Answer
The graphs require computing the time when $v=0,$ in which case,
we use $v=v_{0}+$ $a t^{\prime}=0 .$ Thus,
$x=-\frac{v_{0}^{2}}{2 a}=-\frac{(5.00 \mathrm{\ m} / \mathrm{s})^{2}}{2\left(1.67 \mathrm{\ m} / \mathrm{s}^{2}\right)}=-7.50 \mathrm{m}$
indicates the moment the car was at rest. SI units are understood.
see the image below :
Work Step by Step
The graphs require computing the time when $v=0,$ in which case,
we use $v=v_{0}+$ $a t^{\prime}=0 .$ Thus,
$x=-\frac{v_{0}^{2}}{2 a}=-\frac{(5.00 \mathrm{\ m} / \mathrm{s})^{2}}{2\left(1.67 \mathrm{\ m} / \mathrm{s}^{2}\right)}=-7.50 \mathrm{m}$
indicates the moment the car was at rest. SI units are understood.
See the image below :
