Three persons A, B, and C start running simultaneously on three concentric circular tracks from three collinear points P, Q, and R respectively, which are collinear with the centre and on the same side of the centre. The speeds of A, B, and C are $5\ \text{m/s$, $9\ \text{m/s}$, and $8\ \text{m/s}$ respectively. The lengths of the tracks are:}
A: $400\ \text{m}$, B: $600\ \text{m}$, C: $800\ \text{m}$. A and B run clockwise, and C runs anti-clockwise. Find the first time after they start when A, B, and C are collinear with the centre and on the same side of the centre.
Show Hint
When positions must align radially, focus on half-lap times and find their LCM.
For all three runners to be collinear with the centre, each runner must be either: \[ 0^\circ \quad\text{(starting point)} \quad\text{or}\quad 180^\circ \quad\text{(diametrically opposite)}. \] That is, each must have covered a distance equal to an integer multiple of half their track length.
Step 2: Effective speeds
The direction of running (clockwise or anti-clockwise) does not affect the collinearity condition — only the completion of half-lap multiples matters.
Step 3: Time conditions for each runner
Runner A: Half lap = \(200\) m, speed = \(5\) m/s Time for half lap: \[ \frac{200}{5} = 40 \ \text{s} \] So \(t\) must be a multiple of \(40\).
Runner B: Half lap = \(300\) m, speed = \(9\) m/s Time for half lap: \[ \frac{300}{9} = \frac{100}{3} \ \text{s} \] So \(t\) must be a multiple of \(\frac{100}{3}\).
Runner C: Half lap = \(400\) m, speed = \(8\) m/s Time for half lap: \[ \frac{400}{8} = 50 \ \text{s} \] So \(t\) must be a multiple of \(50\).
Step 4: LCM for first meeting condition
We need: \[ t \ \text{is a common multiple of } \ 40, \ \frac{100}{3}, \ 50 \] First, \(\mathrm{LCM}(40, 50) = 200\). Now find \(\mathrm{LCM}(200, \frac{100}{3})\): \[ 200 = 2^3 \cdot 5^2, \quad \frac{100}{3} = \frac{2^2 \cdot 5^2}{3} \] The LCM is the smallest number divisible by both: \[ \mathrm{LCM} = 400 \ \text{s} \]
