Police Question: How Far Does a Speeder Get Before Being Overtaken?

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SUMMARY

The discussion centers on a physics problem involving a speeder and a police car, where the speeder travels at 30.0 m/s and the police car accelerates from rest at 2.44 m/s². The time taken for the police car to overtake the speeder is calculated to be 24.6 seconds, while the distance covered by the speeder before being overtaken is 738 meters. The solution involves using the equations of motion, specifically the formula for final velocity and distance, leading to a quadratic equation to find the time of intersection.

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Police Question...

Question:

A speeder passes a parked police car at 30.0 m/s. The police car starts from rest with a uniform acceleration of 2.44 m/s^2.

a) How much time passes before the speeder is overtaken my the police car? Ans: 24.6 s

b) How far does the speeder get before being overtaken by the police car?
Ans: 738 m

My Questions/Work:

To the best of my knowledge, we know the following...

Cop Vi: 0.0 m/s
Cop Accel: 2.44 m/s2
Spdr V: 30 m/s

a) Formula: Vf = Vi + AT ?

What should I use as the Cop's Final Velocity?
 
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I presume you were given those answers...

From the formula you give v=u+at, you need to integrate to get one for distance s=ut+at^2/2.

You can then equate this formula using the data for both cars - it will be a quadratic with 2 solutions - one for the time when the cars are together at t=0 and the other for the time when they catch up.

When you've got this time, the distance is easily found using the formula again.

I haven't done this since school - it was a struggle remembering the method :biggrin:
 

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