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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

If two bodies $A$ and $B$ are projected with the same velocity $u$ but with different angles $\theta_1$ and $\theta_2$ respectively with the horizontal such that both have the same range,then the ratio of the times of flight of the bodies $A$ and $B$ is:

The centripetal acceleration of a particle in uniform circular motion is $18 \,m/s^2$. If the radius of the circular path is $50 \,cm$, the change in velocity of the particle in a time of $\frac{\pi}{18} \,s$ is (in $\,m/s$)

$A$ stone is projected vertically upwards with velocity $V$. Another stone of the same mass is projected at an angle of $60^{\circ}$ with the vertical with the same speed $V$. The ratio of their potential energies at the highest points of their journey is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module