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A bus begin to move with an acceleration of $1 m / s^{2}$ . A man who is $48 m$ behind the bus start runs at $10 m / s$ to catch the bus the man will be able to catch the bus after

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8 s

5 s

6 s

7 s

answer is A.

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Detailed Solution

A bus begin to move with an acceleration of

1 m / s^{2}

. A man who is

48 m

behind the bus start runs at

10 m / s

to catch the bus the man will be able to catch the bus after

8 s

.
Let the man will catch after

t

second, initial velocity of the bus

u = 0

and the acceleration of the bus is

a = 1 m / s^{2}

.
So in

t

seconds distance travelled by bus will be,

s = ut + \frac{1}{2} a t^{2}

\Rightarrow s = 0 \times t + \frac{1}{2} \times 1 \times t^{2}

\Rightarrow s = \frac{1}{2} t^{2}

And the Man need to travel

48 + s

after

t

time at

10 m / s

speed then

48 + s = 10 t

\Rightarrow s = 10 t - 48

Then we get form the above equations,

10 t - 48 = \frac{1}{2} t^{2}

Solving the above quadratic equation we get,

t = 8, 12

. So the man can catch the bus after eight seconds.

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