A particle is moving in a circle of radius r with a constant speed V. The change in velocity after the particle has travelled a distance equal to ( (1/8) ) of the circumference of the circle is:

Question

A particle is moving in a circle of radius r with a constant speed V. The change in velocity after the particle has travelled a distance equal to ( (1/8) ) of the circumference of the circle is: (A) zero (B) 0.500 v (C) 0.765 v (D) 0.125 v

Tardigrade Admin · Accepted Answer

Angle traversed by the particle in ( (1/8) ) of the circumference of the circle =( (1/8) )(2π r/r) =(π /4)=450 So, change in velocity, Δ v=√v2+v2-2v2 cos (45o) =√2v2-2v2× (1/√2) =v√2-√2 =0.765 v

Q. A particle is moving in a circle of radius r with a constant speed V. The change in velocity after the particle has travelled a distance equal to $(\frac{1}{8})$ of the circumference of the circle is:

zero

0.500 v

0.765 v

0.125 v

Angle traversed by the particle in $(\frac{1}{8})$ of the circumference of the circle $= (\frac{1}{8}) \frac{2 π r}{r}$ $= \frac{π}{4} = 45^{0}$ So, change in velocity, ___ $Δ v = v^{2} + v^{2} - 2 v^{2} cos (45^{o})$ $= 2 v^{2} - 2 v^{2} \times \frac{1}{2}$ $= v 2 - 2$ $= 0.765 v$

Q. A particle is moving in a circle of radius r with a constant speed V. The change in velocity after the particle has travelled a distance equal to (81​) of the circumference of the circle is: