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Calculating Velocity from a variable force.

  1. Jun 7, 2016 #1
    1. The problem statement, all variables and given/known data
    A body of mass 5 kg is acted on by a force in a straight line. The magnitude of the force after t seconds is given by (2t - 3t^2) kg wt. If the body has an initial velocity of 3ms-1 in the same direction as the force, calculate its velocity after 4 seconds.

    2. Relevant equations
    ∑F = ma
    a = dV/dt

    3. The attempt at a solution
    So what I've done is this.

    m = 5
    F = (2t - 3t^2)
    a = (2t - 3t^2) / 5
    dV/dt = (2t - 3t^2) / 5
    v = ∫(2t - 3t^2) / 5 dt
    v = (t^2 - t^3) / 5 + c

    Using t = 0, v = 3 (from initial velocity), solving for c yields c = 3. So,

    v = (t^2 - t^3)/5 + 3 and substituting t = 4, I get v = -6.6ms-1

    The back of my textbook gives an answer of -91.08ms-1, could someone please tell me where I've gone wrong?
     
  2. jcsd
  3. Jun 7, 2016 #2

    TSny

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    Hello and welcome to PF!

    I think your answer is correct. I don't see how you could get an answer of -91 m/s.
     
  4. Jun 7, 2016 #3

    TSny

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    The wording of the problem is a little odd. "Magnitude" of a force is generally defined to be a positive quantity.
    Yet, the expression 2t - 3t2 takes on negative values during the time interval 2/3 < t < 4.
     
  5. Jun 7, 2016 #4
    I just had a look at all of the question again and realised that the units which the force was given in are kg wt. I forgot to make the conversion between kg wt and N, and that threw me out.
    (2t - 3t^2) kg wt = g(2t - 3t^2) N

    Using F = g(2t - 3t^2), the correct answer of -91.08m/s was obtained.
     
  6. Jun 7, 2016 #5

    TSny

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    OK. Good. I did not notice the kg wt.
     
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