Posted by LadyJ on .
The vector position of a 3.00 g particle moving in the xy plane varies in time according to the following equation.
r1 = <(3i+3j)t + 2jt^2>
At the same time, the vector position of a 5.85 g particle varies according to the following equation.
r2 = <3i  2it^2  6jt>
For each equation, t is in s and r is in cm. Solve the following when t = 2.10
(e) Find the net force exerted on the twoparticle system.
____i ìN
____j ìN

Physics 
Steve,
r1 = 3ti + (2t^2 + 3t)j
r2 = (32t^2)i  6tj
v1 = r1' = 3i + (4t+3)j
v2 = r2' = 4ti  6j
a1 = r1'' = 4j
a2 = r2'' = 4i
F1 = m1a1 = 3*4j = 12j
F2 = m2a2 = 5.85(4i) = 23.4i
Fnet = 23.4i + 12j