Solving Drag Racing Problem: Who Wins and By How Much Time?

[Clutch Cargo]

I have a problem in my D.E. class that is driving me nuts.

Two drivers, A and B are in a race. Beginning from a standing start they both proceed at a constant acceleration. Driver A covers the last 1/4 of the distance in 3 seconds. Driver B covers the last 1/3 of the distance in 4 seconds. Who wins and by how much time?

I am assuming the distance is 1320ft or 1/4 mile.

I know that driver B wins by .594 seconds but I don't know how that number was reached. Can anyone help?

 

[YellowTaxi]

Use s = ut+1/2.a.t.t
separately for each car

s = 1/4mile for both, and u = 0 (initial speed) for both

 

[Leong]

Let the total distance of the race to be x.

Then,

Formula used: $$s = ut + \frac{1}{2}at^2$$

Consider the overall race:

Driver A: $$x = \frac{1}{2}a_at_a^2...(1)$$ since the initial speed, u is zero.

Driver B: $$x = \frac{1}{2}a_bt_b^2...(2)$$ since the initial speed, u is zero.

Next,

Find the velocity of A when reaches 3/4 of the distance (just before entering the last 1/4 of the distance)

Find the velocity of B when reaches 2/3 of the distance (just before entering the last 1/3 of the distance.

Formula used: $$v^2 = u^2 + 2as$$

Consider from the beginning of the race until the above mentioned point.

Driver A: $$v_a^2 = 2a_a(\frac{3}{4}x)$$ since the initial speed, u is zero.
Simplified :
$$v_a^2 = \frac{3a_ax}{2}...(3)$$

Driver B: $$v_b^2 = 2a_b(\frac{2}{3}x)$$ since the initial speed, u is zero.
Simplified :
$$v_b^2 = \frac{4a_bx}{3}... (4)$$

Next,
consider the last part of the race.

Formula used: $$s = ut + \frac{1}{2}at^2$$

Driver A : $$\frac{1}{4}x = \sqrt{\frac{3a_ax}{2}}(3) + \frac{1}{2}a_a(9)$$

Solve for $$a_a \ in \ term \ of \ x: a_a = 0.0039887x; \ \ \ \ 0.77379x...(5)$$

Driver B : $$\frac{1}{3}x = \sqrt{\frac{4a_4bax}{3}}(4) + \frac{1}{2}a_b(16)$$

Solve for $$a_b \ in \ term \ of \ x: a_b = 0.41246x; \ \ \ \ \ 0.0042092x...(6)$$

Next,

put (5) into (1),

$$t_a = 22.3915\ s\ for \ a_a = 0.0039887x; 1.6077 s \ for \ a_a = 0.77379x \ which \ is \ not \ possible \ since \ the \ last \ part \ has\ already \ taken \ 3 s.$$

put (6) into (2),

$$t_b = 21.7979\ s\ for \ a_b = 0.0042092x; 2.2020 s \ for \ a_b = 0.41246x \ which \ is \ not \ possible \ since \ the \ last \ part \ has\ already \ taken \ 4 s.$$

Since $$t_b$$ is shorter, he is the winner. He is faster by 22.3915 - 21.7979 = 0.594 s