Solving Equation of Motion: Find Acceleration After 4.5 sec

[russjai]

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

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 I am not sure what I am doing wrong.

Thanks in advance for help.
Cheers

 

[Dick]

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]

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

 

[Dick]

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

Yes, you would.

 

[russjai]

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]

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]

ok i still can't 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 I am going wrong. Thanks

 

[Ray Vickson]

(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