Answer
t=3.27s
s=6.53m
Work Step by Step
$ds=\frac{vdv}{a}$
$\int _0^s\:ds\:=\int _6^v\:\frac{v}{-1.5v^{\frac{1}{2}}}dv$
$s=\int _6^v\:-0.6667v^{\frac{1}{2}}\:dv$
=$\left(6.532-0.4444v^{\frac{3}{2}}\right)m$
If v = 0,
$\left(6.532-0.4444\left(0\right)^{\frac{3}{2}}\right)m\:=\:6.53m$
The time needed for the particle to stop
$dt=\:\frac{dt}{a}$
$\int _0^t\:dt\:=-\int _6^v\:\frac{dv}{1.5v^{\frac{1}{2}}}$
$t\:=-1.333\cdot \left(v^{\frac{1}{2}}\right)\left(from\:6\:to\:v\right)=\left(3.266-1.333v^{\frac{1}{2}}\right)s$
Thus, when v = 0
=$\left(3.266-1.333\left(0\right)^{\frac{1}{2}}\right)=3.27s$
