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Kinematics>>Equations of Motion

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  1. Feb 18, 2017 #1
    1. The problem statement, all variables and given/known data
    A motorcyclist is travelling at 15.0m/s [forward] and applies brakes. The motorcycle slows down at 5.0m/s [backward].
    a) Determine the motorcycle's breaking distance [ans:23 m [forward]]
    Given:
    Vinitial:
    15.0m/s [forward]
    a: 5.0m/s^2 [backward]
    Vfinal: 0 m/s [forward]

    2. Relevant equations
    d= vi(T) + 1/2a(T^2) >>> vector equation
    vf = vi +aT >>>> vector equation

    3. The attempt at a solution
    1) Find time for breaking distance
    vf = vi + aT
    0 = vi - aT (making acceleration negative so that it's vector is forward)
    0 = 15 - 5T
    -15/-5 =T
    3 = T
    2) Finding breaking distance
    d= vi(T) + 1/2a(T^2)
    d= vi(T) - 1/2a(T^2) (acceleration made forward by making it negative)
    d = 15(3) - 2.5(9)
    d = 75 - 22.5
    d = 52.5 m
    Therefore the braking distance is 52.5 m but the answer is 23m which I think came by ignoring initial velocity but it doesn't make any sense to me why they ignore initial velocity so kindly explain that to me. The picture i uploaded explains why i think initial velocity must be considered. WIN_20170218_20_57_14_Pro.jpg
     
  2. jcsd
  3. Feb 19, 2017 #2

    kuruman

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    Homework Helper
    Gold Member

    15(3) = 45 not 75. :rolleyes:
     
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