A length of path ACB is 1500 m and the length of the path ADB is 2100 m. Two particles start from point A simultaneously around the track ACBDA. One of them travels the track in clockwise sense and other in anticlockwise sense with their respective constant speeds. After 12 s from the start, the first time they meet at the point B. After minimum time (in s) in which they meet first at point B, will they again meet at the point B is time $t_{min} = (12)^{x} s$ . The value of x is __________

Moderate

Solution

$The velocity of first particle is v_{1} = \frac{1500}{12} {ms}^{- 1} and the velocity$
$of second particle is v_{2} = \frac{2100}{12} {ms}^{- 1} . Let after completing n_{1}$
and n₂ trips, they will again meet at the point B.
$\begin{array}{l} ∴ \frac{(1500 + 2100) n_{1}}{v_{1}} = \frac{(1500 + 2100) n_{2}}{v_{2}} = t \\ Or \frac{n_{1}}{n_{2}} = \frac{v_{1}}{v_{2}} = \frac{\frac{1500}{12}}{\frac{2100}{12}} = \frac{1500}{2100} = \frac{5}{7} \\ ∴ t_{min} = \frac{3600 \times 5}{v_{1}} = \frac{3600 \times 5}{1500} \times 12 = 144 s = (12)^{x} \\ ∴ x = 2 \end{array}$

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