Question

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 :

• A. $ \frac{abt}{(a+ b)} $
• B. $ \frac{a^{2}t}{a +b} $
• C. $ \frac{at}{(a +b)} $
• D. $ \frac{b^{2}t}{a +b} $

Answer 1

The Correct Option is A

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 E (1)]
$\Rightarrow t_{1}\left(1+\frac{a}{b}\right)=t$
$\Rightarrow t_{1}=\frac{b t}{a +b}$
$\therefore$ Maximum velocity of car
$v=a t_{1}=\frac{a b t}{a +b}$