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Q. A bus strats to move with an acceleration of $1 \,m / s ^{2}$. A boy who is $48\, m$ behind the bus starts running at $10\, m / s$. The minimum time (in $s$) at which the boy can catch the bus is

Motion in a Straight Line

Solution:

Let $t$ be the time in which the boy catches the bus.
$\therefore 48+\frac{1}{2} \times 1 \times t^{2}=10\, t$
$ \Rightarrow t=8\, s$
or $12 \,s$.
Minimum time, $t=8\, s$