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

The correct option is D $5\text{s}$
Taking rightward direction to be positive,
Displacement of man w.r.t. the bus, $S_{MB}=25\text{m}$
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., $25=10t-\frac{1}{2}\times 2\times t^2$
$\Rightarrow t^2-10t+25=0$
$\Rightarrow t^2-5t-5t+25=0$
$(t-5)(t-5)=0$
$t=5\text{sec}$

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