Question

Find magnitude of average acceleration between the
points P and Q for a particle performing uniform circular motion. Radius of circle is 2m​

Answer 1

We have to find the magnitude of average acceleration between the points P and Q for a particle performing uniform circular motion. where radius of circle is 2 m.

solution : initial velocity, u = 4j

final velocity, v = 4(-cos30° i + sin30° j) = -2√3i + 2j

angular velocity, ω = linear speed/radius = 4/2 = 2 rad/s

angular distance covered by particle, θ = π/3

we know, θ = ωt

⇒π/3 = 2 × t

⇒t = π/6

now acceleration = change in velocity/time

= (final velocity - initial velocity)/time

= (v - u)/t

= (-2√3i + 2j - 4j)/(π/6)

= (-2√3i - 2j)/(π/6)

magnitude of acceleration = √16 × 6/(22/7)

= 24 × 7/22

= 84/11 m/s²

Therefore the average acceleration between points P and Q is 84/11 m/s²

Answer 2

above is good explaination