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Motion in 1 dimension

  1. Sep 8, 2005 #1
    :confused: :confused: Help! I need to know how to work this problem!! A motorcyclist moving with an initial velocity of 8.0 m/s undergoes a constant acceleration for 3.0s, at which time his velocity is 17.0 m/s. What is the acceleration, and how far does he travel in the 3.0 s interval?
     
  2. jcsd
  3. Sep 8, 2005 #2
    aceeleration=change in velocity/change in time so:
    17-8/3=a
    d=at
    so d=a * 3 seconds
     
  4. Sep 8, 2005 #3
    Actually, d=at is incorrect.
    The correct formula is [tex] v^2 - u^2 = 2as [/tex]. u is the initial velocity, in this case 8 m/s. We know [tex] a = 3 m/s^2 [/tex], therefore [tex] 17^2 - 8^2 = 2(3)s [/tex] and so [tex] s = (17^2 - 8^2)/6 [/tex] which turns out to be 37.5 m.
     
  5. Sep 8, 2005 #4
    8+3a=17
    a=3m/s^2
    8*3+0.5*3*3^2=24+13.5=37.5m
     
  6. Sep 10, 2005 #5
    Let me summarize the solution.

    a=acceleration u=initial velocity v=final velocity
    s=displacement t=time taken

    a=(v-u)/t
    a=(17.0-8.0)/3.0
    a=3 m/(s^2)

    s=((u+v)*t)/2
    s=((8.0+17.0)*3.0)/2
    s=37.5 m

    Therefore, the acceleration is 3 m/(s^2) to his original direction and
    the displacement he travelled is 37.5 m to his original direction.
     
    Last edited: Sep 10, 2005
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