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Work, Energy and Power - Variable Acceleration

  1. Feb 23, 2010 #1
    The question is about Mechanics (so a mix of maths and physics)

    1. The problem statement, all variables and given/known data

    A car of mass 900kg moves along a straight horizontal road with its engine working at a constant rate of 20kW. Its speed at A is 10m/s, Assuming that there is no resistance to motion, calculate the time taken for the car the travel from A until its speed reaches 20ms.


    2. Relevant equations

    [tex]\vec{F}_{net} = \Sigma \vec{F} = m \vec{a}[/tex]

    3. The attempt at a solution

    Firstly, I drew a diagram that looked similar to this: http://skrbl.com/130853849 (sorry, scanner broken!)

    Power = Fv therefore F = 20000/v

    FCA F=ma N2L

    Res (->) (20000/v) = 960a
    a = 125/6v

    I know I need to integrate the acceleration formula (due to variable acceleration) but I'm not sure how to do with wrt t. My problem is because there is no t. I do remember my teacher saying something about integrating both sides to find time (school is over here now though so I can't get hold of him). I have attempted to integrate both sides but it still made little sense. I did a little browsing on these forums and came up with this though:

    integral(dv) = integral([125/6v]dt)

    I appreciate any and all help - it's probably something simple that I just can't see! I mainly need hints (would rather not answers) on how to integrate this to give me something in relation to time.

    By the way, according to the textbook the answer is 7.2 seconds.
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Feb 23, 2010 #2


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    P = Fv

    P = m(dv/dt)v = const.
    Pdt = mv(dv)

    Integrate both sides. This is what your teacher meant.
    Last edited: Feb 23, 2010
  4. Feb 23, 2010 #3


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    So you have the formula : a = dv/dt = 125/(6 v).

    You can integrate this differential equation by "separation the variables": collecting everything "v" at one side and "t" at the other side.

    [tex]\int_{10}^{20}{vdv}= \frac{125}{6}\int_0^T{dt}[/tex]

    You integral with respect to v from the initial velocity to the final one, and the upper limit for t is just the time of question.

  5. Feb 23, 2010 #4
    Thanks for the replies. I've tried to understand them as best I can however we have not been taught an awful lot on integration. From ehlid's reply I managed to get the following:

    900 = [tex]{125}\int_0^T{dt}[/tex]

    That's as far as I could get. Do you know of any places where I can find the pure maths that will help me answer this (unless you can give more hints of course)? Such as an example on how to solve something similar?
    Last edited: Feb 23, 2010
  6. Feb 23, 2010 #5


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    You can solve this problem without integration if you are familiar with the work - energy theorem: The change of the KE of a rigid body is equal to the work of all the external forces acting upon it.

    KE2-KE1 = W.

    The initial and final velocities are given, the mass is known, you can calculate the change of kinetic energy.

    You have to get the work. The power is given and it is constant. So the work done in time t is W=Pt. You know everything to calculate the time.

  7. Feb 23, 2010 #6
    Thanks so much for that! I wasn't actually taught that and my teacher says that is the only way.

    It seems so much easier (and it, plus it's right!) Going to be a fun lesson tomorrow
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