Physics Question: Time for Speeder to be Overtaken by Police Car?

[Tmosh]

A speeder passes a parked police car at 27.0 m/s. The police car starts from rest with a uniform acceleration of 2.94 m/s2. cou;d u do this?

question is:
How much time passes before the speeder is overtaken by the police car?


I would really appreciate some help :S I'm stumped!

 

[loonychune]

Try to take this question to pieces...

A speeder passes a parked policar car at 27.0 m/s

Thus the velocity of the speeder is 27.0 m/s and I'm assuming it remains at that speed.

The police car starts from rest with a uniform acceleration of 2.94 m/s^2

The initial velocity of the police car then is u = 0m/s and its acceleration is 2.94m/s^2 and this does not change.

You should be immediately looking to use the SUVAT equations (constant acceleration) and probably want to set up this problem so that both cars are at the exact same point in space when we can consider the police car to have caught up/overtaken the speeder... i.e. they've both traveled the same distance, s.

 

[baba89]

To answer this question, we can use the equation of motion for constant acceleration, which is:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the speeder's initial velocity is 27.0 m/s and the police car's initial velocity is 0 m/s. The acceleration of the police car is 2.94 m/s^2.
We want to find the time it takes for the police car to catch up to the speeder, so we can set v for both vehicles to be equal.
27.0 m/s = 0 m/s + 2.94 m/s^2 * t
Solving for t, we get t = 9.18 seconds.
Therefore, it will take 9.18 seconds for the police car to catch up to the speeder.