# Math Help

1. Oct 5, 2005

### quicknote

I'm suppossed to find the average acceleration of the mass during the time interval from -t to t.
The formula I'm using is $$a(-t,t)=\frac{(v(t)-v(-t))}{(t-(-t))}$$

I've already derived v(t) and v(-t) but I'm not sure how to simplify them after subtracting them.

$$v(t)=-R\omega sin(\omega t)*\hat{x}+R\omega cos(\omega t)\hat{y}$$
$$v(-t)=-R\omega sin(\omega (-t))\hat{x}+R\omega cos(\omega (-t))\hat{y}$$

I got as far as grouping the x and y terms...after that I don't know how else to simplify

$$[(-R \omega \hat{x}) ((sin(\omega t)+sin(\omega (-t))] + [(R \omega \hat{y}) ((cos \omega t) -cos (\omega (-t))]$$

Any help is appreciated.
thanks!

2. Oct 5, 2005

### Gale

from what you have, i get that the velocities all cancel out. $$sin(\omega t) + sin (-\omega t) = sin(\omega t) -sin(\omega t)=0$$ because $$sin(-\omega t)= -sin(\omega t)$$

ok, well just the sine's cancel...

the cosines simplify $$R\omega cos(\omega t) - R\omega cos(-\omega t)= R\omega(cos(\omega t) - cos(-\omega t))= R\omega( cos(\omega t) + cos(\omega t))= 2R\omega cos(\omega t)$$

and thats all in the y hat direction. so all your acceleration is vertical.

Last edited: Oct 5, 2005
3. Oct 5, 2005

### quicknote

Thanks Gale!

I didn't know this rule: sin(-x) = -sin (x)

4. Oct 5, 2005

### Gale

hah, don't feel so bad... i didn't either, but i drew it out. when you look at the picture, obviously they're equal.