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Homework Help: Motion of a body -calc

  1. Sep 1, 2006 #1
    The motion of a body is given by the equation dV(t)/dt = 0.6 - 3V(t)
    where V(t) is the speed (in m/s) at time t (in second). If the body was at rest at t = 0

    1) What is the magnitude of the inital acceleration?
    2) The speed of the body varies with time as

    (A) [tex](1 - e^-^3^t) [/tex]
    (B) [tex]2(1 - e^-^3^t)[/tex]
    (C) [tex]\frac{2}{3}(1 - e^\frac{-3t}{2})[/tex]
    (D) [tex]\frac{2}{3}(1 - e^\frac{-3t}{3})[/tex]

    (B) is the correct answer for Q(2) . But how do you arrive at it? And how did they manage to get a 'e' in the answer?

    Please help.
    Last edited: Sep 1, 2006
  2. jcsd
  3. Sep 2, 2006 #2


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    Homework Helper

    Please show what you've tried.

    You're dealing with a separable differential equation, do you know how to solve one?
  4. Sep 2, 2006 #3
    what is a separable equation? I know basic calculus. but i have no clue on how to arrive at the answer to this question,
  5. Sep 2, 2006 #4


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    I googled for "separable differential equation" and found a decent looking text:

    Applying the above to your problem:

    [tex]\frac{dV(t)}{dt} = 0.6 - 3V(t)[/tex]

    [tex]dV(t) = (0.6 - 3V(t))dt[/tex]

    [tex]\frac{dV(t)}{0.6 - 3V(t)} = dt[/tex]

    [tex]\int_{V_0}^{V}\frac{dV(t)}{0.6 - 3V(t)} = \int_{t_0}^t dt[/tex]

    (You could also use indefinite integral, and solve for the C with the information given in the problem ie. "body was at rest at t = 0")

    Can you manage the rest?

    PS. There's something wrong with the equation or the correct answer. With the given equation you should arrive at:

    To get the given answer (B), the original equation should be:
    [tex]\frac{dV(t)}{dt} = 6 - 3V(t)[/tex]
    Last edited by a moderator: Apr 22, 2017
  6. Sep 2, 2006 #5


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    If you do not know how to solve differential equations, and presumbably aren't expected to here, sSince you are given 4 possible functions, work the other way. Plug each into the equation of motion and see which works. ([itex]\frac{dV}{dt}= 0.6- 3V[/itex] won't work with any of them- as said, it must be 6- 3V.)

    As for part A, that's easy. Just evaluate [itex]\frac{dV}{dt}= 0.6- 3V[/itex] at t= 0. (Of course, you are told V(0).)
  7. Sep 2, 2006 #6
    I am not very sure on how to procede from here.(I have just started learning calculus last week). Anyway should I use
    [tex] \int uv =u \int v - \int \frac{(du)}{(dx)}\int v [/tex] rule?
    Last edited: Sep 2, 2006
  8. Sep 2, 2006 #7


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    Homework Helper

    Here's a formula that should help you:

    [tex]\int \frac{dx}{x} = \ln |x| + C[/tex]
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