Question

A man running uniformly at $8\,m{s^{ - 1}}$ is $16\,m$ behind a bus when it starts accelerating at $2\,m{s^{ - 2}}$. Time taken by him to board the bus is:
(A) $2\,\sec$
(B) $3\,\sec$
(C) $4\,\sec$
(D) $5\,\sec$

Answer 1

Hint
The time taken can be determined by using the equation of motion formula. By using that formula, we can get the result like the equation by solving the equation the time taken can be determined. The solving of the equation is called factorization of the equation.
The equation of the motion is given by,
$\Rightarrow s = ut + \dfrac{1}{2}a{t^2}$
Where, $s$ is the distance between the man and bus, $u$ is the velocity of the man, $t$ is the time taken by the man to board the bus and $a$ is the acceleration of the bus.

Complete step by step answer
Given that, The velocity of the man running is, $u = 8\,m{s^{ - 1}}$,
The distance between the man and the bus is, $s = 16\,m$,
The acceleration of the bus is $a = 2\,m{s^{ - 2}}$.
Now, The equation of the motion is given by,
$\Rightarrow s = ut + \dfrac{1}{2}a{t^2}\,.......................\left( 1 \right)$
By substituting the distance between the man and the bus, velocity of the man and acceleration of the bus in the above equation (1), then the above equation is written as,
$\Rightarrow 16 = 8t - \dfrac{1}{2} \times 2{t^2}\,..................\left( 2 \right)$
By cancelling the same terms in the above equation, then the above equation is written as,
$\Rightarrow 16 = 8t - {t^2}$
By rearranging the terms in the above equation, then the above equation is written as,
$\Rightarrow {t^2} - 8t + 16 = 0$
By taking factorize in the above equation, then the above equation is written as,
$\Rightarrow {\left( {t - 4} \right)^2} = 0$
By taking the square root on both side in the above equation, then
$\Rightarrow t - 4 = 0$
Then the above equation is written as,
$\Rightarrow t = 4\,\sec$
Hence, the option (C) is the correct answer.

Note
The negative sign in the equation (2) which indicates that the acceleration of the bus is going outwards with respect to the velocity of the man. So, the acceleration of the bus is taken as the negative value. By the given information, $4\,\sec$ are required by the man to board the bus.