A particle moves on a circle of radius 2 m. Its angular position is θ(t) = t³ - 3t² + 4t, where θ is in radians and t in seconds. The magnitude of acceleration at t = 2 s is:
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A particle moves on a circle of radius 2 m. Its angular position is θ(t) = t³ − 3t² + 4t, where θ is in radians and t in seconds. The magnitude of acceleration at t = 2 s is:

G1 Physics · Accepted Answer

Angular speed: ω = dθ/dt = 3t² − 6t + 4. At t=2 s, ω=4 rad/s. Angular acceleration: α = dω/dt = 6t − 6, so α=6 rad/s². Tangential acceleration aₜ = Rα = 2×6 = 12 m/s². Radial acceleration a_c = Rω² = 2×16 = 32 m/s². Since aₜ ⟂ a_c, net acceleration a = √(aₜ²+a_c²) = √(12²+32²) = √1168 = 4√73 m/s².