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→=6.0  m/sj^. At t = 5.0 s, it moves through (11.0 m, 0) with velocity v→=−6.0 m/sj^. What is its acceleration at t = 2.5 s?

The motion of the particle is shown in the following figure:Radius, r=11.0−3.02=4.0 m.Magnitude of acceleration, =v2r=6.024.0=9.0  m/s2.at 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 center. We geta→=−9.0 m/s2j^