A particle is moving along $x$-axis, At time $t=0$, its $x$-coordinate is $x=-4 \mathrm{m}$. Its velocity-time equation is $\mathrm{v}=8-2t$  where, $\mathrm{v}$ is in $\mathrm{m}/\mathrm{s}$ and $t$ in seconds.
At how many times, particle is at a distance of $8 \mathrm{m}$ from the origin?

Comparing the given $v-t$equation with $v=u+at$. we have,

$v=8 m/s, a=-2m/s^2=cons\tan t$

Now, motion of the particle is as shown below. 

Now, 8 m distance from origin will be at two coordinates $x=8 m and x=-8m.$From the diagram, we can see that particle will cross these two points three times,$t_1,t_2 and t_3$.