antiderivatives calculus

Question

The acceleration of a car starting from rest is given by a(t) = t^2 + 3t − 7√ t − 2 in m/s^2 . Find the position (relative to the car’s initial position when it was at rest) and velocity of the car after 4 seconds

Answer 1

Okay, since the -2's location is now known, we can solve this. We begin with integrating to go from acceleration to velocity:

v(t) = t^3/3 + 3/2t^2 - 14/3(t)^3/2 -2t + c

We are told the car starts at rest (v(0) = 0). That implies c = 0.

Now integrate again to get the displacement:

x(t) = t^4/12 + 1/2 * t^3 - 14/3(2/5) * t^5/2 - 2t^2 + b

x(t) = 1/12 * t^4 + 1/2 * t^3 - 28/15 * t^5/2 - 2 * t^2 + b

We are asked for the displacement from t = 0, so let's set x(0) = 0. That implies b = 0.

I will leave it to you to plug t = 4 into the equations for v(t) and x(t) (with c and b both = 0)