$x=20t-\large\frac{5}{3}$$t^2$
When the vehicle comes to a stop, velocity =0
Therefore $20-\large\frac{10}{3}$$t=0=>t=6 \;secs$
The distance travelled in $6 \;secs$ is obtained by substituting $t=6$ in (i)
$x=20 \times 6-\large\frac{5}{3}$$ \times 36$
$\quad=120-60=60\; m$
The distance travelled before the vehicle comes to a half $=60 \;m$