Answer:
(a) [tex]\displaystyle s(t)= \frac{t^3}{3}-\frac{t^2}{2}-20t+C\ \ \ \ \forall\ \ 3\leqslant t\leqslant 9[/tex]
(b) 78 m
Explanation:
Physics' cinematics as rates of change.
Velocity is defined as the rate of change of the displacement. Acceleration is the rate of change of the velocity.
[tex]\displaystyle v=\frac{ds}{dt}[/tex]
Knowing that
[tex]\displaystyle v(t)= t^2 - t - 20[/tex]
(a) To find the displacement we need to integrate the velocity
[tex]\displaystyle \frac{ds}{dt}=t^2 - t - 20[/tex]
[tex]\displaystyle ds=(t^2 - t - 20)dt[/tex]
[tex]\displaystyle s(t)= \int(t^2 - t - 20)dt=\frac{t^3}{3}-\frac{t^2}{2}-20t+C\ \ \ \ \forall \ \ \ 3\leqslant t\leqslant 9[/tex]
(b) The displacement can be found by evaluating the integral
[tex]\displaystyle d=\int_{3}^{9} (t^2 - t - 20)dt[/tex]
[tex]\displaystyle d=\left | \frac{t^3}{3}-\frac{t^2}{2}-20t \right |_3^9=\frac{45}{2}+\frac{111}{2}=78\ m[/tex]
