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Math Help - Definite intergrals for an expression of displacement?

  1. #1
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    Definite intergrals for an expression of displacement?

    Hi I'm just getting back into education and I've been given some practice questions but this one kind on thrown me.
    Ive been given the expression
    v=12te^(t/2) +5 for velocity

    I had work out what the velocity is at t= 4.5 s
    i got 10.69m/s

    Then I had to obtain an expression for acceleration and calculate the force acting on a body of 5kg at t = 4.5s so I used product rule and finally got an answer.

    but the last part of the question again asks for an expression this time for displacement s of the body after t seconds, given that when t=0, s=0. Hence calculate the displacement of the body at time t=4.5 seconds
    I was thinking definite intergrals but its been a while since I studied it so any ideas please?
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  2. #2
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    Re: Definite intergrals for an expression of displacement?

    Given that you report finding v(4.5)\approx10.69\,\frac{\text{m}}{\text{s}} I assume we have:

    v(t)=12te^{-\frac{t}{2}}+5

    Since velocity is defined as the time rate of change of displacement, we may write the initial value problem:

    \frac{ds}{dt}=12te^{-\frac{t}{2}}+5 where s(0)=0

    and so integrating with respect to t, using the boundaries as limits, and using dummy variables of integration, we may state:

    \int_0^{s(t)}\,du=\int_0^t 12ve^{-\frac{v}{2}}+5\,dv

    For the first term in the integrand on the right, you will want to use integration by parts. Are you familiar with this method?
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  3. #3
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    Re: Definite intergrals for an expression of displacement?

    I've not touched on any integration for a few years just reading up on it now thanks for the reply
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    Re: Definite intergrals for an expression of displacement?

    Ok I've tried to use integration by parts but keep going down dead ends this is where I'm at:
    I= 12t^-0.5t
    I= 12t x e^-0.5t/-0.5 - { e^-0.5/-0.5 x 12dt
    I= 12te^-0.5t/-0.5. - 12/-0.5 x e^-0.5t/-0.5
    I= 12te-0.5t/-0.5 - 12e^-0.5t/0.25
    Is this correct before I definate integrate and do I add the original +5 back in?
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  5. #5
    MHF Contributor MarkFL's Avatar
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    Re: Definite intergrals for an expression of displacement?

    I would let:

    I_1=12\int_0^t we^{-\frac{w}{2}}\,dw

    I_2=5\int_0^t\,dv=5t (this one we are able to evaluate directly).

    So, we are left with the first term on which to use integration by parts.

    Let:

    u=w\,\therefore\,du=dw

    dv=e^{-\frac{w}{2}}\,dw\,\therefore\,v=-2e^{-\frac{w}{2}}

    and so we have:

    \frac{I_1}{12}=\left(\left[-2we^{-\frac{w}{2}} \right]_0^t+2\int_0^t e^{-\frac{w}{2}}\,dw

    \frac{I_1}{12}=-2te^{-\frac{t}{2}}+2\left[-2e^{-\frac{w}{2}} \right]_0^t

    \frac{I_1}{12}=-2te^{-\frac{t}{2}}+2\left(-2e^{-\frac{t}{2}}+2 \right)

    \frac{I_1}{12}=-2te^{-\frac{t}{2}}-4e^{-\frac{t}{2}}+4

    I_1=48-24e^{-\frac{t}{2}}(t+2)

    Add so, putting everything together, we find:

    s(t)=5t+48-24e^{-\frac{t}{2}}(t+2)
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  6. #6
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    Re: Definite intergrals for an expression of displacement?

    Ok but when t=0 s= 48 and it should equal zero also that's what confused me with my answer?
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  7. #7
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    Re: Definite intergrals for an expression of displacement?

    Ignore that, finger trouble
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