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

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SUMMARY

The problem involves calculating the time it takes for a police car, starting from rest with a uniform acceleration of 2.94 m/s², to overtake a speeder traveling at a constant speed of 27.0 m/s. Utilizing the SUVAT equations for constant acceleration, the distance traveled by both vehicles can be equated to find the time at which the police car catches up to the speeder. The solution requires setting the distance equations equal and solving for time, confirming that both vehicles reach the same point simultaneously.

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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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