A particle moves on a circle of radius 2 m with speed v = 2t² m/s. At t = 1 s, the angle φ between tangential acceleration and net acceleration satisfies:
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A particle moves on a circle of radius 2 m with speed v = 2t² m/s. At t = 1 s, the angle φ between tangential acceleration and net acceleration satisfies:

G1 Physics · Accepted Answer

At t=1 s, v=2 m/s and aₜ=dv/dt=4t=4 m/s². Radial acceleration a_c=v²/R=4/2=2 m/s². When φ is measured from the tangential direction toward the resultant, tanφ = a_c/aₜ = 2/4 = 1/2.