Given functions, determine which has a velocity of zero at t=0 s| University Physics

The expressions below give the position function for an object moving along the x-axis. In which case is the velocity zero at the time $t=0\:s$ . In all cases, $A,\:B,\:\omega$ are constants.

$x\left(t\right)=A\:cos\left(\omega t+\frac{\pi }{2}\right)$
$x\left(t\right)=A\:sin\left(\omega t+\frac{\pi }{2}\right)$
$x\left(t\right)=A\:e^{-\omega t}$
$x\left(t\right)=At^2+Bt$
$x\left(t\right)=\frac{A}{\left(t+\omega ^{-1}\right)}$

ANSWER: $x\left(t\right)=A\:sin\left(\omega t+\frac{\pi }{2}\right)$

To answer this question, we need to get the velocity function by getting the derivative of the given position functions. After getting the derivative, the velocity function with a value of zero at time $t=0$ is the answer.

If we take the function $x\left(t\right)=A\:sin\left(\omega t+\frac{\pi }{2}\right)$ , and find its derivative, the resulting function is

$v\left(t\right)=A\:\omega \:cos\left(\omega t+\frac{\pi }{2}\right)$ .

If we substitute $t=0$ in the velocity function, the result is

$v\left(t\right)=A\:\omega \:cos\left(\frac{\pi }{2}\right)$ .

And we know that the $cos\left(\frac{\pi }{2}\right)$ is equal to zero, which makes the velocity function equal to zero at time zero.