Correct Option (d)
Explanation:
Let the car accelerate for time \(t_{1}\) travelling distance \(S_{1}\) and acquiring maximum velocity \(v\).
$$
\begin{aligned}
&\text { Then, } S_{1}=\frac{1}{2} \alpha t_{1}^{2} \text { [from the equation, } S=u t+\frac{1}{2} a t^{2} \text { ] } \\
&\text { and } \quad \begin{aligned}
v &=\alpha t_{1} \\
\Rightarrow \quad & S_{1}=\frac{1}{2} \alpha \times \frac{v^{2}}{\alpha^{2}} \\
&=\frac{v^{2}}{2 \alpha}
\end{aligned}
\end{aligned}
$$
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 ecuation, \(u=v-a t\) ]
$$
\begin{aligned}
&\Rightarrow \quad S_{2}=\frac{v^{2}}{\beta}-\frac{v^{2}}{2 \beta}=\frac{v^{2}}{2 \beta} \\
&\text { Now }_{1} \quad \begin{array}{c}
t_{1}+t_{2}=t \\
\frac{v}{\alpha}+\frac{v}{\beta}=t
\end{array} \\
&\Rightarrow \quad v=\left(\frac{\alpha \beta}{\alpha+\beta}\right) t
\end{aligned}
$$
Also, total distance travelled,
$$
\begin{aligned}
S &=S_{1}+S_{2} \\
&=\frac{v^{2}}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) \\
&=\frac{1}{2}\left(\frac{\alpha \beta t}{\alpha+\beta}\right)^{2}\left(\frac{\alpha+\beta}{\alpha \beta}\right) \\
&=\frac{1}{2}\left(\frac{\alpha \beta}{\alpha+\beta}\right) t^{2}
\end{aligned}
$$
