Question

A bus begins to move with an acceleration of \[1m/{s^2}\] . A boy who is \[48m\] behind the bus starts running at \[10m/s\] towards the bus. After what time bus will cross the boy \[?\]
A. \[8s\]
B. \[12s\]
C. \[4s\]
D. \[24s\]

Answer 1

Hint: We have to know the definition of acceleration. Acceleration is the rate of change of velocity. Which means the change in speed. But sometimes when any object moves with a constant velocity but in a circular path that is also called acceleration because with the change of its direction the velocity is also changing.

Complete step by step answer:
The bus started with an initial velocity \[0m/s\] and the acceleration \[1m/{s^2}\], hence the formula is,
\[S = u + \dfrac{1}{2}a{t^2}\]
here S is equal to the distance, u is equal to the initial velocity and t is equal to the time and lastly a is equal to the acceleration.
So, now after putting the value of u and a, we can write \[S = \dfrac{{{t^2}}}{2}\] meters.
As the boy must cover \[48m\]additional so, we should have,
\[10t = \dfrac{1}{2}{t^2} + 48\]
\[\Rightarrow{t^2} - 20t + 96 = 0\]
\[\Rightarrow(t - 8)(t - 12) = 0\]
So either \[t = 8\] or \[t = 12\]
So here the bus will cross the boy after \[8\]sec as well as \[12\] sec.
How is that possible? This is so, as initially the boy will catch the bus after \[8\] sec but after \[12\] sec the bus will completely cross the boy. Which means after \[12\] sec the bus will cross its own length.

So the right answer will be option B.

Note:We can get confused between the option B and the option A. but here the question is after which time the bus will cross the boy so here the answer will be option no. B because the bus has to cross the length of its own. But if the question is after how many times the boy will catch the bus then the right option would be option no. A.