Graphs & tables

Average velocity is change in position divided by change in time:

$$v_{\text{avg}}=\frac{x(3)-x(1)}{3-1}$$

From the graph, $x(3)=-2$. To find $x(1)$, use the fact that the first piece is a straight line between $(0,0)$ and $(3,-2)$. The line through these points is

$$x(t)=-\frac{2}{3}t$$

So $x(1)=-\frac{2}{3}$, and the average velocity over $[1,3]$ is

$$\begin{array}{rl}{v}_{\text{avg}}& =\frac{-2-(-\frac{2}{3})}{3-1}=\frac{-\frac{4}{3}}{2}\\ & =-\frac{2}{3}\text{ units/sec}\end{array}$$

A particle is at rest when its velocity is zero. On a position graph, that means the slope is zero, so look for horizontal line segments. These occur over the intervals $3<t<5$ and $7<t<9$.

The particle moves left when $v(t)<0$ and right when $v(t)>0$. On a position graph, look at the sign of the slope on each piece of the graph.

Here is a sign chart for $v(t)$, with each interval matching a piece of the position graph:

From the sign chart, the particle’s movement is as follows:

1. It moves left on the interval $(0,3)$
2. It is at rest on $(3,5)$
3. Moves right on $(5,7)$
4. At rest again on $(7,9)$
5. Continues moving right on $(9,11)$

Look for where $v(t)$ changes sign (from positive to negative or vice versa).

On a velocity graph, the sign comes from whether the graph is above or below the $t$-axis. A sign change happens at a $t$-intercept.

The first $t$-intercept is at $t=3$.

The second one is on the line that passes through $(7,-3)$ and $(10,-2)$. The equation of the line between those two points is

$$y=\frac{5}{3}(t-10)+2$$

So the $t$-intercept occurs at $t=8.8$ minutes.

The third $t$-intercept is at $t=12$.

The sign chart below separates intervals based on these $t$-intercepts.

$v(t)$ changes sign (so the particle changes direction) at $t=3$ and $t=8.8$ minutes only.

Look for when the velocity is constant (so the graph is horizontal and stays at the same value). On a velocity graph, that means the slope is zero.

This occurs on the interval $4<t<7$. Over that time, the velocity is $-3$ meters/minute, meaning the particle moves left at a constant speed of $3$ meters per minute.

A particle speeds up when velocity and acceleration have the same sign, and it slows down when they have opposite signs.

• The sign of velocity comes from whether $v(t)$ is above or below the $x$-axis.
• The sign of acceleration comes from the slope of the velocity graph
  • Positive slope $\Rightarrow a(t)>0$
  • Negative slope $\Rightarrow a(t)<0$
  • Horizontal line $\Rightarrow a(t)=0$

Here’s a sign chart for velocity and acceleration, with intervals separated according to line segments and $t$-intercepts.

Thus, the particle is:

Speeding up on the intervals

• $(0,2)$
• $(3,4)$
• $(8.8,10)$
• $(12,14)$

Slowing down on the intervals

• $(2,3)$
• $(7,8.8)$
• $(10,12)$

Speed is $|v(t)|$, the distance from the velocity graph to the $t$-axis.

The particle hits a maximum speed of $3$ meters/minute. This happens at several times: once at $t=2$ minutes, throughout the interval from $4$ to $7$ minutes (moving left at constant speed), and again at $t=14$ minutes.

$$\begin{array}{rl}{v}_{\text{avg}}& =\frac{x(5)-x(2)}{5-2}=\frac{10-18}{3}\\ & =-\frac{8}{3}\text{ m/s}\end{array}$$

A particle changes direction when $v(t)$ changes sign. From a position table, you can estimate this by looking for where the position switches from increasing to decreasing (or vice versa).

Looking at the table values:

• From $t=0$ to $t=3$, the position increases ($5\to 20$), with a positive average velocity.
• From $t=3$ to $t=4$, the position decreases ($20\to 16$), with a negative average velocity.

So the particle must have changed direction between $3$ and $4$ seconds.

Without a function, the best estimate for $v^{\prime }(2)$ is the average rate of change between the nearby data points $t=1$ and $t=3$.

$$\begin{array}{rl}{a}_{\text{avg}}& =\frac{v(3)-v(1)}{3-1}=\frac{2-5}{2}\\ & =-\frac{3}{2}{\text{ m/s}}^2\end{array}$$

A particle changes direction when its velocity changes sign, which means the velocity must be $0$ at some time.

$v(t)$ is differentiable which implies it’s continuous.

Since the $v(3)$ is positive and $v(4)$ is negative, the Intermediate value theorem guarantees a time in $(3,4)$ when $v(t)=0$. So the particle must have changed direction sometime between $3$ and $4$ seconds.