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Homework Help: Motorcycle physics problem

  1. Aug 18, 2008 #1
    Two motorcycles are traveling due east with different velocities. However, four seconds later, they have the same velocity. During this four-second interval, motorcycle A has an average acceleration of 1.6 m/s2 due east, while motorcycle B has an average acceleration of 3.6 m/s2 due east. By how much did the speeds differ at the beginning of the four-second interval? And which motorcycyle is traveling faster?

    I tried subtracting 1.6 from 3.6, but it didn't accept 2 as the correct answer.
    What am I doing wrong?
  2. jcsd
  3. Aug 18, 2008 #2
    Re: Velocity.

    For motorcycle A, in 4 seconds, it gained an extra 6.4m/s (4)(1.6)
    For motorcycle B, in 4 seconds, it gained an extra 14.4m/s (4)(3.6)
    Since after the 4 second interval, the 2 motorcycles have the same velocity, I can say that
    v_A + 6.4 = v_B + 14.4
    The difference between the 2 bikes' velocity would be v_A - v_B
    By rearranging the above equation, we get v_A - v_B = 8m/s.
    Since v_A > v_B initially, this must mean v_A is travelling faster.

    I think what you did wrong is that when you just did 3.6 - 1.6, it only accounted for a 1-second interval. Over 4 seconds would yield a value of 8, which is what I got.

    I believe this is how to do the question. If that is wrong, I apologize for that.
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