Question

A particle is moving in a plane with constant radial velocity \dot{r} = 2.68 ​r ​˙ ​​ =2.68 m/s, having started at the origin. It also has a constant angular velocity \dot{\theta} = 2.45 ​θ ​˙ ​​ =2.45 rad/s. When the particle is 3.58 m from the origin, what is the magnitude of its velocity in m/s?

Answer 1

Answer:

9.2m/s

Explanation:

We are given that

Radial velocity=$\dot r=2.68m/s$

Angular velocity=$\dot \theta=2.45rad/s$

Displacement=r=3.58 m

We have to find the magnitude of velocity of particle in m/s.

The magnitude of velocity of particle in polar coordinates is given by

$\mid v\mid=\sqrt{(\dot r)^2+r^2(\dot \theta)^2}$

Using the formula

The magnitude of velocity of particle

$\mid v\mid=\sqrt{(2.68)^2+(3.58)^2(2.45)^2}$

$\mid v\mid=9.2m/s$

Hence, the magnitude of velocity of particle=9.2m/s