at t=0, a particle moving in the xy plane with constant acceleration has a velocity of vi=(3.00i-2.00j) m/s and is at the origin. At t=3.00s, the particles velocity is v=(9.00i+7.00j). Find (a) the acceleration of the particle and (b) its coordinates at any time t.

Answer 1

a = ∆v/∆t = ((9i+7j)-(3i-2j)/3 = 2i+3j
v = (2i+3j)t + 3i-2j = (2t+3)i + (3t-2)j
s = 1/2 (2i+3j) t^2 + (3i-2j) t + C
since s(0) = 0, C=0. and
s = 1/2 (2i+3j) t^2 + (3i-2j) t = (t^2+3t)i + (3/2 t^2 - 2t)j

Answer 2

Thanks

Answer 3

2i+1.5j

Answer 4

Its coordinats at any times t

Answer 5

Question

Answer 6

Yes