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Finding the distance from deceleration

  1. Apr 15, 2010 #1
    1. The problem statement, all variables and given/known data

    A lorry is driving at 96 km h^−1. The driver decelerates to a speed of 48 km h^−1. Given that the deceleration is 2.68 m s^−2 find the distance over which the brakes are applied.

    I know the initial speed of the lorry, v = 96 km h^-1 = 26.67 m s^-1
    I know the final speed of the lorry, u = 48 km h^-1 = 13.33 m s^-1
    I know the deceleration rate of the lorry, a = 2.68 m s^-2

    so I'm looking for d (Distance)

    2. Relevant equations

    t=(v-u)/a to find the time t the breaks were applied for

    d = vt + 1/2 x at^2 ?? to find the distance travelled while decelerating?

    3. The attempt at a solution

    Substitute the values for v,u and a to find speed s
    t=(26.67 m s^-1 - 13.33 m s^-1)/2.68 m s^-2

    = 4.98 s

    I'm not sure what the second equation is for working out distance the lorry has travelled based on a decelerating body and the info I now have.

    so I think d= (26.67 m s^-1 x 4.98 s) + 1/2 x 2.68 m s^-2 x 4.98 s^2?

    Also, if this is correct and based on the information that I have, is there any other equation I could use to also find distance? I ask as the question's hint tells me to consider the energies involved and the section just read is also related to energy and force...

    Many thanks
  2. jcsd
  3. Apr 15, 2010 #2


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    You can use energy considerations or you can use the kinematic equation

    2 a (xfinal - xinitial) = vfinal2 - xinitial2

    which is essentially the same thing as the work-energy theorem. Don't forget that the acceleration is a negative number here.
  4. Apr 16, 2010 #3
    Hi Kuruman and thankyou for your answer. I'm not sure I totally understand the equation though. Is this how I would find distance based on the information I have? I'm not totally sure how I would use the work-energy theorem to solve this perticular problem either.
  5. Apr 16, 2010 #4


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    Take the first relevant equation that you quoted,
    t = (vfinal - vinitial)/a
    and substitute for t in
    d = vinitialt + (1/2)at2.
    If you do the algebra correctly, you should get the equation that I quoted. You should use this equation when you don't know (and don't care to know) what the time is.
    The work-energy theorem says
    WNet = ΔK = (1/2)m(vfinal2 - vinitial2)
    Now the net work is the work done by all the force and can be though of as the work done by the net force. If the displacement is d,
    WNet = FNetd
    and since FNet = ma, WNet = mad. So,
    mad = (1/2)m(vfinal2 - vinitial2)
    from which you get
    2ad = (vfinal2 - vinitial2).
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