$$
u=72 km / hr =\frac{72 \times 1000}{3600}=20 m / s
$$
using Newton's equation of uniform acceleration \(v=u+a t \ldots \ldots(1)\)
$$
\begin{aligned}
&a=\frac{v-u}{t} \\
&a=\frac{0-20}{10}=\frac{-20}{10},
\end{aligned}
$$
\(a=-2 m / s ^{2}\) (deceleration)
thus;
$$
\begin{aligned}
&s=u t+1 / 2 a t^{2} \\
&s=20 \times 10+1 / 2(-2) \times 10^{2} \\
&s=200-100=100 m
\end{aligned}
$$
method \(-2\)
using the velocity - time graph below
distance travelled
\(=\) area under the velocity time graph \(=\) area of a triangle
$$
\begin{aligned}
&s=1 / 2 \times b \times h \\
&s=1 / 2 \times 10 \times 20 \\
&=200 / 2=100 m
\end{aligned}
$$