A particle is moving along a circular path with a constant speed of 10 ms-1. What is the magnitude of the change in velocity (in m/s) of the particle, when it moves through an angle of 60° around the centre of the circle?

Question

A particle is moving along a circular path with a constant speed of 10 ms-1. What is the magnitude of the change in velocity (in m/s) of the particle, when it moves through an angle of 60° around the centre of the circle? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Q. A particle is moving along a circular path with a constant speed of $10 m s^{- 1}$ . What is the magnitude of the change in velocity (in $m / s$ ) of the particle, when it moves through an angle of $6 0^{\circ}$ around the centre of the circle?

Change in velocity,
$∣Δ \overset{v}{ˉ} = ∣ v_{1}^{2} + v_{2}^{2} + 2 v_{1} v_{2} cos (π - θ)$
$= 2 v s in \frac{θ}{2} (∵ ∣ v_{1} ∣ = ∣ v_{2} ∣) = v$
$= (2 \times 10) \times s in (3 0^{\circ}) = 2 \times 10 \times \frac{1}{2}$
$= 10 m / s$

Q. A particle is moving along a circular path with a constant speed of 10ms−1. What is the magnitude of the change in velocity (in m/s) of the particle, when it moves through an angle of 60∘ around the centre of the circle?

Change in velocity, ∣Δvˉ=∣v12​+v22​+2v1​v2​cos(π−θ​) =2vsin2θ​(∵∣v1​​∣=∣v2​​∣)=v =(2×10)×sin(30∘)=2×10×21​ =10m/s

Q. A particle is moving along a circular path with a constant speed of $10 m s^{- 1}$ . What is the magnitude of the change in velocity (in $m / s$ ) of the particle, when it moves through an angle of $6 0^{\circ}$ around the centre of the circle?

Change in velocity,
$∣Δ \overset{v}{ˉ} = ∣ v_{1}^{2} + v_{2}^{2} + 2 v_{1} v_{2} cos (π - θ)$
$= 2 v s in \frac{θ}{2} (∵ ∣ v_{1} ∣ = ∣ v_{2} ∣) = v$
$= (2 \times 10) \times s in (3 0^{\circ}) = 2 \times 10 \times \frac{1}{2}$
$= 10 m / s$