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Police Question .

  • #1
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?
 

Answers and Replies

  • #2
J77
1,076
1
I presume you were given those answers...

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

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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