A missile is accelerating at a rate of 4t m/sec^2 from a position at rest in a silo 35m below ground level. How high above the ground will it be after 6 seconds?
So we have height y(t) such that d/dt(dy/dt) = 4t with initial conditions y(0) = -35 m and y'(0) = 0. Integrating both sides we get dy/dt = 2*t^2 + C. Integrating once more we get y = (2/3)*t^3 + C*t + k, where C and k are constants. Now at t = 0 we have dy/dt = C = 0, so y = (2/3)*t^3 + k. Next, at t = 0 we have y = k = -35. Thus height y(t) = (2/3)*t^3 - 35. At t = 6 seconds we get y(6) = (2/3)*6^3 - 35 = 144 - 35 = 109 m.
