Question

A bus starts from rest with an acceleration $1m{s}^{-2}$ . A man who is 48 m behind the bus starts with a uniform velocity of $10m{s}^{-1}$ . Then the minimum rime after which the man will reach the bus

• A. 4 s
• B. 8 s
• C. 10 s
• D. 15 s

Answer 1

Answer: (B)

Let the minimum time after which the man catches the bus be t second. Then,
For bus, $s=ut+\frac{1}{2}a{t}^2$
i.e., $s_b=(0)(t)+\frac{1}{2}(1){(t)}^2=\frac{t^2}{2}$
For man $s_m=(10)(t)+\frac{1}{2}(0)t^2=10t$
From the question $s_m=s_b+48$
$\Rightarrow$ $10t=\frac{t^2}{2}+48$
$\Rightarrow$ $t^2-20t+96=0$
$\Rightarrow$ $(t-8)(t-12)=0$
$\therefore$ $t=8s$