Let the car accelerate for time $t_1$ and decelerate for time $t_2$.

Then, $t_1+t_2=t$

Maximum velocity reached after acceleration phase:

Since the car starts from rest and accelerates at rate $\alpha$, $v=\alpha t_1$

During deceleration, the car comes to rest with retardation $\beta$.

So, $v=\beta t_2$

because final speed becomes zero from speed $v$ in time $t_2$.

Hence, $\alpha t_1=\beta t_2$

Solve for $t_1$ and $t_2$.

From $t_1+t_2=t$ and $\alpha t_1=\beta t_2$

we get $t_2=\frac{\alpha}{\beta}t_1$

Therefore, $t_1+\frac{\alpha}{\beta}t_1=t$ $t_1\left(\frac{\alpha+\beta}{\beta}\right)=t$ $t_1=\frac{\beta t}{\alpha+\beta}$

Similarly, $t_2=\frac{\alpha t}{\alpha+\beta}$

Distance during acceleration:

$s_1=\frac{1}{2}\alpha t_1^2$ $s_1=\frac{1}{2}\alpha\left(\frac{\beta t}{\alpha+\beta}\right)^2$

Distance during deceleration:

Initial speed in this phase is $v=\alpha t_1$ and final speed is $0$.

So average speed is $\frac{v}{2}$, hence $s_2=\frac{v}{2}t_2$

Using $v=\beta t_2$, $s_2=\frac{1}{2}\beta t_2^2$ $s_2=\frac{1}{2}\beta\left(\frac{\alpha t}{\alpha+\beta}\right)^2$

Total distance:

$s=s_1+s_2$ $s=\frac{1}{2}\alpha\left(\frac{\beta t}{\alpha+\beta}\right)^2+\frac{1}{2}\beta\left(\frac{\alpha t}{\alpha+\beta}\right)^2$

$s=\frac{1}{2}\cdot \frac{\alpha\beta^2 t^2+\beta\alpha^2 t^2}{(\alpha+\beta)^2}$

$s=\frac{1}{2}\cdot \frac{\alpha\beta(\alpha+\beta)t^2}{(\alpha+\beta)^2}$

$s=\frac{\alpha\beta}{2(\alpha+\beta)}t^2$

Match with options:

$\boxed{\frac{\alpha\beta}{2(\alpha+\beta)}t^2}$

This is Option C.