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Acceleration problem please check answer

  1. Oct 2, 2004 #1
    A motorist drives along a straight road at a constant speed of 15.0m/s. Just as she passes a parked motocycle police officer, the officer starts to accelerate at 2.00m/s^2 to overtake her. Assuming the officer maintains this acceleration, (a) determine the time it takes the police officer to reach the motorist. Also find (b) the total displacement as he overtakes the motorist.

    Here's what I did:

    Vf=Vi+at
    15=0+2t
    t=7.5s

    d=vt
    d=15*7.5
    d=112.5

    so I did a graph of:
    y=2x^2
    and
    y=15x+112.5

    I found when 2x^2=15x+112.5 to be
    x=12.1
    y=294.5

    So, the time it took is 12.1s and the displacement is 294.5m

    Is this correct?
     
  2. jcsd
  3. Oct 2, 2004 #2
    That is not correct. This is how I did it:

    Distance traveled by motorist = v*t = 15*t
    Distance traveled by cop = .5*a*t^2 = .5*2*t^2 = t^2

    At what time will they have traveled the same distance?

    15*t = t^2
    15 = t

    What is the displacement?

    d = v*t = 15*15 = 225
     
  4. Oct 2, 2004 #3
    The reason why I did the graph of 2x^2=15x+112.5 is because when I did the graph of 2x^2=15x, it gave me x=7.5 and y=112.5...notice...

    Vf=Vi+at
    15=0+2t
    t=7.5s Time it takes the police to catch up to the motorcycle

    d=vt
    d=15*7.5
    d=112.5 distance travled by the motorcycle

    They were the same so I thought that If I added the inital distance of 112.5 meters to the 15x (constant speed of the motorist) then I would have the motorist starting 112.5 meters before the police accelerated, hence 2x^2=15x+112.5

    I thougth I would have gotten the correct answer for this question. Were did I go wrong?
     
  5. Oct 2, 2004 #4

    arildno

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    What you did wrong, is to find the time when their velocities are equal; but that doesn't imply they've traveled the same distance.
     
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