Question

A man is $d$ distance behind a bus. The bus starts from rest and moves away from the man with an acceleration $a$. At the same time man starts running towards the bus with a constant velocity $v$. Which of the following options are correct?

• A. the man catches the bus if $v\ge \sqrt{2ad}$
• B. if man just catches the bus, the time of catching the bus will be $t=v/a$
• C. if man just catches the bus, the time of catching the bus will be $t=2v/a$
• D. the man will catch the bus if $v\ge \sqrt{ad}$

Answer 1

Answer: (A), (B)

The correct options are
A the man catches the bus if $v\ge \sqrt{2ad}$
B if man just catches the bus, the time of catching the bus will be $t=v/a$

Let the man catches the bus after it has travelled distance s

$\therefore$ Total Distance travelled by the man to catch the bus $=d+s$
$\Rightarrow d+s=v\times t$ (1)
For the bus, distance travelled is given by
$s=\frac{1}{2}a{t}^2$ (2)
Put (2) in (1), we get $d+\frac{1}{2}a{t}^2=v\times t$
$\Rightarrow a{t}^2-2vt+2d=0$
On solving this quadratic equation, we get
$t=\frac{2v\pm \sqrt{4v^2-4a\times 2d}}{2a}$ (3)
Man will catch the bus if, $\sqrt{4v^2-4a\times 2d}\ge 0$
$\Rightarrow v\ge \sqrt{2ad}$
If the man has to just catch the bus then, $\sqrt{4v^2-4a\times 2d}=0$
Using equ. (3), we get $t=\frac{v}{a}$