B. Explain why the condition "F is any antiderivative of f on [a,b]" from the Evaluation Theorem is fulfilled by this scenario. If "F" is the antiderivative of "f" and "f" is the velocity, then "F" must be the position function. The derivative of the positions is the velocity. The same reason as Part A applies here. The position of the rocket is one continuous path. It will not jump instantaneously somewhere else; therefore, it is a continuous function of [a,b] The condition "F is any antiderivative of "f, "which is continuous, on [a, b]" from the Evaluation Theorem it would be (s is any antiderivative of v on [a, b] in this situation) is fulfilled by this scenario because s is the position function and is the derivative of v. When we take the antiderivative of v, we will get s. The important part is this could be any antiderivative because we add a constant when we take the antiderivative of a function. Knowing that this means it could be any function because it could be any constant added onto it. (I think of these as functions that work as a family, with the only thing changing being the constant added onto it. We get a specific antiderivative when we have an initial condition. s(t)= [ v(t)dt=S (4.94-3.72t) dt 14.94t-3.72 +c s t =4.94t-1.86t+ C s(t)=f