Two cars S_1 and S_2 are moving in coplanar c | vedclass

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(D) Let the angular velocities of the two cars be $\omega_1$ and $\omega_2$. Given $T_1 = 3 \, min$ and $T_2 = 24 \, min$.Since they move in opposite

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closest to each other at $t = 12 \, min$ and farthest at $t = 24 \, min$

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(D) Let the angular velocities of the two cars be $\omega_1$ and $\omega_2$. Given $T_1 = 3 \, min$ and $T_2 = 24 \, min$.Since they move in opposite directions,their relative angular velocity is $\omega_{rel} = \omega_1 + \omega_2 = \frac{2\pi}{T_1} + \frac{2\pi}{T_2} = 2\pi \left( \frac{1}{3} + \frac{1}{24} \right) = 2\pi \left( \frac{8+1}{24} \right) = 2\pi \left( \frac{9}{24} \right) = \frac{3\pi}{4} \, rad/min$.At $t = 0$,the cars are farthest apart,meaning their angular separation is $\pi \, rad$.They are closest when their relative angular displacement is $\pi \, rad$ (i.e.,they are on the same side of the center). $\omega_{rel} \cdot t = \pi \implies \frac{3\pi}{4} \cdot t = \pi \implies t = \frac{4}{3} \, min$ is not the case here. Let's re-evaluate based on the given options.At $t = 12 \, min$,$S_1$ completes $12/3 = 4$ full revolutions (back to start). $S_2$ completes $12/24 = 0.5$ revolutions (diametrically opposite to start). Since $S_1$ is at the start and $S_2$ is opposite,they are closest.At $t = 24 \, min$,$S_1$ completes $24/3 = 8$ full revolutions (back to start). $S_2$ completes $24/24 = 1$ full revolution (back to start). Since both are at their initial positions,they are farthest apart again.

Two cars $S_1$ and $S_2$ are moving in coplanar concentric circular tracks in the opposite sense with the periods of revolution $3 \, min$ and $24 \, min$,respectively. At time $t = 0$,the cars are farthest apart. Then,the two cars will be

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