Q101 / 200 · NEET 2009 (Screening) · Physics
A bus is moving with a speed of $10 \mathrm{~ms}^{-1}$ on a straight road. A scooterist wishes to overtake the bus in $100 \mathrm{~s}$. If the bus is at a distance of $1 \mathrm{~km}$ from the scooterist, with what speed should the scooterist chase the bus?
- Correct answerA$20 \mathrm{~ms}^{-1}$
- B$40 \mathrm{~ms}^{-1}$
- C$25 \mathrm{~ms}^{-1}$
- D$10 \mathrm{~ms}^{-1}$
Solution
Let $y$ be the relative velocity of scooter(s) w.r.t. bus $(B)$, then

$$ \begin{aligned} \therefore \quad v_S & =\mathrm{v}+\mathrm{v}_{\mathrm{B}} \\ \text { Relative velocity } & =\text { displacement } / \text { time } \\ & =\frac{1000}{100}=10 \mathrm{~ms}^{-1} \end{aligned} $$ Now, substituting the value of in Eq. (i), we get $$ \mathrm{v}_{\mathrm{S}}=10+10=20 \mathrm{~ms}^{-1} $$

$$ \begin{aligned} \therefore \quad v_S & =\mathrm{v}+\mathrm{v}_{\mathrm{B}} \\ \text { Relative velocity } & =\text { displacement } / \text { time } \\ & =\frac{1000}{100}=10 \mathrm{~ms}^{-1} \end{aligned} $$ Now, substituting the value of in Eq. (i), we get $$ \mathrm{v}_{\mathrm{S}}=10+10=20 \mathrm{~ms}^{-1} $$
Hence, the correct answer is option: 'A'.