A particle undergoes uniform circular motion on a horizontal xy plane. At time t = 0, it moves through coordinates (3.0 m, 0) with velocity v→=(6.0m/s)j^. At t = 5.0 s, it moves through (11.0 m, 0) with velocity v→=(-6.0m/s)j^. What is its acceleration at t = 2.5 s?

The motion of the particle is shown in the following figure: Radius, r=(11.0−3.0)2=4.0mMagnitude of acceleration,a=v2r=(6.0)24.0=9.0m/s2at t = 2.5 s, the particle has completed one-fourth of its period and its velocity is parallel to x-axis. Since the centripetal acceleration is directed towards to the centre, We geta→=−9.0m/s2j^