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3-2.Motion in Plane
hard
Three particle $A, B$ and $C$ move in a circle of radius $r = \frac{1}{\pi }\,m$, in anticlockwise direction with speeds $1\, m/s$, $2. 5\, m/s$ and $2\, m/s$ respectively. The initial positions of $A, B$ and $C$ are as shown in figure. The ratio of distance travelled by $B$ and $C$ by the instant $A, B$ and $C$ meet for the first time is
A$3 : 2$
B$5 : 4$
C$3 : 5$
D$3 : 7$
Solution
Let the particles meet at point $P$ after time $t.$
Thus using $\theta^{\prime}=w t$ where $w=\frac{\bar{v}}{r}$
For $A:$ $\theta=\frac{1}{r} t$ $….(1)$
For $B:$ $\theta+\frac{\pi}{2}=\frac{2.5}{r} t$ $…(2)$
For $C:$ $\theta+\pi=\frac{2}{r} t$ $…(3)$
Dividing $(2)$ by $(3),$ we get $\theta=-3 \pi$
Thus, distance moved by $\mathrm{B}, \theta_{B}=-3 \pi+\frac{\pi}{2}=-2.5 \pi$
Distance moved by $\mathrm{C}, \theta_{C}=-3 \pi+\pi=-2 \pi$
$\theta_{B}: \theta_{C}=5: 4$
Thus using $\theta^{\prime}=w t$ where $w=\frac{\bar{v}}{r}$
For $A:$ $\theta=\frac{1}{r} t$ $….(1)$
For $B:$ $\theta+\frac{\pi}{2}=\frac{2.5}{r} t$ $…(2)$
For $C:$ $\theta+\pi=\frac{2}{r} t$ $…(3)$
Dividing $(2)$ by $(3),$ we get $\theta=-3 \pi$
Thus, distance moved by $\mathrm{B}, \theta_{B}=-3 \pi+\frac{\pi}{2}=-2.5 \pi$
Distance moved by $\mathrm{C}, \theta_{C}=-3 \pi+\pi=-2 \pi$
$\theta_{B}: \theta_{C}=5: 4$
Standard 11
Physics
