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A particle moves in the xy plane with constant acceleration

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A particle moves in the xy plane with constant acceleration. At t = 0 the particle is at 1 = (4.4 m) + (3.3 m), with velocity 1. At t = 3 s, the particle has moved to 2 = (11 m) - (1.9 m)and its velocity has changed to 2 = (4.5 m/s) - (5.8 m/s).

(a) Find V1.

(b) What is the acceleration of the particle?

(c) What is the velocity of the particle as a function of time?

(d) What is the position vector of the particle as a function of time?

Reference no: EM13640484

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