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Distance travelled by the particle

  1. Jul 26, 2012 #1

    utkarshakash

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    Gold Member

    1. The problem statement, all variables and given/known data
    A constant power P is applied to a particle of mass m. The distance travelled by the particle when its velocity increases from v1 to v2 is(neglect friction)


    2. Relevant equations
    Work done = ΔKE
    P=dW/dt
    s=(v2^2-v1^2)/2a

    3. The attempt at a solution
    Since work done by all forces equals ΔKE
    Work done = m(v2^2-v1^2)/2
    Now P=dW/dt,
    differentiating W wrt to t, I get
    a=P/m(v2-v1)
    Putting a in last equation I get
    s=m(v2^2-v1^2)(v2-v1)/2P

    but this answer is wrong :(

    The correct answer given in my book is m(v2^3-v1^3)/3P
    I don't know where I've commited the mistake.
     
  2. jcsd
  3. Jul 26, 2012 #2

    PeterO

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    Homework Helper

    Being a cynic about answers like this given in texts, I would consider an equivalent problem with numerical values and check that the "formula" answer does give the correct answer. Perhaps your answer does also, and there is just 2 ways of saying the same thing.

    Use values that should not give a coincidental answer ie avoid a value of 2, since 2 + 2 and 2 x 2 give the same answer.

    EG 14 kg mass travelling at 7 m/s, is accelerated to 11 m/s when power is added at the rate of 5 watts. How far would be covered.

    Your formula says 14(121 - 49)/10
    Their Formla says 14(1331 - 343)/15

    They are not the same answer, so at least one of them is wrong.

    Hint: find the initial and final Kinetic energies. Since the energy was added at a rate of 5 J/s, you can work out how long in time that takes.
    Since the average speed is 9m/s, you can thus calculate the distance covered.
    While doing that numerical calculation, you may even see why your/their answer will be correct.

    Good luck.
    Note: You could repeat the calculation for some different values, to show your agreement was not coincidental.
     
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