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Q.
A car starts from rest, moves with an acceleration $a$ and then decelerates at $a$ constant rate $b$ for sometimes to come to rest. If the total time taken is $t$. The maximum velocity of car is given by :
Let car accelerates for time $t_{1}$ and decelerates for time $t_{2}$ then
$t_{1}+t_{2}=t$ ..(1)
From $v=u +at$
$v=u+ at_{1}$
$\Rightarrow v=at_{1}$
For deceleration
$v =u-a t$
$0 =a t_{1}-b t_{2}\,\,(\because u= v)$
$a t_{1} =b t_{2}$
$\Rightarrow t_{2} =\frac{a t_{1}}{b}$
$\therefore t_{1}+\frac{a t_{1}}{b} =1$ [From Eq. (1)]
$\Rightarrow t_{1}\left(1+\frac{a}{b}\right)=t$
$\Rightarrow t_{1}=\frac{b t}{a +b}$
$\therefore $ Maximum velocity of car
$v=a t_{1}=\frac{a b t}{a +b}$