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#### karush

##### Well-known member

- Jan 31, 2012

- 2,678

For $t \ge 0$ the position of a particle moving along the x-axis is given by $v(t)=\sin t—\cos t$ What is the acceleration of the particle at the point where the velocity is first equal to 0?

$a. \sqrt{2}$

$b. \, —1$

$c. \, 0$

$d. \, 1$

$e. —\sqrt{2}$

Ok well originally it was given as $x(t)$ but I changed it to $v(t)$

So via W|A $v(t)=0$ at

$t = 1/4 (4 π n + π), n \in Z$

So the first 0 would be $\dfrac{5\pi}{4}$

Hopefully so far