Answer
(a) $v(t)=-32t$
(b) $s=-16t^2+540$
(c) $t=\sqrt { \frac{540}{16} }=5.8 sec$
Work Step by Step
Let
$v_f=v(t)=$finial velocity at the instant t
(a)
Given
Initial velocity=$v_i=0$, $a=-32$
Putting in the equation of motion
$v_f=v_i+at$
$v(t)=0-32t=-32t$
(b)
Since
$v=\frac{ds}{dt}$ where s is displacement
$\Longrightarrow$
$ds=vdt$
Putting $v=-32t$
$ds=-32tdt$
Taking integral
$\int ds=\int (-32)tdt$
$s=-32\int tdt$
$s=-32(\frac{t^2}{2})+C$
$s=-16t^2+C$
Put $t=0$, $s=540$ feet
$540=-16(0)+C$
$C=540$
Since
$s=-16t^2+C$
Putting $C=540$
$s=-16t^2+540$
(c)
Since
$s=-16t^2+540$
Putting $s=0$
$0=-16t^2+540$
$16t^2=540$
$t^2=\frac{540}{16}$
$t=\sqrt { \frac{540}{16} }=5.8 sec$
