Question

A bus starts from rest with an acceleration $ 1\,m{{s}^{-2}} $ . A man who is 48 m behind the bus starts with a uniform velocity of $ 10\text{ }m{{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=8\text{ }s $