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Math Help - Rectilinear Motion (Using Integrals)

  1. #1
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    Rectilinear Motion (Using Integrals)

    Ok, i'm a bit confused on what this question is asking. I dont know how to visualize it. Here's the question:

    A particle moves along an s-axis. Use the given information to find the position function of the particle.

    V(t) = 3tē - 2t ; [the given information] s(0) = 1

    Any help would be greatly appreciated, thanks!
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  2. #2
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    What's the s-axis? And s(0)? If s is your time in seconds, is the function s returning time based on acceleration, distance, velocity, what?

    I'm assuming your distance at time 0 is 1. In which case you just take the integral of V(t) and your constant of integration becomes 1.

    D(t) = x^3 + x^2 + 1

    Edit: Well this seems to explain it better than my feeble mind.
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  3. #3
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    Quote Originally Posted by derfleurer View Post
    What's the s-axis? And s(0)? If s is your time in seconds, is the function s returning time based on acceleration, distance, velocity, what?

    I'm assuming your distance at time 0 is 1. In which case you just take the integral of V(t) and your constant of integration becomes 1.

    D(t) = x^3 + x^2 + 1

    Well this seems to explain it better than my feeble mind.
    I'm not sure what the s-axis is, I think its position. Time doesnt matter. I think the function is based off V(t), in which case for what i'm doing is velocity.

    I guess I dont understand why i'm taking the integral. The answer in the solutions part of the book gives the same function you described. Also how do you know 1 is the constant?
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  4. #4
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    Nevremind about the constant, I just rememberd how to do it. So can anyone explain why I have to use an integral?
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  5. #5
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    According to that page, s(t) is your distance. And at t = 0, s = 1.

    V(t) is a velocity-time graph. When you integrate V over a span of dt, you essentially get v * s (or m/s * s), which is just m.

    \int V(t)dt = s(t)

    The integral yeilds s(t) = x^3 + x^2 + c. And again, at t = 0, s = 1.

    1 = (0)^3 + (0)^2 + c

    Thus c = 1.
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  6. #6
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    Quote Originally Posted by derfleurer View Post
    According to that page, s(t) is your distance. And at t = 0, s = 1.

    V(t) is a velocity-time graph. When you integrate V over a span of dt, you essentially get v * s (or m/s * s), which is just m.
    OOOOooooohhhh. Yeah alright, I forgot that velocity isnt just a single variable. That helps a lot, thanks.
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