Explain why the condition “F is any antiderivative of f on [a, b]” from the Evaluation Theorem is fulfilled by this scenario

Question

You are tracking the velocity and position of a rocket-propelled object near the surface of Mars. The velocity is v(t) and the position is s(t), where t is measured in seconds, s in meters, and v in meters per second. It is known that the v(t) = ds/dt = 4.94 – 3.72t and s(0) = 5.

detailed please

Answer 1

Seems like this is requesting that you state why the evaluation theorem works in this situation.

We have observations of specific positions and associated velocities at these positions from the description.

Continuity is required F'(t) = f(t) in the interval is required

You have data for the position intervals, instantaneous velocity is the time derivative of the position or slope of the time vs position plot. The position function is continuous between the values you have noted for the positions. It is also known that the velocity function is continuous. The position is the anti derivitive of the velocity in the intervals observed for the displacements and associated velocities. QED.

Basically we then have S(t) - S(0) = 4.94 x t -3.72/2 x t² for all t in our observations.