Question

A car starts from rest, moves with an acceleration $a$ and then decelerates at a constant rate $b$ for some time 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)

From equation of motion, we have
$v=u+at$
Since, body starts from rest $u=0$ .
Let $t_1$ be time when body accelerates and $t_2$
when it decelerates. $\therefore$ $t=t_1+t_2$
$\Rightarrow$ $t_2=t-t_1$
$\therefore$ $v=0+a{t}_1$ ...(1)
When, car finally comes to rest
$v=0$
$\therefore$ $0=v-b(t-t_1)$ ...(2)
From Eqs. (1) and (2), we get
$t_1=\frac{b}{a+b}t,v=\frac{ab}{a+b}t$