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Math Help - Need help with a kinematics question

  1. #1
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    Need help with a kinematics question

    Hi, I have a kinematics question which i can't solve and i need help. Here it is.

    A particle moves in a straight line passing a fixed point O with a velocity of 3m/s. It moves in such a manner that t seconds after passing O, its velocity is given by v= at²+b. If the particle is again at O after 3 seconds, find its speed at that instant. Find the total distance traveled between t=0 and t=3.

    This is what i have tried so far :
    v= at²+b

    subst. t=0, v=3.
    3 = a(0)² + b
    b=3

    ∴ subst. b=3, t=3 into v= at²+b
    v = a(3)² + 3
    v = 9a + 3
    ----------------------------------------------------------------------------

    I couldn't continue after arriving here, can't get the unknown,a. I only need help here. I understand how to get the answer for the distance part. Thanks
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  2. #2
    Senior Member Sambit's Avatar
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    As far as I see, the problem involves 2 unknowns with 1 equation!!
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  3. #3
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    Ya! I've been thinking for a long time to get the other unknown
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  4. #4
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    I don't understand the quoted part of your question:

    Quote Originally Posted by Unagi9 View Post
    ...
    A particle moves in a straight line passing a fixed point O

    If the particle is again at O
    ...
    In my opinion the particle can pass a certain place several times if it moves on a closed curve (circle, ellipse, ...). But then it doesn't move in a straight line.

    Or: If the particle moves in a straight line it passes a certain place only once.
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  5. #5
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    Quote Originally Posted by Unagi9 View Post
    A particle moves in a straight line passing a fixed point O with a velocity of 3m/s. It moves in such a manner that t seconds after passing O, its velocity is given by v= at²+b. If the particle is again at O after 3 seconds, find its speed at that instant. Find the total distance traveled between t=0 and t=3.
    the particle changes direction as it moves in a straight line ...

    v(0) = 3 \, m/s

    v(t) = at^2 + b

    b = 3

    since the particle returns to point O at t = 3 , then its displacement from t = 0 to t = 3 is zero ...

    \displaystyle \int_0^3 at^2 + 3 \, dt = 0

    \left[\dfrac{at^3}{3} + 3t\right]_0^3 = 0<br />

    9a + 9 = 0

    a = -1

    v(t) = 3 - t^2

    v(3) = 3 - 9 = -6 , so its speed at t = 3 is |-6| = 6 \, m/s

    total distance = \displaystyle \int_0^3 |3-t^2| \, dt = \int_0^{\sqrt{3}} 3 - t^2 \, dt - \int_{\sqrt{3}}^3 3 - t^2 \, dt

    you can finish up by finding the total distance.
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  6. #6
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    Thanks skeeter! I didn't know that you could use definite integrals.
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