Question

A speeding car traveling at 41 m/s passes a parked police car. One second after getting passed, the police car begins pursuit. The police car accelerates at a rate of 7.5 m/s/s. The police car catches up after 12.8 seconds and the police car travels 527 meters.

What is the velocity of the police car when it catches up to the speeding car?

Answer 1

To solve this problem, we can use the equation:

distance = initial velocity x time + 1/2 x acceleration x time^2

First, we need to find the initial distance between the two cars. The speeding car travels for 1 second before the police car begins pursuit, so its initial distance from the parked police car is:

initial distance = 41 m/s x 1 s = 41 m

Now we can use the equation to find the time it takes for the police car to catch up to the speeding car:

distance = initial velocity x time + 1/2 x acceleration x time^2

527 m = 0 m/s x t + 1/2 x 7.5 m/s^2 x t^2

Simplifying:

t = sqrt((2 x 527 m) / 7.5 m/s^2) = 12.92 s

So the police car catches up to the speeding car after 12.92 seconds. Now we can use the equation:

final velocity = initial velocity + acceleration x time

to find the velocity of the police car when it catches up to the speeding car:

final velocity = 0 m/s + 7.5 m/s^2 x 12.92 s = 96.9 m/s

Therefore, the velocity of the police car when it catches up to the speeding car is 96.9 m/s.

Step-by-step explanation: