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ta123

  • 4 years ago

need help on the attachment

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  1. ta123
    • 4 years ago
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  2. m1m1c3ntr4l.h3rn4nd3z
    • 4 years ago
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    idk

  3. XianForce
    • 4 years ago
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    Okay so it tells us that the function y(t) = B * cos(wt) represents position. Recall that finding the derivative tells us the slope of a line at any given point. So if y(t) is position, and t is time, the slope would be position / time (rise over run). position over time is velocity, which is one of the things it asks us to find!

  4. ta123
    • 4 years ago
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    ok

  5. XianForce
    • 4 years ago
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    So we can take the derivative of that function and say: y'(t) = B * (-sin(wt)) * w. We can do this because w is said to be a constant. It's an application of the chain rule with the derivative of a trig function. The derivative of cos(u) = -sin(u) * du/dx.

  6. XianForce
    • 4 years ago
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    So that's our velocity equation. So now we have y'(t) is velocity and t is still time, so if we take the derivative of that, we should have velocity over time, which is acceleration. So y''(t) = B * (-cos(wt)) * w * w. Does that all make sense?

  7. ta123
    • 4 years ago
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    so how did you know taking the derivative twice would be the same as acceleration? doesn't acceleration came before velocity?

  8. XianForce
    • 4 years ago
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    No. Position / time = velocity; velocity / time = acceleration; acceleration / time = jerk. Think about it in terms of units. Position is measured in meters, time in seconds. position / time would be measured in meters per second, which is what velocity is measured in. Consequently, acceleration is m/s^2, and jerk is m/x^3 :)

  9. ta123
    • 4 years ago
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    oh, ok

  10. ta123
    • 4 years ago
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    Thanks

  11. XianForce
    • 4 years ago
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    No problem :)

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