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

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To determine how long it takes for the police car to overtake the speeder, the problem involves the speeder traveling at a constant speed of 27.0 m/s while the police car accelerates from rest at 2.94 m/s². The SUVAT equations for constant acceleration are applicable here, allowing for the calculation of the distance each vehicle travels until they meet. By setting the distances equal, the time can be calculated when both vehicles are at the same position. The discussion emphasizes breaking down the problem using these equations to find the solution effectively.
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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!
 
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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.
 
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