Question

A body starting from rest, accelerates at a constant rate $a\, ms ^{-2}$ for some time after which it decelerates at a constant rate $b\, ms ^{-2}$ to come to rest finally. If the total time elapsed is $t \sec$, the maximum velocity attained by the body is given by

• A. $\frac{a b}{a+b} t\, ms ^{-1}$
• B. $\frac{a b}{a-b} t\, ms ^{-1}$
• C. $ \frac{2ab}{a+b}t\,ms^{-1} $
• D. $ \frac{2ab}{a-b}t\,ms^{-1} $

Answer 1

Answer: (A)

When the body accelerates initial velocity $u =0$
acceleration $= a$ and time $=t_{1}$
then final velocity $v=0+a t_{1} \ldots$ (i)
Now, as the body decelerates initial velocity, $= v$
deceleration $=- b$ and time $=\left(t-t_{1}\right)$
$[ t =$ total time elapsed $]$
final velocity here $v_{2}=0$
Again $0=v-b\left(t-t_{1}\right) \ldots$ (ii)
From Eqs. (i) and (ii), we get
$a t_{1}=b\left(t-t_{1}\right)$
or $(a+b) t_{1}=b t$
$\Rightarrow t_{1}=\frac{b t}{(a+b)}$
Substituting the value of $t_{1}$ in Eq. (i), we get
$v=\frac{a b t}{(a+b)} \,m s^{-1}$