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

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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.
At what time do the cars collide?

Find the speed of car 1 right before it crashes into car 2.

Physics

1 answer:

inna [3.1K] · Answer 1 · 2026-04-01T16:17:46+03:00

Hello! We are aware that: The distance separating the vehicles at t=0 is D. Car 2 begins with a velocity of v0 and has no acceleration. Car 1 starts from rest with an acceleration of ax, which begins at t=0. We can derive the acceleration for car 1 by integrating it over time, yielding v(t) = ax*t, where there’s no integration constant because car 1 has an initial velocity of zero. Given that the vehicles are approaching each other, we need to find out at what time t they will collide; this is effectively determining: position of car 1 + position of car 2 = D. Hence, we can disregard integration constants. For each vehicle's position, we integrate again: P1(t) = (1/2)ax*t² and P2(t) = v0t. Thus, v0t + (1/2)ax*t² = D, simplifying to v0t + (1/2)ax*t² - D = 0. This can be solved for t using the Bhaskara formula, but we cannot derive a solution without knowing the specific values for v0, D, and ax. However, you can substitute these values into the equation to find the time, ensuring to select the positive solution (as this quadratic equation presents two options). Now, we seek the velocity of car 1 just prior to the collision; this can be calculated by substituting the time we just determined into the velocity equation for car 1.