A car is travelling on a straight road. The maximum velocity the car can attain is ${\text{24\hspace{0.17em}ms}}^{-1}$ . The maximum acceleration and deceleration it can attain are ${\text{1\hspace{0.17em}\hspace{0.17em}ms}}^{\text{-2}}{\text{\hspace{0.17em}and\hspace{0.17em}4\hspace{0.17em}ms}}^{\text{-2}}$  respectively. The shortest time the car takes from rest to rest in a distance of  $200 m$ is,

Assuming the car accelerates for time t₁ and then decelerates for time t₂.

Let the maximum velocity reached is v, then

time of acceleration is ${\mathrm{t}}_1=\frac{\mathrm{v}}{\mathrm{a}}=\frac{\mathrm{v}}{1}$

time of deceleration is ${\mathrm{t}}_2=\frac{\mathrm{v}}{\mathrm{d}}=\frac{\mathrm{v}}{4}$

Area under the v - t graph gives the total distance travelled by the car in shortest time

$\Rightarrow 200=\frac{1}{2}\mathrm{v}({\mathrm{t}}_1+{\mathrm{t}}_2)=\frac{1}{2}\mathrm{v}\left(\frac{5\mathrm{v}}{4}\right)$

$\Rightarrow \mathrm{v}=8\sqrt{5}$

Total time is 

$\mathrm{T}={\mathrm{t}}_1+{\mathrm{t}}_2=\frac{5\mathrm{v}}{4}=22.4 \mathrm{s}$