Question

suppose the position of a particle along the $x$-axis is given by $s(t)$ where $t$ is measured in seconds. suppose that- $s(t) > 0$ on $(-2,0) \cup (3,5) \cup (5,\infty)$,- $s(t) < 0$ on $(-\infty,-2) \cup (0,3)$, and- $s(-2) = s(0) = s(3) = s(5) = 0$.which of the following statements are true? there may be more than one true statement.$square$ the total distance traveled by the particle on $2,4$ is given by $|s(4) - s(2)|$.$square$ the displacement of the particle on $2,4$ is given by $s(4) - s(2)$.$square$ the total distance traveled by the particle on $0,2$ is given by $|s(2) - s(0)|$.$square$ the displacement of the particle on $0,2$ is given by $s(2) - s(0)$.

Explanation:

Step1: Analyze sign of $s'(t)$ on $[2,4]$

On $[2,3]$, $s'(t)<0$; on $[3,4]$, $s'(t)>0$.

Step2: Evaluate total distance on $[2,4]$

Total distance = $|s(3)-s(2)| + |s(4)-s(3)|
eq |s(4)-s(2)|$.

Step3: Evaluate displacement on $[2,4]$

Displacement is always $s(\text{end})-s(\text{start}) = s(4)-s(2)$.

Step4: Analyze sign of $s'(t)$ on $[0,2]$

On $[0,2]$, $s'(t)<0$ (particle moves left).

Step5: Evaluate total distance on $[0,2]$

Since direction is constant, total distance = $|s(2)-s(0)|$.

Step6: Evaluate displacement on $[0,2]$

Displacement is always $s(\text{end})-s(\text{start}) = s(2)-s(0)$.

Answer:

• The displacement of the particle on $[2,4]$ is given by $s(4) - s(2)$.
• The total distance traveled by the particle on $[0,2]$ is given by $|s(2) - s(0)|$.
• The displacement of the particle on $[0,2]$ is given by $s(2) - s(0)$.