Question

A man is $25\text{m}$ behind a bus, when the bus starts accelerating at $2{\text{m/s}}^2$ and man starts moving with constant velocity of $10\text{m/s}$. Time taken by him to board the bus is

• A. $2\text{s}$
• B. $3\text{s}$
• C. $4\text{s}$
• D. $5\text{s}$

Answer 1

The correct option is D $5\text{s}$

At the time when the man boards the bus,
Displacement of man = displacement of bus + 25
i.e. $10t=\frac{1}{2}\times 2\times t^2+25$
$\Rightarrow 10t=t^2+25$
$\Rightarrow t^2-10t+25=0$
$\Rightarrow t^2-5t-5t+25=0$
$\Rightarrow (t-5{)}^2=0$
Solving this equation we get $t=5\text{s}$

Alternate Sol:

Displacement of man w.r.t the bus, $S_{MB}=25$
Initial velocity of man w.r.t the bus, $u_{MB}=10\text{m/s}$
Acceleration of man w.r.t the bus, $a_{MB}=-2{\text{m/s}}^2$
Using second equation of motion,
$S_{MB}=u_{MB}t+\frac{1}{2}{a}_{MB}{t}^2$
i.e., $25=10t-\frac{1}{2}\times 2\times t^2$
$\Rightarrow t^2-10t+25=0$
Solving the equation, we get $t=5\text{sec}$