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Math Help - Net Displacement (physics)

  1. #1
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    Net Displacement (physics)

    Net Displacement (physics)-f131g2_r.jpg
    the velocity-versus-time graph for a basketball player traveling up and down the court in a straight-line path. Find the net displacement of the player for the 10 s shown on the graph.

    I am not too sure where to even start. Or even how someone achieves a velocity of -2m/s. Am i supposed to find his displacement on the X axis? if so, am i supposed to use this equation: X=X_{xo}+V_{xo}+\frac{1}{2}a_{x}t^2
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  2. #2
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    Quote Originally Posted by mode View Post
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    the velocity-versus-time graph for a basketball player traveling up and down the court in a straight-line path. Find the net displacement of the player for the 10 s shown on the graph.

    I am not too sure where to even start. Or even how someone achieves a velocity of -2m/s. Am i supposed to find his displacement on the X axis? if so, am i supposed to use this equation: X=X_{xo}+V_{xo}+\frac{1}{2}a_{x}t^2


    The equation \Delta x = v_0 t + \frac{1}{2}at^2 is only valid if acceleration is constant. The given velocity graph clearly shows that acceleration is different over each lettered interval.

    FYI, on a velocity vs. time graph, displacement is the area between the graph and the t-axis. Area above the t-axis is positive displacement, area below is negative displacement. Algebraically sum the positive and negative displacements to get the total displacement.
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  3. #3
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    so i just sum up the velocity at each t interval?
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  4. #4
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    Quote Originally Posted by mode View Post
    so i just sum up the velocity at each t interval?
    one more time ...

    on a velocity vs. time graph, displacement is the area between the graph and the t-axis. Area above the t-axis is positive displacement, area below is negative displacement. Algebraically sum the positive and negative displacements to get the total displacement.
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