Question

1. the graph shows the position s(t) of an object moving along a straight line.

(i) when is the object stationary?
(ii) when is the object moving forwards?
(iii) sketch the graph of the velocity function y = v(t).
(iv) determine the time intervals when the object is speeding up.

Explanation:

Step1: Recall velocity - position relationship

Velocity $v(t)$ is the derivative of position $s(t)$. The object is stationary when $v(t)=0$, which means the slope of the $s(t)$ - graph is 0.

Step2: Find stationary intervals

Looking at the graph of $s(t)$, the slope is 0 in the interval $[3,5]$. So the object is stationary when $3\leq t\leq5$.

Step3: Determine forward - moving intervals

The object is moving forwards when $v(t)>0$, i.e., the slope of $s(t)$ is positive. From the graph, this occurs when $0 < t<2$ and $6 < t<7$.

Step4: Sketch velocity function

• For $0 < t<2$, $v(t)>0$ and is increasing (since the slope of $s(t)$ is increasing).
• At $t = 2$, $v(t)=0$ (slope of $s(t)$ is 0).
• For $2 < t<3$, $v(t)<0$ and is decreasing.
• For $3\leq t\leq5$, $v(t)=0$.
• For $5 < t<6$, $v(t)<0$ and is decreasing.
• At $t = 6$, $v(t)=0$.
• For $6 < t<7$, $v(t)>0$ and is increasing.

Step5: Find speeding - up intervals

The object is speeding up when $v(t)$ and $a(t)$ (acceleration, derivative of $v(t)$) have the same sign.

• For $0 < t<1$, $v(t)>0$ and $a(t)>0$ (slope of $v(t)$ is positive).
• For $5 < t<6$, $v(t)<0$ and $a(t)<0$ (slope of $v(t)$ is negative).

Answer:

(i) $3\leq t\leq5$
(ii) $0 < t<2$ and $6 < t<7$
(iii) Sketch as described above.
(iv) $0 < t<1$ and $5 < t<6$