The vector position of a 3.85 g particle moving in the xy plane varies in time according to r1 = (3î + 3ĵ)t + 2ĵt2 where t is in seconds and r is in centimeters. At the same time, the vector position of a 5.40 g particle varies as r2 = 3î − 2ît2 − 6ĵt. (a) Determine the vector position (in cm) of the center of mass of the system at t = 2.80 s. rcm = cm (b) Determine the linear momentum (in g · cm/s) of the system at t = 2.80 s. p = g · cm/s (c) Determine the velocity (in cm/s) of the center of mass at t = 2.80 s. vcm = cm/s (d) Determine the acceleration (in cm/s2) of the center of mass at t = 2.80 s. acm = cm/s2 (e) Determine the net force (in µN) exerted on the two-particle system at t = 2.80 s. Fnet =  µN

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