Help! physics prob!?

Question

a speeder passes a parked police car at a constant speed of 24 m/s. At that instant, the police car starts from rest with a uniform acceleration of 2.8 m/s^2.

How much time t passes before the speeder get before being overtaken by the police car?

and

how far in distance does the speeder get before being overtaken by the police car?

Answer 1

the speeder covers a distance v * t before being overtaken

the police covers the same distance s = v * t

we know s = u*t + .5*a*t^2

            

v*t  = 0 + .5 * 2.8 * t^2

                

24  =  1.4 * t [v= 24m/s]

                  

t = 24/1.4

                    

= 17.142 s

Distance = v * t

                

= 24 * 17.142

                

= 411.43 m

Answer 2

To overtake the speeder, let's say the police car needs to reach a final velocity of 25 m/s.  Use vf^2 = v1^2 + 2ad and switch it around to solve for the distance it will take to accelerate to that velocity.  (vf^2-v1^2/2a = d).  Then take the speeder's velocity and divide by the same distance.  That will give you the time in seconds it takes for the police car to overtake the speeder.

Answer 3

Let

Ds = distance travelled by the speeder before police catches up

Dp = distance travelled by the police to catch up with the speeder

Ds = 24T

Dp = (1/2)(2.8)T^2

where T = time police catches up with speeder.

When the police catches up with the speeder, Ds = Dp, hence

24T = (1/2)(2.8)T^2

T = 48/2.8

T = 17.14 sec.

**************************************...

Ds = 24(17.14)

Ds = 411.36 meters