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

Answer: (A)

Let car accelerates for time $t_1$ and decelerates for time $t_2$ then
$t_1+t_2=t$ ..(1)
From $v=u+at$
$v=u+a{t}_1$
$\Rightarrow v=a{t}_1$
For deceleration
$v=u-at$
$0=a{t}_1-b{t}_2(\because u=v)$
$a{t}_1=b{t}_2$
$\Rightarrow t_2=\frac{a{t}_1}{b}$
$\therefore t_1+\frac{a{t}_1}{b}=1$ [From Eq. (1)]
$\Rightarrow t_1\left(1+\frac{a}{b}\right)=t$
$\Rightarrow t_1=\frac{bt}{a+b}$
$\therefore$ Maximum velocity of car
$v=a{t}_1=\frac{abt}{a+b}$