Problem: Find the position of a body that is dropped from the point with coordinates (0,10) and experiences the acceleration

a⃗(t) = ⟨12t, −32⟩.

(The units are the English system. In addition to gravity, there is a horizontal wind of linearly increasing strength.)

Solution: The velocity vector must be

v⃗(t) = ⟨6t^{2} + C_{1}, −32t + C_{2}⟩.

“Dropped” implies that the initial velocity was zero, so C_{1} and C_{2} are 0. Thus the position is

x⃗(t) = (2t^{3} + D_{1}, −16t^{2} + D_{2})

for some constants D_{1} and D_{2} . Comparing with the initial data, one gets

x⃗(t) = (2t^{3}, −16t^{2} + 10).