Calculus Question

Question

An object has a constant acceleration of 40 ft/sec², an initial velocity of -20 ft/sec, and an initial position of 10 ft. Find the position function, s(t), describing the motion of the object

Answer 1

Since acceleration is the second derivative of position, we have: a(t) = s''(t) = 40.

Integrating once, we get velocity: v(t) = s'(t) = 40t + C.

Since we know the initial velocity, we have v(0) = -20, so C = -20, and our equation for velocity is v(t) = 40t-20.

Integrating a second time, we have s(t) = 20t²-20t+C (we get a new constant of integration.

From the initial position, we have s(0) = 10, so C = 10 here.

Putting all of this together, our position function is s(t) = 20t²-20t+10.

Hope this helps!