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Simple Relative Motion Police Car Problem

  1. Sep 25, 2004 #1
    Ok, here’s the question:

    A motorist traveling at 89 km/hr is being chased by a police car at 115 km/hr. If the police car starts from 3.5 km back, how long does it take to catch the motorist? Leave time in hours.

    Now I’m looking at this as a relative motion problem since there is no acceleration, therefore time should be equal to the distance between the cars, divided by the difference in the velocity between the two cars.

    Therefor; t = (3.5 km) / [(115 km/hr) - (89 km/hr)]

    Therefore; t = 0.13 hr

    However it’s apparently wrong, so does anybody know what I’m doing wrong and how to fix it? :surprised Thanks
  2. jcsd
  3. Sep 25, 2004 #2
    I don't see what is wrong with this.
  4. Sep 25, 2004 #3
    M moves 1.5km every minute
    P moves 1.9km every minute
    M moves 0.02km per second
    P moves 0.03km per second

    M+8mins = 15.4
    P+8mins = 15.4

    M x 8 = 11.9 + 3.5 = 15.4
    P x 8 = 15.3

    M+8mins = 15.4 + m4secs = 0.1 = 15.5
    P+8mins = 15.3+ p4secs = 0.12 = 15.5

    I did this the long way because this is the first time I tried a question like this.
    so I get 8mins and 4 secs with a total distance of 15.46kms
    Does it sound right?

    But just because it takes that much time to be at the same distance, the question is whether or not the motorist stops. :grumpy:

    EDIT I truncated your spurious digits! You must learn to use only the digits which have meaning!
    Last edited by a moderator: Sep 26, 2004
  5. Sep 25, 2004 #4
    Yes, that is the same answer as mwells got using his method. Mwells, I think there's a mistake in your solutions manual or where ever the answer is from. Your answer is correct.
  6. Sep 25, 2004 #5
    What method is that? I would love to understand it.
  7. Sep 25, 2004 #6
    Ill give you a hint, and Im drunk right now, so that makes this problem easy. Use relative velocity. Subtrqact their welocities from each other and then use the equation [tex] d = {V_o}t + \frac{1}{2} at^2[/tex]
  8. Sep 26, 2004 #7


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    Except he needs t to solve this formula. But t is what he's looking for. a = 0 in this formula, so everything after the + disappears, and he's left with the simple d=vt formula which can be rewritten as t=d/v. But he needs d to solve this.

    I also get the same answer. Is this one of those computer-graded homework assignments? It might not like your significant digits. It may or may not want you to put the word "hours" after your answer.
  9. Sep 26, 2004 #8


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    Gold Member

    Let d be the distance traveled by the police car.
    Let Vp = the speed of the police car
    Let D be the distance traveled by the fleeing car.
    Let Vc = the speed of the fleeing car.

    We must have D + 3.5 = d and the time of travel (t) for each car be the same.

    D = Vct

    Vct+3.5 = Vpt

    Now solve for t,

    [tex] t = \frac {3.5} {V_p - V_c}[/tex]

    This is the solution given in the first post. As suggested you may need to play with the number of significant digits. (should be 2)
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