Question

A trooper is moving due south along the freeway at a speed of 23 m/s. at time t = 0, a red car passes the trooper. the red car moves with constant velocity of 31 m/s southward. at the instant the trooper's car is passed, the trooper begins to speed up at a constant rate of 2.3 m/s2. what is the maximum distance ahead of the trooper that is reached by the red car?

Answer 1

t = 8/1.7 = 4.706 s <= time at which max separation between cars occurs 
red car distance from pass point after time(t) = 36(4.706) = 169.4 m 
trooper's car distance from pass point after time(t) = 28(4.706) + 0.85(4.706)² = 150.6 m 
max separation distance = 169.4 - 150.6 = 18.8 m ANS
Hope this Helps! :D

Answer 2

Answer: The maximum distance is 13.9m

Explanation: The initial velocity of the tropper is 23m/s, The initial velocity of the red car is 31m/s and both of them move to the south, so this is a one dimensional problem.

The acceleration of the tropper starts at t = 0s, and is equl to 2.3m/s^2.

Now, the position equations for both cars are obtained by integrating in time. (We assume that the initial position of both cars, at t= 0, is the same, so we do not have constants of integration in the position)

red car:

v(t) = 31m/s

p(t) = 31m/s*t

tropper:

A(t) = 2.3m/s^2

V(t) = 2.3m/s^2*t + 23m/s

P(t) = 1.15m/s^2*t^2 + 23m/s*t

We want to find the time at wich the equation difference between p(t) and P(t) is bigger. This would be at the point in wich the velocity of the tropper is equal at the velocity of the red car, and after that point the velocity of the tropper is bigger than the one of the car, so after that point the distance between the cars will start to decrease, so first we have:

2.3m/s^2*t + 23m/s = 31m/s

2.3m/s^2*t = 8m/s

t = (8/2.3)s = 3.48s

Now, the maximum distance will be equal to:

p(3.48s) - P(3.48s) = (31*3.48 - 1.15*3.48^2 - 23*3.48)m = 13.9m