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Deceleration. My answer is different from the books and it's driving me crazy

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data
    The driver of a pickup truck going 100 km/h applies the brakes, giving the truck a uniform deceleration of 6.50 m/s^2 while it travels 20.0 m.

    a) what is the speed of the truck in kilometers per hour at the end of this distance?

    b) How much time has elapsed?

    2. Relevant equations

    3. The attempt at a solution

    The first thing I did was convert 100 km/h into m/s and got 277.7 m/s.

    After converting I tried using V^2= V(initial)^2 x 2(a)x, where a = acceleration and x = horizontal distance.

    I got

    V^2 = (277.7 m/s)^2 + 2 (-6.50 m/s^2) (20.0-0)
    V^2 = 77117.29 + (-260)
    V^2 = 76857.29
    V= 277.23 m/s which equates to 99.8 km/hr which makes no sense.

    I then tried x = x(initial) + 1/2 at^2 to try and acquire time first.

    I ended up getting 83 seconds. Which is way off. So I checked the answers in the back of the book and it read:

    for part a) 81.4 km/h
    b) .794 s

    can someone please tell me what I'm doing wrong? I'm obviously not asking for the answer since I've got the book.
  2. jcsd
  3. Sep 16, 2009 #2
    You have an error in your conversion in m/s. It's more like 27 m/s.
    Just divide 100 km/h by 3.6.
  4. Sep 16, 2009 #3


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    I think that you're off by a decimal place.
  5. Sep 16, 2009 #4
    x = x(initial) + v(initial)(t) + 1/2at^2....correct?
  6. Sep 16, 2009 #5

    Sorry your very first step of converting kmph to mps is wrong. A small mistake in decimal point .

    1km= 1000m
    1 hr= 3600s

    100kmph= (100*1000)/3600 = 1000/36
  7. Sep 16, 2009 #6
    Sorry your very first step of converting kmph to mps is wrong. A small mistake in decimal point.

    1km= 1000m
    1 hr= 3600s

    100kmph= (100*1000)/3600 = 1000/36

    http://arasaraja.blogspot.com/" [Broken]
    Last edited by a moderator: May 4, 2017
  8. Sep 17, 2009 #7
    You need to use the equation d = Vot + at^2/2 (with d = 20), you know Vo and a so if you solve it you'll receive 2 solutions for t, only one is valid
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