Question

A man is $d$ distance behind a bus. The bus starts moving away from the man with an acceleration $a$. At the same time man starts running towards the bus with a constant velocity $v$.

• 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 $t=\frac{v}{a}$
• C. If man just catches the bus, the time of catching the bus will be $t=\frac{2v}{a}$
• D. The man will catch the bus if $v\le \sqrt{ad}$

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 $t=\frac{v}{a}$



Let the man catches the bus after time $t$. Distance travelled by the bus in time $t$
$s_1=\frac{1}{2}a{t}^2$
Distance travelled by the man in time $t$
$s_2=vt$
The passenger will catch the bus if,
$d+s_1=s_2$
$d+\frac{1}{2}a{t}^2=vt$
$\frac{1}{2}a{t}^2-vt+d=0$
Solution of above equation is
$t=\frac{v\pm \sqrt{v^2-2ad}}{a}$
this $t$ should be real for man to catch the bus so
$v^2-2ad>0\Rightarrow v>\sqrt{2ad}$
If the man just catches the bus
then relative velocity of man wrt bus is zero, at that instant.
$V_{mb}=U_{mb}+a_{mb}t$
$0=v-at$
$\therefore t=\left(\frac{v}{a}\right)$