# Car chase question

1. Sep 23, 2007

### anglum

1. The problem statement, all variables and given/known data
A speeder traveling 110 km/h races past a cop car. The cop car is at rest and pursues at acceleration of 9km/h until it reaches its maximum speed of 170km/h until it catches up with the speeder. How long does it take the cop car to catch the speeder answer in units of seconds

3. The attempt at a solution

i know the distance the 2 travel is equal

so for the speeder d = 110/t

and for the cop car i should use what formula so that i can relate them to d=d

so that 11/t = the distance formula for the cop car

2. Sep 23, 2007

### anglum

3. Sep 23, 2007

### anglum

can someone point me in thee right direction so i can start tryin to solve this the correct way

4. Sep 23, 2007

### anglum

learningphysics can you help me

5. Sep 23, 2007

### learningphysics

I recommend first find out the distance between the two cars, when the cop car has reached 170km/hr...

How long does it take the cop car to reach 170km/h? Over what distance does this happen?

How long has the speeder travelled in this time?

6. Sep 23, 2007

### anglum

ok so the time for the cop car to get to 170 km/h is

170/9 = 18.8888 is that in seconds then?

so then in 18.88 seconds the otehr car would have gone .577133 km?

Last edited: Sep 23, 2007
7. Sep 23, 2007

### learningphysics

What distance has the cop travelled in 18.888 seconds? What distance has the speeder driven in 18.888s?

8. Sep 23, 2007

### anglum

i am gettin screwed up becuz the speeds are in km/h and the time is in seconds GRRR i suck

9. Sep 23, 2007

### anglum

ok so am i right in my sayin it takes the cop 18.88 seconds to reach 170 km/h

and then i converted 170 to km/s aand got .0472 km/s
and i converted 110 km/h to km/s and got .03055 km/s

then the distance the speeder went was .57713 km
and the distance for the cop is what???

10. Sep 23, 2007

### learningphysics

Is the acceleration 9km/h per second? What exactly is the acceleration given?

11. Sep 23, 2007

### anglum

yes that is the acceleration

12. Sep 23, 2007

### learningphysics

For the cop, you'd use:

d = v1*t + (1/2)at^2. here v1 = 0. convert 9km/hr per second into km/s^2...

13. Sep 23, 2007

### anglum

the distance the cop travels then is.445568 km after 18.88 seconds

so now i knwo they are .131562 km apart at that time

so now what do i do to solve for the time of the cop car to catch the speeder

14. Sep 23, 2007

### anglum

so i then get this

cop ---- d+.131562 = .0472km/s (t)
speeder ---- d = .03055km/s (t)

so then i can say that .0472t - .131562 = .03055t ?????

15. Sep 23, 2007

### learningphysics

yes. that looks right... at the end remember to add 18.888s to whatever you get here to get total time.

16. Sep 23, 2007

### anglum

so the total time it takes to catch teh speeder is 26.7896 seconds

now to find the total distance each car traveled i simply take

d = .03055km/s (26.7896) or d +.131562 = .0472km/s (26.7896) ??????

17. Sep 23, 2007

### anglum

cuz if i do that the distances are different

18. Sep 23, 2007

### learningphysics

Take the speeder's velocity times the total time.

19. Sep 23, 2007

### anglum

one more question regarding car chases learning if u dont mind

speeder goin 30.9 m/s

cop accelerates at 2.49m/s squared

how long until he catches speeder

20. Sep 23, 2007

### anglum

so again the distances and times are equal for the two cars correct?