Question

Two stunt drivers drive directly toward each other. At time t=0 the two cars are a distance D apart, car 1 is at rest, and car 2 is moving to the left with speed v_0. Car 1 begins to move at t=0, speeding up with a constant acceleration a_x. Car 2 continues to move with a constant velocity.

Find the speed of car 1 just before it collides with car 2.

Note: The correct answer does not depend on the variable: t

Answer 1

Two stunt drivers drive directly toward each other. At time t=0 the two cars are a distance D apart, car 1 is at rest, and car 2 is moving to the left with speed Vo. Car 1 begins to move at t=0, speeding up with a constant acceleration Ax. Car 2 continues to move with a constant velocity.

Find the speed of car 1 just before it collides with car 2.

D = the distance between them at t = 0.
V2 = the constant velocity of car 2.
A = the constant acceleration of car 2.
x = the time of travel of car 2 to impact with catr 1.
t = the time of travel to impact.

1--t = x/V2 or x = tV2
2--D - x = At^2/2
3--D - x + x = D = At^2/2 + tV2
4--At^2/2 + tV2 - D = 0
5--At^2 + 2tV2 - 2D = 0
6--Using the quadratic equation,
---t = [sqrt(V2^2 + 2DA)- V2]/a
7--From Vf = Vo + At, the final velocity of car 1 at impact is
---Vf1 = sqrt(V2^2 + 2DA) - V2

Example:
D = 1000 miles
V2 = 100 miles/hr
A = 10 miles/hr^2

After a few guesses, x = 732 miles from which t = 7.32 hr, from which (D - x) = 268 miles which should equal AT^2/2 = 10(7.32)^2/2 = 268 miles and Vf = At = 73.2 mph.

Using Vf1 = sqrt(V2^2) + 2DA0) - V2,
Vf1 = sqrt((100^2) 2(1000)10 - 100 = 73.2mph.

Therefore, Vf1 = sqrt(V2^2 + 2DA) - V2