A man is running with constant speed along a | vedclass

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

Question

(B) The man is running with a constant speed along a circular path. In uniform circular motion,the speed of the object remains constant at all points

Vedclass · Accepted Answer

$\frac{2 \sqrt 2 \pi}{5} \, m/s$

Vedclass · Answer

(B) The man is running with a constant speed along a circular path. In uniform circular motion,the speed of the object remains constant at all points on the path. The radius of the path is $r = 2 \sqrt 2 \, m$. The time taken to complete one round (period) is $T = 10 \, s$. The distance covered in one round is the circumference of the circle,$C = 2 \pi r$. $C = 2 \pi (2 \sqrt 2) = 4 \sqrt 2 \pi \, m$. The constant speed $v$ is given by the ratio of total distance to total time: $v = \frac{C}{T} = \frac{4 \sqrt 2 \pi}{10} = \frac{2 \sqrt 2 \pi}{5} \, m/s$. Since the speed is constant,the instantaneous speed at any time $t$,including $t = 2.5 \, s$,is the same as the average speed. Therefore,the instantaneous speed is $\frac{2 \sqrt 2 \pi}{5} \, m/s$.

$A$ man is running with constant speed along a circular path of radius $2 \sqrt 2 \, m$. He completes $1$ round in $10 \, s$. Find the instantaneous speed at $2.5 \, s$.

Similar Questions

Can a moving object have a constant speed but a changing velocity?

$A$ particle is performing $U.C.M.$ along a circle of radius $R$. In half the period of revolution, its displacement and distance covered are respectively

Find the angular speed of the hour hand of a clock.

In uniform circular motion,

Motion of a particle in $x-y$ plane is described by a set of following equations $x=4 \sin \left(\frac{\pi}{2}-\omega t\right) \text{ m}$ and $y=4 \sin (\omega t) \text{ m}$. The path of the particle will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module