The driver of a Cadillac is traveling at a constant 60mi/hr. A police car is 20m behind the perpetrator when it too reaches 60 mi/hr and at that moment the officer floors the gas pedal.

If the police car roars up to the rear of the Cadillac 2.0s later what was its acceleration, assuming it to be constant?

V=60mi/h * 1600m/mi*1h/3600s.=26.67m/s

d2 = d1 + 20 m
Vo*t + 0.5a*t^2 = V1*t + 20
26.67*2 + 0.5a*2^2 = 26.67*2 + 20
53.33 + 2a = 73.33
2a = 20
a = 10 m/s^2.

To find the acceleration of the police car, we can use the kinematic equation:

v = u + at

Where:
v = final velocity of the police car (60 mi/hr)
u = initial velocity of the police car (0 mi/hr)
a = acceleration of the police car
t = time taken (2.0 s)

First, let's convert the velocities from miles per hour (mi/hr) to meters per second (m/s). We know that:

1 mile = 1609.34 meters
1 hour = 3600 seconds

So, the initial velocity of the police car (u) is 0 mi/hr, which is 0 m/s.

The final velocity of the police car (v) is 60 mi/hr, which is (60 * 1609.34) / 3600 m/s = 26.8222 m/s.

Substituting the known values into the kinematic equation:

26.8222 = 0 + a * 2.0

Rearranging the equation:

a = (26.8222 - 0) / 2.0
a = 13.4111 m/s^2

Therefore, the acceleration of the police car is 13.4111 m/s^2.