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The equation of motion is given by s=sin2πt (m) . Find the acceleration after 4.5 sec 
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#1
Jun711, 11:14 PM

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1. The equation of motion is given by s=sin2πt (m) . Find the acceleration after 4.5 seconds
3. The attempt at a solution Im getting confused on whether the above equation is for velocity or distance. I know the derivative of distance is velocity and that the derivative of velocity is acceleration. If i take the derivative of the above equation i end up with Sa=cos 2π This eliminates T , So im not sure what im doing wrong. Thanks in advance for help. Cheers 


#2
Jun711, 11:20 PM

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Yes, s is distance. But the derivative d/dt sin(2*pi*t) definitely isn't cos(2*pi). Use the chain rule. What's the derivative of sin(t)? Does the t disappear?



#3
Jun711, 11:34 PM

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Ok would i be correct in saying the derivative of
S=sin2πt is Sv=2π cos 2πt ? 


#4
Jun711, 11:35 PM

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The equation of motion is given by s=sin2πt (m) . Find the acceleration after 4.5 sec



#5
Jun711, 11:43 PM

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So to get the derivative of Sv=2π cos 2πt to find acceleration . Would i just use the chain rule again?
If so does this look correct ? Sa=2π^2 sin 2πt? thanks 


#6
Jun711, 11:48 PM

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#7
Jun811, 12:05 AM

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ok i still cant seem to get the correct answer. with this equation Sv=2π cos 2πt
Im taking "2π cos " to be the outer function & 2πt to be the inner function so the derivative of "2π cos" would be just "sin" And the derivative of "2πt" would be "2π" When i put it together i end up with is "sin 2πt * 2π" or " Sa=2π sin 2πt Can you please let me know where im going wrong. Thanks 


#8
Jun811, 02:13 AM

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(d/dt)[2*pi*cos(2*pi*t)] = (2*pi)*(d/dt) cos(2*pi*t). You really do have to be more careful!
RGV 


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