Question

A bus starts from rest with an acceleration of $ 1\,m/s^{2}. $ A man who is $48 \,m$ behind the bus starts with a uniform velocity of $10\, m/s$. Then the minimum time after which the man will catch the bus?

• A. 4 s
• B. 10 s
• C. 12 s
• D. 8 s

Answer 1

Answer: (D)

Bus starts from rest hence initial velocity is zero.
From equation of motion, we have
$s=u t+\frac{1}{2} a t^{2}$
where, $s$ is displacement,
$u$ is initial velocity,
$a$ is acceleration and $t$ is time.
Since, bus starts from rest, $u =0$
$ \therefore s_{b}=+\frac{1}{2} \times 1 \times t^{2}$
Man is $48 m$ behind the bus
$\therefore s_{m}=10 \times t+\frac{1}{2} \times 0 \times t^{2}$
$s_{m}=10\, t$
When man catches the bus
$s_{m}=s_{b}+4810 t=\frac{t^{2}}{2}+48 $
$\Rightarrow (t-8)(t-12)=0$
Hence, minimum time after which man will catch the bus is $8\, s$.