If a car at rest accelerates uniformly to a speed of 144 km/h in 20 sec, it covers a distance of

$\begin{array}{l}\mathrm{v}=[144\times 1000/(60\times 60\left)\right]\mathrm{m}/\sec \\ \mathrm{v}=\mathrm{u}+\mathrm{ft}\end{array}$
or (144x 1000) / (60 x 60) = 0 + f x 20
$\therefore \mathrm{f}=\frac{144\times 1000}{60\times 60\times 20}=2\mathrm{m}/{\sec }^2$
Now $\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{ft}}^2$
$=0+\frac{1}{2}\times 2\times (20{)}^2=400\mathrm{m}$