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need help on the attachment

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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idk
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!

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ok
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.
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?
so how did you know taking the derivative twice would be the same as acceleration? doesn't acceleration came before velocity?
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 :)
oh, ok
Thanks
No problem :)

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