I have done the first part. The second appears to require integrating v. Could I have some help?

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- Dec 10th 2011, 06:46 AMStuck Manacceleration
I have done the first part. The second appears to require integrating v. Could I have some help?

- Dec 10th 2011, 06:57 AMskeeterRe: acceleration
start by taking the derivative on the far right side w/r to s ...

$\displaystyle \frac{d}{ds}\left(\frac{v^2}{2}\right) = v \cdot \frac{dv}{ds} = \frac{ds}{dt} \cdot \frac{dv}{ds} = \frac{dv}{dt} = a$ - Dec 10th 2011, 07:00 AMsbhatnagarRe: acceleration
Note that

$\displaystyle a=a\cdot\frac{ ds}{ds}= {dv \over dt}\cdot\frac{ ds}{ds}={ds \over dt}\cdot\frac{ dv}{ds}=v\left(\frac{dv}{ds} \right)$