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Quick Motion Question - Rejogging memory after a while.

  1. Nov 23, 2014 #1
    1. The problem statement, all variables and given/known data
    A car travelling with a constant speed of 80km/hr passes a stationary motorcycle policeman. The policeman sets off in pursuit, accelerating uniformly to 80km/hr in 10 seconds and reaching a constant speed of 100km/hr in a further 5 secs. At what time will the policeman catch up with the car.

    2. Relevant equations
    N/A

    3. The attempt at a solution
    I figure if they catch up to each other, then assuming they began from the same point then their displacement will be equal to each other. Thus I did, the speeding car was doing 80km in one hour, and the cop was doing 100 km in one hour. Thus the cop would reach 80km displacement in 48 mins. (0.8 x 60 mins)
     
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  3. Nov 23, 2014 #2

    haruspex

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    How did you determine that 80km was the appropriate displacement?
    Hint: how much lead does the motorist have when the cop first reaches 100km/h?
     
  4. Nov 23, 2014 #3
    15 secs before he reaches 100km/hr. So then what?
     
  5. Nov 23, 2014 #4

    haruspex

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    I mean lead as a distance.
     
  6. Nov 23, 2014 #5
    333m. As 80km/hr = 22m/s (80/3.6) = 22m/s.
    Thus, 15 secs = 333.3 (infinite) meters.
     
  7. Nov 23, 2014 #6

    haruspex

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    That's how far the car goes in 15 seconds, but the cop is not standing still.
     
  8. Nov 23, 2014 #7
    The car has a 333m lead over the policeman. Thats what you asked?
     
  9. Nov 23, 2014 #8

    haruspex

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    333m is the lead he would have had if the policeman had not given chase. Where is the cop after 15 seconds?
     
  10. Nov 23, 2014 #9
    After 15 seconds he is travelling at 100 km/hr and has travelled 360m. As for the first 10 seconds he covers 222m as he is travelling at 80km/hr, and for the next 5 seconds he travels at 100km/hr covering 138m thus after 15 secs the cop is 360meters from the starting position.
     
  11. Nov 23, 2014 #10

    haruspex

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    No he isn't - read the question again.
     
  12. Nov 23, 2014 #11
    Sorry, he covers 416 meters in the 15 seconds. As 100/3.6 = 27.8 x15 = ~416m.
     
  13. Nov 23, 2014 #12

    haruspex

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    Pay attention! The cop is not moving at uniform speed. For the first 10 seconds he accelerates uniformly from 0 to 80km/h. Do you know the SUVAT equations? These can be used for uniform (i.e. constant) acceleration.
     
  14. Nov 23, 2014 #13
    Yes, i know them. How are they linked though. I reliaze he accelerates at different rates from 0 to 10 then 10 to 15 seconds. Hence why I said, After 15 seconds he is travelling at 100 km/hr and has travelled 360m. As for the first 10 seconds he covers 222m as he is travelling at 80km/hr, and for the next 5 seconds he travels at 100km/hr covering 138m thus after 15 secs the cop is 360meters from the starting position.
     
  15. Nov 23, 2014 #14

    haruspex

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    No he isn't! Why do you keep asserting that when I have already pointed out that it is wrong?
    He starts at 0 km/h, and only reaches 80km/h at the end of the 10 seconds. He does not travel at 80km/h for 10 seconds.
    In the usual form, there are five SUVAT equations, each involving four variables. Typically, you know the value of three, and you want to find a fourth, so pick the equation that involves those four variables.
    What three values do you know here? What other variable do you want to determine? Which equation should you use?
     
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