#### A car accelerates from rest at a constant rate for sometime after which it decelerates at a constant rate to come to rest. If the total time elapsed is t seconds, the total distance travelled is :Option: 1Option: 2Option: 3Option: 4

Let the car accelerate for time $t_{1}$ traveling distance $S_{1}$ and acquire maximum velocity v.
Then,$S_{1}=\frac{1}{2} \alpha t_{1}^{2}$ [from the equation, $S=u t+\frac{1}{2} a t^{2}$ ]
and $v=\alpha t_{1} \quad$  [from the equation, v=u+a t ]

$\Rightarrow \quad S_{1}=\frac{1}{2} \alpha \times \frac{v^{2}}{\alpha^{2}} =\frac{v^{2}}{2 \alpha} ......(1)$
After this car decelerates for time $t_{2}$ to come to rest
Hence, $\quad S_{2}=v t_{2}-\frac{1}{2} \beta t_{2}^{2}$       [from the equation, $S=u t-\frac{1}{2} a t^{2}$ ]
and $0=v-\beta t_{2}$         [from the equation, u=v-a t ]
$\Rightarrow \quad S_{2}=\frac{v^{2}}{\beta}-\frac{v^{2}}{2 \beta}=\frac{v^{2}}{2 \beta} \ldots (ii)$

$\begin{array}{lrl}\text { Now, } & t_{1}+t_{2} & =t \\ \\ \Rightarrow & \frac{v}{\alpha}+\frac{v}{\beta}=t \\ \\ \Rightarrow & v=\left(\frac{\alpha \beta}{\alpha+\beta}\right) t\end{array}$

Also, total distance traveled,

\begin{aligned} S &=S_{1}+S_{2} \\ &=\frac{v^{2}}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) \\ &=\frac{1}{2}\left(\frac{\alpha \beta t}{\alpha+\beta}\right)^{2}\left(\frac{\alpha+\beta}{\alpha \beta}\right) \\ &=\frac{1}{2}\left(\frac{\alpha \beta}{\alpha+\beta}\right) t^{2} \end{aligned}