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m1m1c3ntr4l.h3rn4nd3z Group TitleBest ResponseYou've already chosen the best response.0
idk
 2 years ago

XianForce Group TitleBest ResponseYou've already chosen the best response.1
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!
 2 years ago

XianForce Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

XianForce Group TitleBest ResponseYou've already chosen the best response.1
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?
 2 years ago

ta123 Group TitleBest ResponseYou've already chosen the best response.1
so how did you know taking the derivative twice would be the same as acceleration? doesn't acceleration came before velocity?
 2 years ago

XianForce Group TitleBest ResponseYou've already chosen the best response.1
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 :)
 2 years ago

XianForce Group TitleBest ResponseYou've already chosen the best response.1
No problem :)
 2 years ago
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