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Math Help - kinematics problem

  1. #1
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    kinematics problem

    i have like 5 questions the same style and i cant do them, so if somebody could work through this 1 ill be really grateful.

    Two trains A and B run on parallel striaght tracks. Initially both are at rest in the station and level with each other. At time t=0, A starts to move. It moves with constant acceleration for 12 sec up to a speed of 30m/s, and then moves at a constant speed of 30m/s. Train B starts to move in the same direction as A when t=40, where t is measure in seconds. It accelerates with the same initial acceleration as A, up to a speed of 60m/s. it then moves at a constant speed of 60m/s. Train B overtakes A after both have reached their maximum speed. Train B overtakes A when t = T.

    We had to draw the speed time graphs of both trains for 0 ≤ t ≤ T.
    The second part is find the value of T.
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  2. #2
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    hopefully, you know why it helps to draw a velocity vs. time graph.

    the area between the graph and the time axis is displacement.

    displacement of train A = \frac{1}{2}(12)(30) + 30(T-12)

    displacement of train B = \frac{1}{2}(24)(60) + 60(T-64)

    the two trains have the same displacement when B passes A.
    Last edited by skeeter; November 9th 2008 at 12:52 PM. Reason: fixed latex
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  3. #3
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    Quote Originally Posted by ellie_h View Post
    i have like 5 questions the same style and i cant do them, so if somebody could work through this 1 ill be really grateful.

    Two trains A and B run on parallel striaght tracks. Initially both are at rest in the station and level with each other. At time t=0, A starts to move. It moves with constant acceleration for 12 sec up to a speed of 30m/s, and then moves at a constant speed of 30m/s. Train B starts to move in the same direction as A when t=40, where t is measure in seconds. It accelerates with the same initial acceleration as A, up to a speed of 60m/s. it then moves at a constant speed of 60m/s. Train B overtakes A after both have reached their maximum speed. Train B overtakes A when t = T.

    We had to draw the speed time graphs of both trains for 0 ≤ t ≤ T.
    The second part is find the value of T.
    Quote Originally Posted by skeeter View Post
    hopefully, you know why it helps to draw a velocity vs. time graph.

    the area between the graph and the time axis is displacement.

    displacement of train A = \frac{1}{2}(12)(30) + 30(T-12)

    displacement of train B = \frac{1}{2}(24)(60) + 60(T-64)

    the two trains have the same displacement when B passes A.


    so then do i have 2 simultanious equations to solve or something like that?
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  4. #4
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    all set up for you ...

    displacement of train A = \frac{1}{2}(12)(30) + 30(T-12)

    displacement of train B = \frac{1}{2}(24)(60) + 60(T-64)

    displacement of train A = displacement of train B

    set the two equal and solve for T.
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  5. #5
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    Quote Originally Posted by skeeter View Post
    all set up for you ...

    displacement of train A = \frac{1}{2}(12)(30) + 30(T-12)

    displacement of train B = \frac{1}{2}(24)(60) + 60(T-64)

    displacement of train A = displacement of train B

    set the two equal and solve for T.

    i know that i thught i understood but i actually dont see how you got the two equations.

    you see i missed school an want to catch up, but i dont understand how to do these problems, so i thought if somebody could explain it exactly ill be able to do the other problems.

    i know that this is really rude but do you mind explaining it in a bit more depth?
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