Question

An unmarked police car traveling a constant 85 km/h is passed by a speeder. Precisely 2.00 s after the speeder passes, the police officer steps on the acceleratorIf the police car accelerates uniformly at 2.00 m/s2 and overtakes the speeder after accelerating for 9.00 s , what was the speeder's speed?.

Answer 1

To find the speeder's speed, we need to determine the distance the police car and the speeder traveled before they met. By using the equation for uniformly accelerated motion, we can find the distance traveled by the police car. The time it took for the speeder to travel that distance can then be used to calculate the speeder's speed.

Step-by-step explanation:

To find the speeder's speed, we need to determine the distance the police car and the speeder traveled before they met.

First, let's calculate the distance the police car traveled.

Using the equation for uniformly accelerated motion, we can find the distance traveled:

d = vot + 0.5at^2

Let's assume the distance the police car traveled is d.

Since the police car accelerated for 9.00 s at an acceleration of 2.00 m/s^2, we can substitute the values into the equation:

d = 0 + 0.5(2.00)(9.00^2)

Simplifying this equation gives us:

d = 0 + 0.5(2.00)(81.0)

d = 0 + 81.0 m

d = 81.0 m

So, the distance the police car traveled before catching the speeder was 81.0 meters.

Now, let's calculate the time it took for the speeder to travel that distance.

The speeder passed the police car and 2.00 seconds later, the police car started accelerating. Therefore, the time it took for the police car to catch up with the speeder is 2.00 seconds.

We can now find the speeder's speed by dividing the distance traveled by the time taken:

speeder's speed = distance / time

speeder's speed = 81.0 m / 2.00 s

speeder's speed = 40.5 m/s

Therefore, the speeder's speed was 40.5 m/s.