The rotor shown is subject to a constant angular acceleration of $14ra\frac{d}{s^2}.$ The time needed to reach a rotor speed of $2200re\frac{v}{m}in$ is best given by

a$.16\mathrm{.}5$ s

$.)157$ s

given $=14$

c$.8\mathrm{.}23$ s

d$.70\mathrm{.}3$ s

$=$

$2800=0+16t$

$\frac{2200}{14}=15$