"A trooper is moving due south along the freeway at a speed of 28 m/s. At time t = 0, a red car passes the trooper. The red car moves with constant velocity of 40 m/s southward. At the instant the trooper's car is passed, the trooper begins to speed up at a constant rate of 2.9 m/s2. What is the maximum distance ahead of the trooper that is reached by the red car?"

24.83 m

Explanation:

Applying the equation of motion;

d = vt + 0.5at^2 1

Where;

d = distance

v = velocity

t = time

a = acceleration

For the trooper;

v = 28 m/s

a = 2.9 m/s^2

Substituting into equation 1;

d1 = 28t + 0.5(2.9t^2)

d1 = 28t + 1.45t^2

For the red car;

v = 40 m/s

a = 0

Substituting into equation 1

d2 = 40t

The difference in distance is;

d = d2 - d1

d = 40t - (28t + 1.45t^2)

d = 12t - 1.45t^2

The maximum distance is at d(d)/dt = 0

differentiating d;

d' = 12 - 2.9t = 0

2.9t = 12

t = 12/2.9 = 4.137931034482

t = 4.138 s

Substituting t into function d;

d(max) = 12(4.138) - 1.45(4.138^2)

d(max) = 24.8275862 = 24.83 m

the maximum distance ahead of the trooper that is reached by the red car is 24.83 m

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