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At what time will the policeman catch up with the car?

  1. Dec 18, 2007 #1
    A car travelling with a constant speed of 80km/hr passes a stationary motorcycle policeman. The policeman sets off in pursuit, accelerating to 80km/hr in 10 seconds and reaching a constant speed of 100km/hr after a further 5 seconds. At what time will the policeman catch up with the car?

    The past questions I have been doing in the same section have been about graphs, I think that this is the same but I'm stumped. I then thought about putting two equations together and cancelling but this worked to no avail as way. The working for the equations is below but I wasn't sure about how to put the working for that here

    Any help would be greatly appreciated:smile:

    x=ut+1/2at^2
    and
    x=vt-1/2at^2
    I put these two together and got
    ut+1/2at^2=vt-1/2at^2
    I rearranged to get
    u2t/4at^4
    That is where I got stuck because I wasn't sure how to rearrange
     
    Last edited: Dec 18, 2007
  2. jcsd
  3. Dec 18, 2007 #2
    You are taking the correct approach, but you are missing something:

    First, notice that the car has a constant velocity; therefore, its acceleration is...???

    (Once you figure that out, the rest of the problem should fall apart.)

    Second, your algebra is messed up; I'm not sure how you got the [tex]t^{4}[/tex] term, but that is incorrect.

    (Anyway, once you correct the acceleration of the car, the algebra should be easy.

    HTH,

    jIyajbe
     
  4. Dec 18, 2007 #3

    rl.bhat

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    Homework Helper

    First of all convert km/hr to m/s. Now calculate the acceleration of the motercycle during 10 second and distance travelled during that period. Repeat the same thing for 5 second period. Note down the total distance travelled by motercycle and car in 15 second. At that instant find the distance between them. Now both are moving with constant speed. Knowing the velocity of motercycle, car and distance between them, you can find the time taken by policeman to catch the car.
     
  5. Dec 18, 2007 #4
    Thanks I'll try it
     
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