Answer
See explanations.
Work Step by Step
a. With the given equation $x(t)=At^2+Bt+C$, we can get $v(t)=x'(t)=2At+B$. The average velocity over $[t_1,t_2]$ can be found as $\bar v=\frac{s(t_2)-x(t_1)}{t_2-t_1}=\frac{A(t_2^2-t_1^2)+B(t_2-t_1)}{t_2-t_1}=A(t_2+t_1)+B$. On the other hand, the midpoint of the time interval is $t_m=\frac{t_2+t_1}{2}$ and the instantaneous velocity at this time is $v(t_m)=2A\frac{t_2+t_1}{2}+B=A(t_2+t_1)+B$. Thus we have $\bar v=v(t_m)$.
b. If we graph the position as a function of time, the average velocity over $[t_1,t_2]$ will be represented by the secant or average slope of this interval. On the other hand, the instantaneous velocity is represented by the slopes of tangent lines to the curve and the results in part (a) indicates that the secant will be equal to the slope of the tangent line at the midpoint of the interval.
