Question

A particle is moving on a circular path with
constant speed v. What is the change in its
velocity after it has described an angle of 60º ?

Answer 1

heya...

since A) Since the particle has moved by an angle of 60o, if we join the initial and final positions of the particles with the center, the figure so formed will be an equilateral triangle. Since all the sides of an equilateral triangle are equal, the displacement would be equal to the radius of the circle.



Average velocity of the particle is = total displacent/time

Let the radius of the circle be R

Displacement = R

Now,
total time taken by the particle to complete 1 rotation i.e to cover 360o⇒T=2πRv

So, time taken to cover 60o⇒T1=16T
T1=πR3v

So, average velocity =RπRv
=vπ

hope helped..

Answer 2

Answer: V

Explanation:

Now resolve vector Vo on vector V and you will get:

Vo Cos 120° = V

Vo (1/2) = V

Vo = 2V

So the difference = Vo - V

= 2V - V = V