## ta123 4 years ago need help on the attachment

1. ta123

2. m1m1c3ntr4l.h3rn4nd3z

idk

3. XianForce

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

ok

5. XianForce

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

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

so how did you know taking the derivative twice would be the same as acceleration? doesn't acceleration came before velocity?

8. XianForce

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

oh, ok

10. ta123

Thanks

11. XianForce

No problem :)

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