How Far Can a Car Travel Before a Collision?

[rgalvan2]

Here's the problem:
A certain automobile can decelerate at |a1| = 1.7 m/s2. Traveling at a constant v1 = 30 m/s, this car comes up behind a car traveling at a constant v2 = 9 m/s.

a) What is the speed of the faster car relative to the slower car?
b) How close to the slower car can the driver of the faster car come before applying his brakes and still avoid a collision?
c) At what time does the inevitable collision of the two cars occur?
d) How far beyond its position at t = 0 does the slower car get before it is hit?

I figured out a) was 21 m/s and b) was 129.7 m. For part c) is says to find the relative speed first which was my answer for a) but I cannot figure out what to do from here. I also need help with part d). Thanks!

 

[jbsteele]

a) The speed of the faster car relative to the slower car is 21 m/s. b) The driver of the faster car can come up to 129.7 m before applying the brakes and still avoid a collision. c) The inevitable collision of the two cars will occur when the distance between them has decreased by 129.7 m. This can be calculated by solving the equation 0.5*a1*t^2 + v1*t - 129.7 = 0, where t is the time after which the collision occurs. d) The slower car will get 129.7 m beyond its position at t = 0 before it is hit.

 

[baba89]

a) The relative speed of the faster car relative to the slower car can be calculated by subtracting their velocities: v_rel = v1 - v2 = 30 m/s - 9 m/s = 21 m/s.

b) To calculate how close the faster car can come to the slower car before applying its brakes and avoiding a collision, we can use the equation s = v^2 / (2a), where s is the stopping distance, v is the initial velocity, and a is the deceleration. Plugging in the given values, we get s = (30 m/s)^2 / (2*1.7 m/s^2) = 129.7 m. This means that the driver of the faster car can come as close as 129.7 m to the slower car before applying the brakes and still avoid a collision.

c) To find the time of the inevitable collision, we can use the equation v = v0 + at, where v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time. In this case, v = 9 m/s (since both cars will have the same final velocity at the time of collision), v0 = 30 m/s, and a = -1.7 m/s^2 (since the cars are decelerating). Solving for t, we get t = (v - v0) / a = (9 m/s - 30 m/s) / -1.7 m/s^2 = 12.35 seconds. Therefore, the inevitable collision will occur after 12.35 seconds.

d) To find how far beyond its initial position the slower car will get before it is hit, we can use the equation s = v0t + (1/2)at^2, where s is the displacement, v0 is the initial velocity, a is the acceleration, and t is the time. In this case, s = ?, v0 = 9 m/s, a = -1.7 m/s^2, and t = 12.35 seconds (from part c). Solving for s, we get s = (9 m/s * 12.35 seconds) + (0.5 * -1.7 m/s^2 * (12.35 seconds)^2) = 81.9 m. Therefore, the slower car will travel 81.