Answer to Question #168958 in Calculus for Shein

Question

An athlete doing agility training starts at point A and runs to point B and then turns and runs back to point A and turns again and runs back to point B. The position for the athlete at time t is given by

𝑠(𝑡) = 𝑡³ − 6𝑡² + 9𝑡 − 2

1. On which intervals is the velocity positive? In which direction is the athlete running during these

time intervals?

2. On which intervals is the velocity negative? In which direction is the athlete running during these

time intervals?

3. At which points is velocity zero? What is the athlete’s movement at these points?

Answer 1

1. The velocity is: $s'(t)=v(t)=3t^2-12t+9=3(t^2-4t+3)=3(t-1)(t-3)$ . It is positive for $t\in(-\infty,1)\cup(3,+\infty)$. $s(1)=2$; $s(3)=-1$. During time $(-\infty,1)$ the athlete is running in the positive direction. During time $(3,+\infty)$ the athlete starts running in the negative direction and then he runs in the positive direction.
2. The velocity is negative for $t\in(1,3).$ The athlete runs in the positive direction and then in the negative direction.
3. The velocity is zero at $t=1$ and $t=3$. At $t=1$ the athlete runs in the positive direction and at $t=3$ the athlete runs in the negative direction.