Distance and Displacement with Integrals

Question

An object has a constant acceleration of 30ft/sec^2, an initial velocity of -10ft/sec, and an initial position of 4 ft. Find the position function, s(t), describing the motion of the object.

Answer 1

Acceleration = change in velocity / change in time

a = dv/dt

a * dt = dv

Integrate both sides

a * t + C = v

v = a * t + C

Substitute -10 ft/s for v and 0 for t.

-10 = 30 * 0 + C

-10 = C

v = 30t - 10

Velocity = change in distance / change in time

v = ds/dt

v * dt = ds

ds = (30t - 10) * dt

Integrate both sides

s = 15t^2 - 10t + C

Substitute 4 ft for s and 0 for t.

4 = 15 * 0^2 - 10 * 0 + C

4 = C

s = 15t^2 - 10t + 4