AP Physics problem I need some help with...?

Question

A police car at rest, passed by a speeder traveling at a constant 110 km/h, takes off in hot pursuit. The police officer catches up to the speeder in 700m, maintaining a constant acceleration.

- Calculate how long it took the police officer to overtake the speeder

- Calculate the required police car acceleration

- Calculate the speed of the police car at the overtaking point.

Answer 1

1)  Given:

     Vi = 0, initial velocity of police car

     Vs = 110 km/hr, constant speed of speeder

       d = 700 m, distance traveled by police car when it catches up with

             speeder

       a = constant acceleration of police car

  

     Find:  t = time it took before the police car overtook the speeder

    

     Solution:  By the time the police car overtook the speeder, both of

     them would have traveled the same distance from the point where

     the police car was at rest when speeder passed it.  Therefore,

     d = Vs(t)

     0.700 km = (110 km/hr)t

     t = 0.700 km/(110 km/hr)

     t = 0.00636hr or 0.00636 hr x 3600 sec/1hr = 22.9 sec   ANSWER

2)  d = Vit + (1/2)at^2

    700m = (0)(22.9sec) + (1/2)(a)(22.9sec)^2

    a = 700(2)/22.9^2

    a = 2.67 m/sec^2   ANSWER

3)  2ad = Vf^2 - Vi^2

     Vf = sqrt(2ad)

     Vf = sqrt[(2)(2.67)(700)]

     Vf = 61.1 m/sec or (61.1m/sec)(1km/1000m)(3600sec/1hr) = 220 km/hr

Hope this helps.

teddy boy

Answer 2

The speeder's car is moving at 110 km/h

The police's car starts just when the speeder's car passes the police's car.

The police's car moves for 700 m and catches the speeder's car.

Is the above correct? If no, then let me know the actual question.

The police's car catches the speeder's car after travelling for 700 m

Speed of speeder's car = 110 km/h = 110 * 1000/3600 = 30.56 m/s

Time taken by the speeder's car to cover 700 m = 700/30.56 s = 22.91 s

So, the police's car catches up to the speeder in 22.91 s

For the police's car

Initial velocity u = 0

Displacement s = 700 m

Time t = 22.91 s

Acceleration a = ?

s = ut + 1/2 at^2

700 = 0 + 1/2 * a * 22.91^2

Or, a = 700 * 2/22.91^2 = 2.67 m/s^2

Let the speed of the police's car when it overtakes the speeder = v

v = u + at

v = 0 + 2.67 * 22.91 = 61.17 m/s = 61.17 * 3600/1000 km/h = 220 km/h

Ans:

22.91 second

2.67 m/s^2

220 km/h