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Homework Help: Car chase question

  1. Sep 23, 2007 #1
    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. jcsd
  3. Sep 23, 2007 #2
    what should my next steps be?? please help me
  4. Sep 23, 2007 #3
    can someone point me in thee right direction so i can start tryin to solve this the correct way
  5. Sep 23, 2007 #4
    learningphysics can you help me
  6. Sep 23, 2007 #5


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    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?
  7. Sep 23, 2007 #6
    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
  8. Sep 23, 2007 #7


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    What distance has the cop travelled in 18.888 seconds? What distance has the speeder driven in 18.888s?
  9. Sep 23, 2007 #8
    i am gettin screwed up becuz the speeds are in km/h and the time is in seconds GRRR i suck
  10. Sep 23, 2007 #9
    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???
  11. Sep 23, 2007 #10


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    Is the acceleration 9km/h per second? What exactly is the acceleration given?
  12. Sep 23, 2007 #11
    yes that is the acceleration
  13. Sep 23, 2007 #12


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    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...
  14. Sep 23, 2007 #13
    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
  15. Sep 23, 2007 #14
    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 ?????
  16. Sep 23, 2007 #15


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    yes. that looks right... at the end remember to add 18.888s to whatever you get here to get total time.
  17. Sep 23, 2007 #16
    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) ??????
  18. Sep 23, 2007 #17
    cuz if i do that the distances are different
  19. Sep 23, 2007 #18


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    Take the speeder's velocity times the total time.
  20. Sep 23, 2007 #19
    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
  21. Sep 23, 2007 #20
    so again the distances and times are equal for the two cars correct?
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