Question

A particle is moving in a circle in a radius 'r'
with a consatnt speed 'v'. then change in velocity after the particle has travelled a
distance equal to
(1/8)th of the circumference
of the circle is what?​

Answer 1

Let,

Initial position of particle be A velocity be   $\bold{v_A = vi+0j}$

After it has travelled (1/8)th of circle ⇒ (360/8 = 45)

Therefore θ = 45°

So,

Velocity, $\bold{v_B=v\:cos45\:i-v\:sin45\:j}$

So,

• Change in velocity in x direction = $\bold{v\:cos45-v=-0.292v}$

• Change in velocity in y direction = $\bold{-v\:sin45-0=-0.707v}$

Net change in velocity

= $\bold{-0.292\:vi-0.707\:vj}$

= $\bold{\sqrt{(0.292)^2+(0.707)^2}}$

= 0.765 V

$\boxed{\bold{\therefore{\:Change \:in \:Velocity=0.765\:V}}}$

Answer 2

Probably you had got your answer.

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