly   

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