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Working with two moving targets ?

  1. Mar 4, 2008 #1
    Working with two moving targets ??

    Hey guys, Here is the question I am working on

    "A car traveling at a constant speed of 80kmh passes as stationary motorcycle policeman. The policeman sets off in pursuit, accelerating to 80kmh in 10 seconds and reaching a constant speed of 100kmh in another 5 seconds. At what time will the policeman catch up with the car. "

    I have worked this out so far

    80 km h–1 = 22.222 m s–1
    and 100 km h–1 = 27.778 m s–1

    Let the required time = t.

    The distance traveled by the policeman in the first 10.0 s
    = 0.5(22.222 m s–1)(10 s) = 111.111 m,
    and in the next 5.0 s he travels a distance of 0.5(22.222 m s–1 + 27.778 m s–1)(5.0 s)
    = 125 m.

    When the policeman catches up with the car, it is

    (22.222 m s–1)t = 236.111 m + (27.778 m s–1)(t – 15 s)

    and so t = 32.5 s.

    So I have the correct answer but to find the answer I just used trial an error till I got the correct time. What I want to know is surely there is an easier way to work out the answer without having to take a stab in the dark. I am sure this is probably a simple maths thing but it just keeps eluding me. Any help you can offer would be great.

  2. jcsd
  3. Mar 4, 2008 #2


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    Welcome to PF!

    Hi maca! Welcome to PF! :smile:

    hmm … it's good that you can see you did something wrong :smile: … your answer just doesn't look neat, does it, and that's usually a sign that it could be improved!

    But you didn't use trial and error - you used an equation for t, which is what you're supposed to do:
    and solved it! :smile:

    (btw, I haven't actually checked whether it's right)

    The "something wrong" is that you put in an unnecessary step - when you wrote:
    you could just as easily have written:
    and in the next t s he travels a distance of ….

    That would have given you a very similar equation, which already has t in it, giving you the answer more quickly and more neatly! :smile:

    Write it again, with that change. And then … can you see an alternative way of calculating the second (steady-speeds) patt of the distance? :smile:
  4. Mar 4, 2008 #3


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    For the bike to catch up to the car means they've travelled the same distance in the same time. The only difference is in how they did it. The car was travelling at a constant speed while the bike had to accelerate and then travel at a constant speed to catch up.

    So if you work out how far the bike went while accelerating and how far the car went in the same time. You can then use [itex] s = x_0 + ut [/itex] for both the car and bike, set them equal and solve for time, then add that to the time when the bike was accelerating. This is pretty much what you've done.
    Last edited: Mar 4, 2008
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