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Math Help - Rectilinear Motion

  1. #1
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    Rectilinear Motion

    The acceleration function and the initial velocity for a particle moving along a line are: a(t)= 2t + 3, v(0)= -4, 0<t<3

    Find the distance traveled during the time interval


    10, my answer, isn't correct. I tried subtracting v(0) from v(3), but it doesn't yield the correct answer. How should I approach this?
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  2. #2
    Senior Member eumyang's Avatar
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    They want the distance traveled. v(t) is the velocity function. So evaluating v(3) - v(0) would not be correct.

    Integrate a(t)
    \displaystyle \int 2t + 3 \,dt = t^2 + 3t + C = v(t)

    To find C, plug in the initial condition:
    v(0) = 0^2 + 3(0) + C = -4 \rightarrow C = -4
    So
    v(t) = t^2 + 3t - 4

    I'll call the position function s(t). Integrate v(t):
    \displaystyle \int t^2 + 3t - 4 \,dt = \dfrac{t^3}{3} + \dfrac{3t^2}{2} - 4t + C = s(t)

    Can you take it from here?


    EDIT: Typos. TT_TT
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  3. #3
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    d(t) = \int v(t)~dt

    So integrate twice.
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  4. #4
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    Quote Originally Posted by eumyang View Post

    I'll call the position function s(t). Integrate v(t):
    \displaystyle \int 2t + 3 \,dt = \dfrac{t^3}{3} + \dfrac{3t^2}{2} - 4t + C = s(t)
    Little bit of a typo copy/paste mistake here..
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