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The 2.8-kg particle moves in the horizontal x-y plane and has the velocity shown at time t = 0. If the force F = 2.80 + 0.65t² newtons, where t is time in seconds, is applied to the particle in the y-direction beginning at time t = 0, determine the velocity v of the particle 3.6 seconds after F is applied.

у
F
--x
2.8 kg
4
7.0 m/s

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