Question

A bus starts from rest with an acceleration of $1m/s^2.$ A man who is $48m$ behind the bus starts with a uniform velocity of $10m/s$. 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
$s=ut+\frac{1}{2}a{t}^2$
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 $48m$ behind the bus
$\therefore s_m=10\times t+\frac{1}{2}\times 0\times t^2$
$s_m=10t$
When man catches the bus
$s_m=s_b+4810t=\frac{t^2}{2}+48$
$\Rightarrow (t-8)(t-12)=0$
Hence, minimum time after which man will catch the bus is $8s$.