"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

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.