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

by russjai
Tags: algebra, calculus, confusing
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russjai
#1
Jun7-11, 11:14 PM
P: 7
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
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Dick
#2
Jun7-11, 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?
russjai
#3
Jun7-11, 11:34 PM
P: 7
Ok would i be correct in saying the derivative of
S=sin2πt is Sv=2π cos 2πt ?

Dick
#4
Jun7-11, 11:35 PM
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The equation of motion is given by s=sin2πt (m) . Find the acceleration after 4.5 sec

Quote Quote by russjai View Post
Ok would i be correct in saying the derivative of
S=sin2πt is Sv=2π cos 2πt ?
Yes, you would.
russjai
#5
Jun7-11, 11:43 PM
P: 7
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
Dick
#6
Jun7-11, 11:48 PM
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Quote Quote by russjai View Post
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
Sure, use the chain rule again. But try to do it right. The derivative of cos(t) isn't sin(t), it's -sin(t). And the number in front of the sin(2*pi*t) isn't 2pi^2. Take another try.
russjai
#7
Jun8-11, 12:05 AM
P: 7
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
Ray Vickson
#8
Jun8-11, 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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