Question
A bus is moving with a speed of $$10\,m{s^{ - 1}}$$ on a straight road. A scooterist wishes to overtake the bus in $$100\,s.$$ If the bus is at a distance of $$1\,km$$ from the scooterist, with what speed should the scooterist chase the bus?
A.
$$20\,m{s^{ - 1}}$$
B.
$$40\,m{s^{ - 1}}$$
C.
$$25\,m{s^{ - 1}}$$
D.
$$10\,m{s^{ - 1}}$$
Answer :
$$20\,m{s^{ - 1}}$$
Solution :
Let $$v$$ be the relative velocity of scooter $$\left( s \right)$$ w.r.t. bus $$\left( B \right),$$ then
$$v = {v_s} - {v_B}$$
$$\therefore {v_s} = v + {v_B}\,......\left( {\text{i}} \right)$$
$${\text{Relative velocity = }}\frac{{{\text{Displacement}}}}{{{\text{Time}}}}$$
$$ = \frac{{1000}}{{100}} = 10\,m{s^{ - 1}}$$
Now, substituting the value of $$v$$ in Eq. (i), we get
$${v_s} = 10 + 10 = 20\,m{s^{ - 1}}$$
Let $$v$$ be the relative velocity of scooter $$\left( s \right)$$ w.r.t. bus $$\left( B \right),$$ then
$$v = {v_s} - {v_B}$$
$$\therefore {v_s} = v + {v_B}\,......\left( {\text{i}} \right)$$
$${\text{Relative velocity = }}\frac{{{\text{Displacement}}}}{{{\text{Time}}}}$$
$$ = \frac{{1000}}{{100}} = 10\,m{s^{ - 1}}$$
Now, substituting the value of $$v$$ in Eq. (i), we get
$${v_s} = 10 + 10 = 20\,m{s^{ - 1}}$$
