A scooterist sees a bus 1km ahead of him moving with a velocity of $10\,m/s$. With what speed the scooterist should move to overtake the bus in 100 seconds?
(A) \[10\,m/s\]
(B) \[20\,m/s\]
(C) \[50\,m/s\]
(D) \[30\,m/s\]

Hint: In this solution, we will use the concepts of relative velocity. The scooter should have such an absolute velocity that the relative velocity of the scooter with respect to the bus should allow him to cover a distance of 1 kilometre in 100 seconds.
Formula used: In this solution, we will use the following formula:
$v = \dfrac{d}{t}$ where $v$ is the velocity, $d$ is the distance, and $t$ is the time.

Complete step by step answer:
We’ve been given that a scooterist wants to cover a distance of 1 kilometre that exists between him and the bus in 100 seconds. Let us assume the absolute velocity of the scooterist to be $x$. Then the relative velocity of the scooterist with the bus will be
$v = x - 10$
This velocity should be such that the scooterist can cover a distance of 1 kilometre in 100 seconds, so we can write
$v = \dfrac{1}{{100}}$
Which gives us
$v = 0.01\,km/s$
Or $v = 10\,m/s$
Now that we know the relative velocity of the scooterist, we can calculate its absolute velocity from the relation $v = x - 10$ as
\[x = v + 10\]

\[ \Rightarrow x = 20\,m/s\] which corresponds to option (B).

Note: While using the concepts of relative velocity, we have to assume that the scooter is moving faster than the bus which is the only way it can overtake the bus. Also, since the bus and the scooterist are moving in the same direction, their relative velocity will be the difference of their absolute velocities.
Another way of solving this problem is to calculate the distance travelled by the bus in 100 seconds which will be
$d = 10 \times 100$
$ \Rightarrow d = 1000\,m$
Since the bus started 1 kilometre or 1000 metres in front of the scooterist, the net distance he will have to cover in 100 seconds will be $1000 + 1000 = 2000\,m$ for which it will require a velvety of
$x = \dfrac{{2000}}{{100}} = 20\,m/s$.