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

[arizona_cards_11]

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?

 

[J77]

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

 

[kyle8921]

I would recommend using the formula d = vi*t + 1/2*a*t^2 to calculate the distance traveled by both the speeder and the police car. In this case, the initial velocity for the police car is 0 m/s, and the acceleration is 2.44 m/s^2. The initial velocity for the speeder is 30 m/s, and we can solve for the time it takes for the police car to catch up to the speeder using the formula vf = vi + at. We can then plug in the calculated time into the distance formula to determine how far the speeder has traveled before being overtaken by the police car. This will provide a more accurate answer than just using the final velocity for the police car.