Q122 / 160 · AP EAMCET 2021 (25 Aug Shift 1) · Physics
Two towns $A$ and $B$ are connected by a regular bus service with a bus leaving in either direction every $T$ min. A man cycling with a speed of $20 \mathrm{~km} / \mathrm{h}$ from $A$ to $B$ notices that a bus travelling in the direction of his motion goes past him every $18 \mathrm{~min}$ and every $6 \mathrm{~min}$ he notices a bus travelling in the opposite direction go past him. Assuming that, the buses travel with a constant speed. Find $T$ and the constant speed of the buses.
- A$\frac{2}{27} \mathrm{~h}$ and $38 \mathrm{~km} / \mathrm{h}$
- B$\frac{5}{8} \mathrm{~h}$ and $40 \mathrm{~km} / \mathrm{h}$
- Correct answerC$\frac{3}{20} \mathrm{~h}$ and $40 \mathrm{~km} / \mathrm{h}$
- D$\frac{2}{3} \mathrm{~h}$ and $28 \mathrm{~km} / \mathrm{h}$
Solution
Let, the speed of bus be $v_B$ and that of cyclist be $v_C$.
Case 1 Bus moving from $A$ to $B$,
Relative speed of bus $=v_B-v_C=v_B-20$
Distance covered $=$ Velocity $\times$ Time
$$
d=\left(v_B-20\right) \times 18...(i)
$$
Case 2 Bus moving from $B$ to $A$,
Relative speed of bus $=v_B+v_C$
$$
=v_B+20
$$
Distance covered $=$ Velocity $\times$ Time
$$
d=\left(v_B+20\right) \times 6...(ii)
$$
From Eqs. (i) and (ii), we get
$$
\begin{aligned}
& \left(v_B-20\right) \times 18=\left(v_B+20\right) \times 6 \\
& \Rightarrow \quad v_B=40 \mathrm{~km} / \mathrm{h} \\
& \text { From Eq. (ii), }\left(v_B+20\right) \times 6=v_B \times T \\
& \Rightarrow \quad(40+20) \times 6=40 \times T \\
& \Rightarrow \quad T=9 \min =\frac{9}{60} \mathrm{~h}=\frac{3}{20} \mathrm{~h} . \\
&
\end{aligned}
$$
Hence, the correct answer is option: 'C'.