If an object moves along the $y-$axis $($marked in feet$)$ so that its position at time $x($in seconds$)$ is given by $f(x)=152x-19x^2,$ find the following.

$($A$)$ The instantaneous velocity function $v=f^{\text{'}}(x)$

$($B$)$ The velocity when $x=0$ and $x=4sec$

$($C$)$ The time$($s$)$ when $v=0$

$($A$)v(x)=$