A particle is moving in xy plane on a circular track of radius 2 m with constant speed v = 3 m/s as shown in the figure. If at time t = 0, particle is at point A, then acceleration of the particle (in ms-2)
at time t = π s will be
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Answers
acceleration of the particle at time t = πs will be (3√2/π) m/s² or 1.35 m/s² .
first of all, find time period of particle.
i.e ., T = 2πr/v
= 2π(2m)/(3m/s)
= 4π/3 sec
angle made by particle after time (t = πs), θ = π/(4π/3) × 2π = 3/2 π = 270°
now, acceleration = change in velocity/time
= √(v1² + v2² - 2v1v2cos270°)/(π s)
as particle moves with constant speed.
so, v1 = v2 = 3m/s
so, acceleration = √{(3)² + (3)²}/π
= (3√2/π) m/s²
= 1.35 m/s²
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first of all, find time period of particle.
i.e ., T = 2πr/v
= 2π(2m)/(3m/s)
= 4π/3 sec
angle made by particle after time (t = πs), θ = π/(4π/3) × 2π = 3/2 π = 270°
now, acceleration = change in velocity/time
= √(v1² + v2² - 2v1v2cos270°)/(π s)
as particle moves with constant speed.
so, v1 = v2 = 3m/s
so, acceleration = √{(3)² + (3)²}/π
= (3√2/π) m/s²
= 1.35 m/s²