Thank you for reporting, we will resolve it shortly
Q.
A car accelerates from rest at a constant rate $\alpha$ for some time after which it decelerates at a constant rate $\beta$ to come to rest. If the total time elapsed is $t$ seconds, the total distance travelled is:
$v _{0}=\alpha t _{1}$ and $0= v _{0}-\beta t _{2}$
$ \Rightarrow v _{0}=\beta t _{2}$
$v _{0}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)= t$
$t _{1}+ t _{2}= t$
$\Rightarrow v _{0}=\frac{\alpha \beta t }{\alpha+\beta}$
Distance $=$ area of $v - t$ graph
$=\frac{1}{2} \times t \times v _{0}=\frac{1}{2} \times t \times \frac{\alpha \beta t }{\alpha+\beta}=\frac{\alpha \beta t ^{2}}{2(\alpha+\beta)}$
