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Math Help - Velocity and Position Problem

  1. #1
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    Velocity and Position Problem

    A particle moves along the x-axis so that at any time t>0, its velocity is given by v(t)=4-6t^2. If the particle is at position x=7 at time t=1, what is the position of the particle at time t=2?
    (a) -10
    (b) -5
    (c) -3
    (d) 3
    (e)17

    How would I figure out the position of the particle when I'm only given the velocity equation and not the position equation? I tried integrating the velocity equation to get the position equation, but when I plug back in t=1, the position equation that I get from integrating doesn't give me x=7.
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  2. #2
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    You have to remember your constant of integration (that + c). The question gives you the one point (1, 7) so that you can calculate that constant.

    4t - 2t^3 + c

    4(1) - 2(1)^3 + c = 7
    c = 5

    4t - 2t^3 + 5
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  3. #3
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    Quote Originally Posted by summermagic View Post
    A particle moves along the x-axis so that at any time t>0, its velocity is given by v(t)=4-6t^2. If the particle is at position x=7 at time t=1, what is the position of the particle at time t=2?
    (a) -10
    (b) -5
    (c) -3
    (d) 3
    (e)17

    How would I figure out the position of the particle when I'm only given the velocity equation and not the position equation? I tried integrating the velocity equation to get the position equation, but when I plug back in t=1, the position equation that I get from integrating doesn't give me x=7.
    x(2) - x(1) = \int_1^2 v(t) \, dt

    so ...

    x(2) = 7 + \int_1^2 4 - 6t^2 \, dt
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  4. #4
    MHF Contributor Calculus26's Avatar
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    Integrate v(t) to get the position s(t)

    s(t) = 4t - 2*t^3 +c yes the int constant is imporant!

    s(1) = 7 so 4-2+c = 7 so c= 5

    s(t) = 4t - 2*t^3 + 5

    now finish
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