A car starts from the rest and moves along the $\mathrm{x}-$axis with a constant acceleration of $5 \mathrm{m}/{\mathrm{s}}^2$for$8\mathit{ sec}$. If it continues with constant velocity, what distance will the car cover in $12 \mathrm{seconds}$ since it started from the rest?

The total distance covered by the car is $320\mathit{ m}$.
Given, 
The acceleration of the car $\mathrm{a}=5 \mathrm{m}/{\mathrm{s}}^2$
Time taken $\mathrm{t}=8 \mathrm{s}$
From the second equation of motion we have,
$\Rightarrow \mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^{2\mathrm{ }}$
$\mathrm{s}$ is the distance covered by car
$u$ is the velocity of car at rest
The initial velocity$\mathrm{u} =0$, the time taken to complete the distance,$\mathrm{t} =\mathrm{ }8 \mathrm{s}$, the acceleration,$\mathrm{a} =\mathrm{ }5 \mathrm{m}/{\mathrm{s}}^2$, then the distance:
${\Rightarrow s}_1=0+\frac{1}{2}\times 5\times 8\times 8$
         $=160\mathit{ m}$
The velocity, $\mathrm{v}=\mathrm{u}+\mathrm{at}$
                          $=0+5\times 8$
                          $=40 \mathrm{m}/\mathrm{s}$
The distance covered in the next 4 s:
${\Rightarrow \mathrm{s}}_2=\mathrm{vt}$
         $=40\times 4$
         $=160\mathit{ m}$
Therefore, the total distance travelled is
$\Rightarrow \mathrm{s}={\mathrm{s}}_1+{\mathrm{s}}_2$
        $=160+160$
        $=320 \mathrm{m}$