Question

A car is travelling on a straight road. The maximum velocity the car can attain is $24\text{m/s}$. The maximum acceleration and deceleration it can attain are $1{\text{ms}}^{-2}\text{and}4{\text{ms}}^{-2}$ respectively. The shortest time( in $\text{s}$) the car takes from rest to rest in a distance of $480\text{m}$ is:

Answer 1

Using
$t=\frac{1}{v}+\frac{v}{2}(\frac{1}{\alpha }+\frac{1}{\beta })$
$=\frac{480}{24}+\frac{24}{2}(\frac{1}{1}+\frac{1}{4})$
$=20+12\times \frac{5}{4}=20+15=35\text{s}$

OR

Distance covered to attain maximum possible speed with maximum acceleration is $S_1=\frac{V^2}{2a}=\frac{{24}^2}{2}=288\text{m}$
Time taken $t_1=\frac{v}{a}=24\text{s}$
Distance covered when car reaches from maximum possible speed to rest is $S_2=\frac{V^2}{2a}=\frac{{24}^2}{2\left(4\right)}=72\text{m}$
Time taken $t_2=\frac{v}{a}=24/4=6\text{s}$
$S_1+S_2=360\text{m}$

Given distance is $480\text{m}$, hence remaining $120\text{m}$ is covered at maximum speed,
For which time taken is $t_3=\frac{120}{24}=5\text{s}$

Total time taken is $t_1+t_2+t_3=35\text{s}$