4.2.2 Graphs & tables

Achievable AP Calculus AB

4. Contextual uses

4.2. Straight-line motion

Our AP Calculus AB course is currently in development and is a work-in-progress.

Graphs & tables

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What you’ll learn:

How to interpret position and velocity from graphs or tables

In addition to finding functions that model an object’s movement, you may be asked to analyze motion from a position, velocity, or acceleration graph. Here are the main ideas to use when reading these graphs.

Position graph:

The slope of the tangent line at a point on a position graph represents the instantaneous velocity.

An increasing graph means velocity is positive, so the object is moving right/forward/up.
A decreasing graph means velocity is negative.
Horizontal tangents or horizontal line segments (zero slope) mean the object is at rest.

Velocity graph:

The slope of the tangent line at a point on a velocity graph represents the instantaneous acceleration.

Check whether the graph is above or below the $x$ -axis to determine the sign of the velocity.
Combine the sign of velocity with the sign of acceleration to decide whether the object is speeding up or slowing down over a time interval.

A particle moves along the number line with its position $p (t)$ plotted as a function of time (in seconds).

Graph of p(t)

a) What is its average velocity from $t = 1$ to $t = 3$ seconds?
b) When is the particle at rest?
c) When is it moving left and when is it moving right?

Solutions

a) What is its average velocity from $t = 1$ to $t = 3$ seconds?

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

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

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

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

So $p (1) = - \frac{2}{3}$ , and

$v_{avg} = \frac{- 2 - ( - \frac{2}{3} )}{3 - 1}$

$= \frac{- \frac{4}{3}}{2}$

$= - \frac{2}{3}$

The average velocity over $[1, 3]$ is $- \frac{2}{3} units/sec$ . It’s negative because the position is decreasing over this interval, so the particle moves left.

b) When is the particle at rest?

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

c) When is it moving left and when is it moving right?

The particle moves left when $v (t) < 0$ and right when $v (t) > 0$ . On a position graph, this comes from 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 (as if $p (t)$ were written as a piecewise function):

Interval	Sign of $v (t)$
$(0, 3)$	$-$
$(3, 5)$	$0$
$(5, 7)$	$+$
$(7, 9)$	$0$
$(9, 11)$	$+$

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

It moves left on $0 < t < 3$
Stops on $3 < t < 5$
Moves right on $5 < t < 7$
Stops again on $7 < t < 9$
Continues moving right on $9 < t < 11$ > 2. The velocity of a particle moving along a number line, in meters per minute, is displayed in the graph below.

Graph of v(t)

a) At what time(s) does the particle change direction?
b) When is it moving at constant speed?
c) When is it speeding up and when is it slowing down?
d) What is the particle’s maximum speed, and when does this occur?

Solutions

a) At what time(s) does the particle change direction?

(spoiler)

The particle changes direction when the velocity changes sign (from positive to negative or from negative to positive).

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

The first $x$ -intercept is $(3, 0)$ .

The second intercept isn’t labeled exactly, but the line segment passes through $(7, - 3)$ and $(10, - 2)$ . Using these two points, the line has slope

$m = \frac{- 2 - ( - 3 )}{10 - 7} = \frac{1}{3} .$

Using point-slope form with $(10, - 2)$ gives

$v (t) = \frac{1}{3} (t - 10) - 2.$

The $x$ -intercept occurs when $v (t) = 0$ :

$0 = \frac{1}{3} (t - 10) - 2 \Rightarrow \frac{1}{3} (t - 10) = 2 \Rightarrow t = 16.$

The third $x$ -intercept is at $(12, 0)$ .

The organized sign chart shown below separates intervals based on the $x$ -intercepts.

Interval	Sign of $v (t)$
$(0, 3)$	$+$
$(3, 8.8)$	$-$
$(8.8, 12)$	$+$
$(12, 14)$	$+$

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

b) When is it moving at constant speed?

(spoiler)

A particle moves at constant speed when its velocity is constant (so the graph is horizontal). On a velocity graph, that means the slope is zero.

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

c) When is it speeding up and when is it slowing down?

(spoiler)

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 $\Rightarrow a (t) = 0$ ).

Here’s a sign chart for with intervals separated according to line segments and $x$ -intercepts.

Interval	Sign of $v (t)$	Sign of $a (t)$
$(0, 2)$	$+$	$+$
$(2, 3)$	$+$	$-$
$(3, 4)$	$-$	$-$
$(4, 7)$	$-$	$0$
$(7, 8.8)$	$-$	$+$
$(8.8, 10)$	$+$	$+$
$(10, 12)$	$+$	$-$
$(12, 14)$	$+$	$+$

The particle is

Speeding up in:

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

Slowing down in:

$(2, 3)$
$(7, 8.8)$
$(10, 12)$

d) What is the particle’s maximum speed, and when does this occur?

(spoiler)

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

A particle’s position along the $x$ -axis is tracked at specific points in time and recorded in the following table:

Time $t$ (seconds) Position (meters)

$0$ $5$

$1$ $13$

$2$ $18$

$3$ $20$

$4$ $16$

$5$ $10$

a) What is the average velocity from $t = 2$ to $t = 5$ seconds?
b) Estimate when the particle changes direction.

Time $t$ (seconds)	Position (meters)
$0$	$5$
$1$	$13$
$2$	$18$
$3$	$20$
$4$	$16$
$5$	$10$

a) Average velocity over $[2, 5]$

Solution

(spoiler)

$v_{avg} = \frac{p ( 5 ) - p ( 2 )}{5 - 2}$

$= \frac{10 - 18}{3}$

$= - \frac{8}{3} m/s$

b) Estimate when the particle changes direction.

Solution

(spoiler)

A particle changes direction when its velocity 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 ( $5 \to 20$ ), so the particle moves forward.
From $t = 3$ to $t = 4$ , the position decreases ( $20 \to 16$ ), so the particle moves backward.

So the particle must have changed direction at about $3$ seconds.

A particle’s moves along the $x$ -axis with its velocity tracked at specific points in time and recorded in the following table:

Time $t$ Velocity (m/s)

$0$ $0$

$1$ $5$

$2$ $7$

$3$ $2$

$4$ $- 3$

$5$ $- 9$

a) Estimate the rate of change of velocity at $t = 2$ seconds.
b) Estimate when the particle changes direction.

Time $t$	Velocity (m/s)
$0$	$0$
$1$	$5$
$2$	$7$
$3$	$2$
$4$	$- 3$
$5$	$- 9$

a) Rate of change of velocity at $t = 2$ seconds

Solution

(spoiler)

The rate of change of velocity is acceleration. To estimate the acceleration at $t = 2$ , use a symmetric difference quotient with the surrounding data points at $t = 1$ and $t = 3$ .

$a_{avg} = \frac{v ( 3 ) - v ( 1 )}{3 - 1}$

$= \frac{2 - 5}{2}$

$= - \frac{3}{2} m/s^{2}$

b) Estimate when the particle changes direction.

Solution

(spoiler)

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

Since the velocity at $t = 3$ seconds is positive and at $t = 4$ seconds it’s negative, the Intermediate value theorem implies there is a time in $(3, 4)$ when the velocity is $0$ . So the particle changes direction sometime between $3$ and $4$ seconds.

Interpreting position graphs

Slope of tangent = instantaneous velocity
Increasing graph: velocity positive (moving right/forward/up)
Decreasing graph: velocity negative; horizontal segments: object at rest

Interpreting velocity graphs

Slope of tangent = instantaneous acceleration
Above/below $x$ -axis: sign of velocity
Combine velocity and acceleration signs to determine speeding up/slowing down

Position graph example (problem 1)

Average velocity formula: $v_{avg} = \frac{p ( 3 ) - p ( 1 )}{3 - 1}$
Particle at rest: intervals with horizontal segments $(3, 5)$ and $(7, 9)$
Moving left: negative slope $(0, 3)$ ; moving right: positive slope $(5, 7)$ and $(9, 11)$

Velocity graph example (problem 2)

Changes direction: when $v (t)$ crosses $x$ -axis (at $t = 3$ and $t \approx 8.8$ )
Constant speed: horizontal segments on velocity graph (here, $(4, 7)$ )
Speeding up: velocity and acceleration same sign; slowing down: opposite signs
Maximum speed: greatest $∣ v (t) ∣$ (here, $3$ m/min at $t = 2$ , $4 \leq t \leq 7$ , $t = 14$ )

Position table example (problem 3)

Average velocity: $v_{avg} = \frac{p ( 5 ) - p ( 2 )}{5 - 2}$
Change in direction: where position switches from increasing to decreasing (at $t \approx 3$ )

Velocity table example (problem 4)

Rate of change of velocity (acceleration): use symmetric difference quotient, e.g., $\frac{v ( 3 ) - v ( 1 )}{3 - 1}$
Change in direction: when velocity changes sign (between $t = 3$ and $t = 4$ )

Graphs & tables

What you’ll learn:

How to interpret position and velocity from graphs or tables

Position graph:

The slope of the tangent line at a point on a position graph represents the instantaneous velocity.

An increasing graph means velocity is positive, so the object is moving right/forward/up.
A decreasing graph means velocity is negative.
Horizontal tangents or horizontal line segments (zero slope) mean the object is at rest.

Velocity graph:

The slope of the tangent line at a point on a velocity graph represents the instantaneous acceleration.

Check whether the graph is above or below the $x$ -axis to determine the sign of the velocity.
Combine the sign of velocity with the sign of acceleration to decide whether the object is speeding up or slowing down over a time interval.

A particle moves along the number line with its position $p (t)$ plotted as a function of time (in seconds).

a) What is its average velocity from $t = 1$ to $t = 3$ seconds?
b) When is the particle at rest?
c) When is it moving left and when is it moving right?

Solutions

a) What is its average velocity from $t = 1$ to $t = 3$ seconds?

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

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

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

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

So $p (1) = - \frac{2}{3}$ , and

$v_{avg} = \frac{- 2 - ( - \frac{2}{3} )}{3 - 1}$

$= \frac{- \frac{4}{3}}{2}$

$= - \frac{2}{3}$

The average velocity over $[1, 3]$ is $- \frac{2}{3} units/sec$ . It’s negative because the position is decreasing over this interval, so the particle moves left.

b) When is the particle at rest?

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

c) When is it moving left and when is it moving right?

The particle moves left when $v (t) < 0$ and right when $v (t) > 0$ . On a position graph, this comes from 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 (as if $p (t)$ were written as a piecewise function):

Interval	Sign of $v (t)$
$(0, 3)$	$-$
$(3, 5)$	$0$
$(5, 7)$	$+$
$(7, 9)$	$0$
$(9, 11)$	$+$

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

It moves left on $0 < t < 3$
Stops on $3 < t < 5$
Moves right on $5 < t < 7$
Stops again on $7 < t < 9$
Continues moving right on $9 < t < 11$ > 2. The velocity of a particle moving along a number line, in meters per minute, is displayed in the graph below.

a) At what time(s) does the particle change direction?
b) When is it moving at constant speed?
c) When is it speeding up and when is it slowing down?
d) What is the particle’s maximum speed, and when does this occur?

Solutions

a) At what time(s) does the particle change direction?

(spoiler)

The particle changes direction when the velocity changes sign (from positive to negative or from negative to positive).

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

The first $x$ -intercept is $(3, 0)$ .

The second intercept isn’t labeled exactly, but the line segment passes through $(7, - 3)$ and $(10, - 2)$ . Using these two points, the line has slope

$m = \frac{- 2 - ( - 3 )}{10 - 7} = \frac{1}{3} .$

Using point-slope form with $(10, - 2)$ gives

$v (t) = \frac{1}{3} (t - 10) - 2.$

The $x$ -intercept occurs when $v (t) = 0$ :

$0 = \frac{1}{3} (t - 10) - 2 \Rightarrow \frac{1}{3} (t - 10) = 2 \Rightarrow t = 16.$

The third $x$ -intercept is at $(12, 0)$ .

The organized sign chart shown below separates intervals based on the $x$ -intercepts.

Interval	Sign of $v (t)$
$(0, 3)$	$+$
$(3, 8.8)$	$-$
$(8.8, 12)$	$+$
$(12, 14)$	$+$

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

b) When is it moving at constant speed?

(spoiler)

A particle moves at constant speed when its velocity is constant (so the graph is horizontal). On a velocity graph, that means the slope is zero.

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

c) When is it speeding up and when is it slowing down?

(spoiler)

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 $\Rightarrow a (t) = 0$ ).

Here’s a sign chart for with intervals separated according to line segments and $x$ -intercepts.

Interval	Sign of $v (t)$	Sign of $a (t)$
$(0, 2)$	$+$	$+$
$(2, 3)$	$+$	$-$
$(3, 4)$	$-$	$-$
$(4, 7)$	$-$	$0$
$(7, 8.8)$	$-$	$+$
$(8.8, 10)$	$+$	$+$
$(10, 12)$	$+$	$-$
$(12, 14)$	$+$	$+$

The particle is

Speeding up in:

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

Slowing down in:

$(2, 3)$
$(7, 8.8)$
$(10, 12)$

d) What is the particle’s maximum speed, and when does this occur?

(spoiler)

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

A particle’s position along the $x$ -axis is tracked at specific points in time and recorded in the following table:

Time $t$ (seconds) Position (meters)

$0$ $5$

$1$ $13$

$2$ $18$

$3$ $20$

$4$ $16$

$5$ $10$

a) What is the average velocity from $t = 2$ to $t = 5$ seconds?
b) Estimate when the particle changes direction.

Time $t$ (seconds)	Position (meters)
$0$	$5$
$1$	$13$
$2$	$18$
$3$	$20$
$4$	$16$
$5$	$10$

a) Average velocity over $[2, 5]$

Solution

(spoiler)

$v_{avg} = \frac{p ( 5 ) - p ( 2 )}{5 - 2}$

$= \frac{10 - 18}{3}$

$= - \frac{8}{3} m/s$

b) Estimate when the particle changes direction.

Solution

(spoiler)

A particle changes direction when its velocity 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 ( $5 \to 20$ ), so the particle moves forward.
From $t = 3$ to $t = 4$ , the position decreases ( $20 \to 16$ ), so the particle moves backward.

So the particle must have changed direction at about $3$ seconds.

A particle’s moves along the $x$ -axis with its velocity tracked at specific points in time and recorded in the following table:

Time $t$ Velocity (m/s)

$0$ $0$

$1$ $5$

$2$ $7$

$3$ $2$

$4$ $- 3$

$5$ $- 9$

a) Estimate the rate of change of velocity at $t = 2$ seconds.
b) Estimate when the particle changes direction.

Time $t$	Velocity (m/s)
$0$	$0$
$1$	$5$
$2$	$7$
$3$	$2$
$4$	$- 3$
$5$	$- 9$

a) Rate of change of velocity at $t = 2$ seconds

Solution

(spoiler)

The rate of change of velocity is acceleration. To estimate the acceleration at $t = 2$ , use a symmetric difference quotient with the surrounding data points at $t = 1$ and $t = 3$ .

$a_{avg} = \frac{v ( 3 ) - v ( 1 )}{3 - 1}$

$= \frac{2 - 5}{2}$

$= - \frac{3}{2} m/s^{2}$

b) Estimate when the particle changes direction.

Solution

(spoiler)

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

Achievable AP Calculus AB

4. Contextual uses

4.2. Straight-line motion

Our AP Calculus AB course is currently in development and is a work-in-progress.

Graphs & tables

8 min read

Font

Discuss

Feedback

What you’ll learn:

How to interpret position and velocity from graphs or tables

Position graph:

The slope of the tangent line at a point on a position graph represents the instantaneous velocity.

An increasing graph means velocity is positive, so the object is moving right/forward/up.
A decreasing graph means velocity is negative.
Horizontal tangents or horizontal line segments (zero slope) mean the object is at rest.

Velocity graph:

The slope of the tangent line at a point on a velocity graph represents the instantaneous acceleration.

Check whether the graph is above or below the $x$ -axis to determine the sign of the velocity.
Combine the sign of velocity with the sign of acceleration to decide whether the object is speeding up or slowing down over a time interval.

A particle moves along the number line with its position $p (t)$ plotted as a function of time (in seconds).

Graph of p(t)

a) What is its average velocity from $t = 1$ to $t = 3$ seconds?
b) When is the particle at rest?
c) When is it moving left and when is it moving right?

Solutions

a) What is its average velocity from $t = 1$ to $t = 3$ seconds?

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

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

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

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

So $p (1) = - \frac{2}{3}$ , and

$v_{avg} = \frac{- 2 - ( - \frac{2}{3} )}{3 - 1}$

$= \frac{- \frac{4}{3}}{2}$

$= - \frac{2}{3}$

The average velocity over $[1, 3]$ is $- \frac{2}{3} units/sec$ . It’s negative because the position is decreasing over this interval, so the particle moves left.

b) When is the particle at rest?

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

c) When is it moving left and when is it moving right?

The particle moves left when $v (t) < 0$ and right when $v (t) > 0$ . On a position graph, this comes from 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 (as if $p (t)$ were written as a piecewise function):

Interval	Sign of $v (t)$
$(0, 3)$	$-$
$(3, 5)$	$0$
$(5, 7)$	$+$
$(7, 9)$	$0$
$(9, 11)$	$+$

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

It moves left on $0 < t < 3$
Stops on $3 < t < 5$
Moves right on $5 < t < 7$
Stops again on $7 < t < 9$
Continues moving right on $9 < t < 11$ > 2. The velocity of a particle moving along a number line, in meters per minute, is displayed in the graph below.

Graph of v(t)

a) At what time(s) does the particle change direction?
b) When is it moving at constant speed?
c) When is it speeding up and when is it slowing down?
d) What is the particle’s maximum speed, and when does this occur?

Solutions

a) At what time(s) does the particle change direction?

(spoiler)

The particle changes direction when the velocity changes sign (from positive to negative or from negative to positive).

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

The first $x$ -intercept is $(3, 0)$ .

The second intercept isn’t labeled exactly, but the line segment passes through $(7, - 3)$ and $(10, - 2)$ . Using these two points, the line has slope

$m = \frac{- 2 - ( - 3 )}{10 - 7} = \frac{1}{3} .$

Using point-slope form with $(10, - 2)$ gives

$v (t) = \frac{1}{3} (t - 10) - 2.$

The $x$ -intercept occurs when $v (t) = 0$ :

$0 = \frac{1}{3} (t - 10) - 2 \Rightarrow \frac{1}{3} (t - 10) = 2 \Rightarrow t = 16.$

The third $x$ -intercept is at $(12, 0)$ .

The organized sign chart shown below separates intervals based on the $x$ -intercepts.

Interval	Sign of $v (t)$
$(0, 3)$	$+$
$(3, 8.8)$	$-$
$(8.8, 12)$	$+$
$(12, 14)$	$+$

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

b) When is it moving at constant speed?

(spoiler)

A particle moves at constant speed when its velocity is constant (so the graph is horizontal). On a velocity graph, that means the slope is zero.

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

c) When is it speeding up and when is it slowing down?

(spoiler)

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 $\Rightarrow a (t) = 0$ ).

Here’s a sign chart for with intervals separated according to line segments and $x$ -intercepts.

Interval	Sign of $v (t)$	Sign of $a (t)$
$(0, 2)$	$+$	$+$
$(2, 3)$	$+$	$-$
$(3, 4)$	$-$	$-$
$(4, 7)$	$-$	$0$
$(7, 8.8)$	$-$	$+$
$(8.8, 10)$	$+$	$+$
$(10, 12)$	$+$	$-$
$(12, 14)$	$+$	$+$

The particle is

Speeding up in:

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

Slowing down in:

$(2, 3)$
$(7, 8.8)$
$(10, 12)$

d) What is the particle’s maximum speed, and when does this occur?

(spoiler)

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

A particle’s position along the $x$ -axis is tracked at specific points in time and recorded in the following table:

Time $t$ (seconds) Position (meters)

$0$ $5$

$1$ $13$

$2$ $18$

$3$ $20$

$4$ $16$

$5$ $10$

a) What is the average velocity from $t = 2$ to $t = 5$ seconds?
b) Estimate when the particle changes direction.

Time $t$ (seconds)	Position (meters)
$0$	$5$
$1$	$13$
$2$	$18$
$3$	$20$
$4$	$16$
$5$	$10$

a) Average velocity over $[2, 5]$

Solution

(spoiler)

$v_{avg} = \frac{p ( 5 ) - p ( 2 )}{5 - 2}$

$= \frac{10 - 18}{3}$

$= - \frac{8}{3} m/s$

b) Estimate when the particle changes direction.

Solution

(spoiler)

A particle changes direction when its velocity 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 ( $5 \to 20$ ), so the particle moves forward.
From $t = 3$ to $t = 4$ , the position decreases ( $20 \to 16$ ), so the particle moves backward.

So the particle must have changed direction at about $3$ seconds.

A particle’s moves along the $x$ -axis with its velocity tracked at specific points in time and recorded in the following table:

Time $t$ Velocity (m/s)

$0$ $0$

$1$ $5$

$2$ $7$

$3$ $2$

$4$ $- 3$

$5$ $- 9$

a) Estimate the rate of change of velocity at $t = 2$ seconds.
b) Estimate when the particle changes direction.

Time $t$	Velocity (m/s)
$0$	$0$
$1$	$5$
$2$	$7$
$3$	$2$
$4$	$- 3$
$5$	$- 9$

a) Rate of change of velocity at $t = 2$ seconds

Solution

(spoiler)

The rate of change of velocity is acceleration. To estimate the acceleration at $t = 2$ , use a symmetric difference quotient with the surrounding data points at $t = 1$ and $t = 3$ .

$a_{avg} = \frac{v ( 3 ) - v ( 1 )}{3 - 1}$

$= \frac{2 - 5}{2}$

$= - \frac{3}{2} m/s^{2}$

b) Estimate when the particle changes direction.

Solution

(spoiler)

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

Interpreting position graphs

Slope of tangent = instantaneous velocity
Increasing graph: velocity positive (moving right/forward/up)
Decreasing graph: velocity negative; horizontal segments: object at rest

Interpreting velocity graphs

Slope of tangent = instantaneous acceleration
Above/below $x$ -axis: sign of velocity
Combine velocity and acceleration signs to determine speeding up/slowing down

Position graph example (problem 1)

Average velocity formula: $v_{avg} = \frac{p ( 3 ) - p ( 1 )}{3 - 1}$
Particle at rest: intervals with horizontal segments $(3, 5)$ and $(7, 9)$
Moving left: negative slope $(0, 3)$ ; moving right: positive slope $(5, 7)$ and $(9, 11)$

Velocity graph example (problem 2)

Changes direction: when $v (t)$ crosses $x$ -axis (at $t = 3$ and $t \approx 8.8$ )
Constant speed: horizontal segments on velocity graph (here, $(4, 7)$ )
Speeding up: velocity and acceleration same sign; slowing down: opposite signs
Maximum speed: greatest $∣ v (t) ∣$ (here, $3$ m/min at $t = 2$ , $4 \leq t \leq 7$ , $t = 14$ )

Position table example (problem 3)

Average velocity: $v_{avg} = \frac{p ( 5 ) - p ( 2 )}{5 - 2}$
Change in direction: where position switches from increasing to decreasing (at $t \approx 3$ )

Velocity table example (problem 4)

Rate of change of velocity (acceleration): use symmetric difference quotient, e.g., $\frac{v ( 3 ) - v ( 1 )}{3 - 1}$
Change in direction: when velocity changes sign (between $t = 3$ and $t = 4$ )

Graphs & tables

What you’ll learn:

How to interpret position and velocity from graphs or tables

Position graph:

The slope of the tangent line at a point on a position graph represents the instantaneous velocity.

An increasing graph means velocity is positive, so the object is moving right/forward/up.
A decreasing graph means velocity is negative.
Horizontal tangents or horizontal line segments (zero slope) mean the object is at rest.

Velocity graph:

The slope of the tangent line at a point on a velocity graph represents the instantaneous acceleration.

Check whether the graph is above or below the $x$ -axis to determine the sign of the velocity.
Combine the sign of velocity with the sign of acceleration to decide whether the object is speeding up or slowing down over a time interval.

A particle moves along the number line with its position $p (t)$ plotted as a function of time (in seconds).

a) What is its average velocity from $t = 1$ to $t = 3$ seconds?
b) When is the particle at rest?
c) When is it moving left and when is it moving right?

Solutions

a) What is its average velocity from $t = 1$ to $t = 3$ seconds?

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

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

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

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

So $p (1) = - \frac{2}{3}$ , and

$v_{avg} = \frac{- 2 - ( - \frac{2}{3} )}{3 - 1}$

$= \frac{- \frac{4}{3}}{2}$

$= - \frac{2}{3}$

The average velocity over $[1, 3]$ is $- \frac{2}{3} units/sec$ . It’s negative because the position is decreasing over this interval, so the particle moves left.

b) When is the particle at rest?

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

c) When is it moving left and when is it moving right?

The particle moves left when $v (t) < 0$ and right when $v (t) > 0$ . On a position graph, this comes from 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 (as if $p (t)$ were written as a piecewise function):

Interval	Sign of $v (t)$
$(0, 3)$	$-$
$(3, 5)$	$0$
$(5, 7)$	$+$
$(7, 9)$	$0$
$(9, 11)$	$+$

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

It moves left on $0 < t < 3$
Stops on $3 < t < 5$
Moves right on $5 < t < 7$
Stops again on $7 < t < 9$
Continues moving right on $9 < t < 11$ > 2. The velocity of a particle moving along a number line, in meters per minute, is displayed in the graph below.

a) At what time(s) does the particle change direction?
b) When is it moving at constant speed?
c) When is it speeding up and when is it slowing down?
d) What is the particle’s maximum speed, and when does this occur?

Solutions

a) At what time(s) does the particle change direction?

(spoiler)

The particle changes direction when the velocity changes sign (from positive to negative or from negative to positive).

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

The first $x$ -intercept is $(3, 0)$ .

The second intercept isn’t labeled exactly, but the line segment passes through $(7, - 3)$ and $(10, - 2)$ . Using these two points, the line has slope

$m = \frac{- 2 - ( - 3 )}{10 - 7} = \frac{1}{3} .$

Using point-slope form with $(10, - 2)$ gives

$v (t) = \frac{1}{3} (t - 10) - 2.$

The $x$ -intercept occurs when $v (t) = 0$ :

$0 = \frac{1}{3} (t - 10) - 2 \Rightarrow \frac{1}{3} (t - 10) = 2 \Rightarrow t = 16.$

The third $x$ -intercept is at $(12, 0)$ .

The organized sign chart shown below separates intervals based on the $x$ -intercepts.

Interval	Sign of $v (t)$
$(0, 3)$	$+$
$(3, 8.8)$	$-$
$(8.8, 12)$	$+$
$(12, 14)$	$+$

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

b) When is it moving at constant speed?

(spoiler)

A particle moves at constant speed when its velocity is constant (so the graph is horizontal). On a velocity graph, that means the slope is zero.

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

c) When is it speeding up and when is it slowing down?

(spoiler)

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 $\Rightarrow a (t) = 0$ ).

Here’s a sign chart for with intervals separated according to line segments and $x$ -intercepts.

Interval	Sign of $v (t)$	Sign of $a (t)$
$(0, 2)$	$+$	$+$
$(2, 3)$	$+$	$-$
$(3, 4)$	$-$	$-$
$(4, 7)$	$-$	$0$
$(7, 8.8)$	$-$	$+$
$(8.8, 10)$	$+$	$+$
$(10, 12)$	$+$	$-$
$(12, 14)$	$+$	$+$

The particle is

Speeding up in:

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

Slowing down in:

$(2, 3)$
$(7, 8.8)$
$(10, 12)$

d) What is the particle’s maximum speed, and when does this occur?

(spoiler)

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

A particle’s position along the $x$ -axis is tracked at specific points in time and recorded in the following table:

Time $t$ (seconds) Position (meters)

$0$ $5$

$1$ $13$

$2$ $18$

$3$ $20$

$4$ $16$

$5$ $10$

a) What is the average velocity from $t = 2$ to $t = 5$ seconds?
b) Estimate when the particle changes direction.

Time $t$ (seconds)	Position (meters)
$0$	$5$
$1$	$13$
$2$	$18$
$3$	$20$
$4$	$16$
$5$	$10$

a) Average velocity over $[2, 5]$

Solution

(spoiler)

$v_{avg} = \frac{p ( 5 ) - p ( 2 )}{5 - 2}$

$= \frac{10 - 18}{3}$

$= - \frac{8}{3} m/s$

b) Estimate when the particle changes direction.

Solution

(spoiler)

A particle changes direction when its velocity 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 ( $5 \to 20$ ), so the particle moves forward.
From $t = 3$ to $t = 4$ , the position decreases ( $20 \to 16$ ), so the particle moves backward.

So the particle must have changed direction at about $3$ seconds.

A particle’s moves along the $x$ -axis with its velocity tracked at specific points in time and recorded in the following table:

Time $t$ Velocity (m/s)

$0$ $0$

$1$ $5$

$2$ $7$

$3$ $2$

$4$ $- 3$

$5$ $- 9$

a) Estimate the rate of change of velocity at $t = 2$ seconds.
b) Estimate when the particle changes direction.

Time $t$	Velocity (m/s)
$0$	$0$
$1$	$5$
$2$	$7$
$3$	$2$
$4$	$- 3$
$5$	$- 9$

a) Rate of change of velocity at $t = 2$ seconds

Solution

(spoiler)

The rate of change of velocity is acceleration. To estimate the acceleration at $t = 2$ , use a symmetric difference quotient with the surrounding data points at $t = 1$ and $t = 3$ .

$a_{avg} = \frac{v ( 3 ) - v ( 1 )}{3 - 1}$

$= \frac{2 - 5}{2}$

$= - \frac{3}{2} m/s^{2}$

b) Estimate when the particle changes direction.

Solution

(spoiler)

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