3.2 Graphical analysis of motion

Achievable AP Physics 1

3. Kinematics

Our AP Physics 1 course is currently in development and is a work-in-progress.

Graphical analysis of motion

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This sub-chapter shows you how to analyze motion using graphs. We’ll focus on four tools:

Position time graphs
Velocity time graphs
Acceleration time graphs
Algebraic analysis

Position-time graphs

A position-time graph plots an object’s position $s$ on the vertical axis versus time $t$ on the horizontal axis. From this kind of graph, you can read position directly, and you can derive other quantities from the shape of the curve.

Definitions

Displacement: The change in position is given by the difference in the vertical values at two times:
$Δ s = s_{f} - s_{i} .$
Average velocity: This is the displacement divided by the elapsed time, or “rise over run.”

Sidenote

Average velocity note

You can estimate the average velocity by drawing a straight line between two points on the graph and calculating its slope. The displacement is the vertical difference between the initial and final positions.

Consider the position-time curve below. Compare the starting point at $t = 0$ and the ending point at $t = 4$ . The vertical distance between these points is the displacement.

The slope of the tangent line at a point gives the instantaneous velocity at that time.
The slope of the secant line between two points gives the average velocity over that time interval.

Quantities derived from a position-time graph

Example problem 1

An object moves according to the function

$s (t) = - t^{2} + 3 t + 4, t \in [0, 4] .$

a) Using the graph, determine the displacement from $t = 0$ s to $t = 4$ s.

b) Explain how you would estimate the average velocity over this interval.

Solution:

(spoiler)

Write down the position function.

$s (t) = - t^{2} + 3 t + 4.$
Evaluate the position at $t = 0$ and $t = 4$ .

$s (0) = - 0^{2} + 3 \cdot 0 + 4 = 4, s (4) = - 4^{2} + 3 \cdot 4 + 4 = - 16 + 12 + 4 = 0.$
Find the displacement from $t = 0$ to $t = 4$ .

$Displacement = s (4) - s (0) = 0 - 4 = - 4 (units) .$

The negative sign means the final position is 4 units below the initial position on the $s$ -axis.
Estimate the average velocity over the interval $[0, 4]$ .

Use displacement over elapsed time:

$v_{avg} = \frac{Δ s}{Δ t} = \frac{s ( 4 ) - s ( 0 )}{4 - 0} = \frac{- 4}{4} = - 1 (units/s) .$

On the graph, this is the slope of the secant line connecting $(0, s (0))$ and $(4, s (4))$ .

Velocity-time graphs

A velocity-time graph shows velocity on the vertical axis and time on the horizontal axis. Two key ideas come directly from the geometry of the graph:

Definitions

Acceleration: The slope of the graph (change in velocity over time).
Displacement: The area under the curve represents the displacement over the given time interval. To approximate this area, you can break it into basic shapes such as rectangles and trapezoids.

For example, if the graph is a straight line, the area under the line between two time points forms a trapezoid.

In the graph below, the slope $Δ v /Δ t$ corresponds to acceleration, and the shaded area under the curve represents the displacement over that time interval.

Example problem 2

An object moves along a straight line with the following velocity-time profile:

$v (t) = 2, 6 - t, if 0 \leq t < 4, if 4 \leq t \leq 8.$

Use the velocity-time graph to:

Determine the displacement from $t = 0$ to $t = 4$ seconds.

Determine the displacement from $t = 4$ to $t = 8$ seconds.

Find the net displacement from $t = 0$ to $t = 8$ seconds.

Find the total distance traveled over the entire 8-second interval.

Solution:

(spoiler)

Velocity-time graph setup:
- From $t = 0$ to $t = 4$ s, the velocity is constant at $+ 2$ m/s.
- From $t = 4$ to $t = 8$ s, the velocity decreases linearly from $+ 2$ m/s to $- 2$ m/s.
The velocity crosses zero between 4 and 8 seconds, which indicates a change in direction.

Displacement from $t = 0$ to $t = 4$ s:

$Δ s_{0 \to 4} = 2 m/s \times 4 s = 8 m .$
Displacement from $t = 4$ to $t = 8$ s:

The area under the velocity curve from $t = 4$ to $t = 8$ is a trapezoid with bases $+ 2$ and $- 2$ and a width of 4 s:

$Δ s_{4 \to 8} = \frac{( 2 + ( - 2 ))}{2} \times 4 = 0 m .$

Thus, the net displacement from $(t = 4)$ to $(t = 8)$ is zero (the object returns to the same position it had at $(t = 4)$ , but not necessarily to the same position it started at $(t = 0)$ ).

Net displacement from $t = 0$ to $t = 8$ s:

$Δ s_{0 \to 8} = 8 + 0 = 8 m .$
Total distance traveled from $t = 0$ to $t = 8$ s:

Net displacement can be zero even when the object moves, because motion in opposite directions cancels in the signed area. To find total distance, split the interval $[4, 8]$ at the instant where $v = 0$ .
- $t_{turn}$ is the time when velocity crosses zero.
- From $t = 4$ to $t_{turn}$ , the velocity is positive, so the object moves forward.
- From $t_{turn}$ to $t = 8$ , the velocity is negative, so the object moves backward.
- From $t = 0$ to $t = 4$ : Distance = 8 m.
- From $t = 4$ to $t = 8$ : Distance = $(2 + 2)$ m.
- Total distance $= 8 + 2 (2) = 12$ m.

Acceleration-time graphs

An acceleration-time graph plots acceleration on the vertical axis versus time on the horizontal axis.

For a constant acceleration, the graph is a horizontal line.
The change in velocity over a time interval is determined by the area under the curve. When acceleration varies, approximate this area using basic geometric shapes (rectangles or trapezoids).

Below is an acceleration-time graph showing a constant acceleration of $a (t) = 3 m/s^{2}$ . The shaded area between $t = 1$ and $t = 3$ s represents the change in velocity over that interval.

Example problem 3

Consider the acceleration function:

$a (t) = 4 - t, t \in [0, 4] .$

a) Estimate the change in velocity between $t = 1$ s and $t = 3$ s by approximating the area under the acceleration-time graph.

b) Explain what the sign of the change in velocity indicates about the object’s motion.

Solution:

(spoiler)

At $t = 1$ , $a (1) = 4 - 1 = 3 m/s^{2}$ and at $t = 3$ , $a (3) = 4 - 3 = 1 m/s^{2}$ .

Approximating the area under the graph as a trapezoid:

$Δ v \approx \frac{3 + 1}{2} \times (3 - 1) = 2 \times 2 = 4 m/s .$
A positive change in velocity means the object’s velocity increases by $4 m/s$ over the interval.

Example problem 4

An object moves along a straight line for the time interval $0 \leq t \leq 6$ seconds. The following two graphs describe the motion:

a) From the position-time graph, determine the displacement of the object between $t = 0$ s and $t = 3$ s.

b) From the velocity-time graph, calculate the net displacement of the object over the entire interval $t = 0$ s to $t = 6$ s.

c) Using the velocity-time graph, determine the total distance traveled by the object from $t = 0$ s to $t = 6$ s.

d) At $t = 2$ s, use the slope of the position-time graph to find the instantaneous velocity.

e) From the velocity-time graph, find the average acceleration of the object during the first 3 seconds.

Solution:

(spoiler)

(a) Displacement from $t = 0$ to $t = 3$ s (Position-time graph):

At $t = 0$ , the position is $s (0) = 0$ m.
At $t = 3$ , the position is $s (3) = 6$ m.
Displacement: $6 - 0 = 6$ m.

(b) Net displacement from $t = 0$ to $t = 6$ s (Velocity-time graph):

From $t = 0$ to $t = 3$ : $v = + 2$ m/s, area = $2 \times 3 = 6$ m.
From $t = 3$ to $t = 4$ : $v = 0$ m/s, area = 0 m.
From $t = 4$ to $t = 6$ : $v = - 2$ m/s, area = $- 2 \times 2 = - 4$ m.
Net displacement: $6 + 0 - 4 = 2$ m.

Distance from $0$ to $t = 3$ : 6 m.
Distance from $t = 4$ to $t = 6$ : $∣ - 4 ∣ = 4$ m.
Total distance: $6 + 0 + 4 = 10$ m.

(d) Instantaneous velocity at $t = 2$ s (Position-time graph):

The slope between $t = 0$ and $t = 3$ is $\frac{6 - 0}{3 - 0} = 2$ m/s.
Thus, $v (2) = 2$ m/s.

(e) Average acceleration from $t = 0$ to $t = 3$ s (Velocity-time graph):

Since the velocity is constant at $2$ m/s from $t = 0$ to $t = 3$ , $Δ v = 2 - 2 = 0$ m/s.
Therefore, $a_{avg} = \frac{0}{3} = 0 m/s^{2}$ .

Position-time graphs\

Vertical difference between two times = displacement ( $Δ s = s_{f} - s_{i}$ )
Slope of tangent = instantaneous velocity; slope of secant = average velocity
Average velocity = displacement / elapsed time ( $v_{avg} = \frac{Δ s}{Δ t}$ )

Velocity-time graphs\

Slope = acceleration ( $a = \frac{Δ v}{Δ t}$ )
Area under curve = displacement over interval
- Net (signed) area = net displacement
- Sum of absolute areas = total distance traveled

Acceleration-time graphs\

Horizontal line = constant acceleration
Area under curve = change in velocity ( $Δ v$ )
- For varying acceleration, approximate area with geometric shapes

Algebraic analysis\

Use position, velocity, or acceleration functions to calculate:
- Displacement: $s (t_{2}) - s (t_{1})$
- Average velocity: $\frac{s ( t _{2} ) - s ( t _{1} )}{t _{2} - t _{1}}$
- Change in velocity: area under $a (t)$ curve

Key problem-solving strategies\

Read displacement directly from position-time graph; use slope for velocities
Find displacement and distance from velocity-time graph areas
Use area under acceleration-time graph for velocity changes
For average acceleration: $\frac{Δ v}{Δ t}$ from velocity-time graph

Summary\

Displacement: vertical difference (position-time), area (velocity-time)
Average/instantaneous velocity: slope (position-time)
Acceleration: slope (velocity-time), area (acceleration-time)
Total distance: sum of absolute areas (velocity-time)

Next | 3.3 Equations of motion and free fall

Graphical analysis of motion

This sub-chapter shows you how to analyze motion using graphs. We’ll focus on four tools:

Position time graphs
Velocity time graphs
Acceleration time graphs
Algebraic analysis

Position-time graphs

Definitions

Displacement: The change in position is given by the difference in the vertical values at two times:
$Δ s = s_{f} - s_{i} .$
Average velocity: This is the displacement divided by the elapsed time, or “rise over run.”

Sidenote

Average velocity note

Consider the position-time curve below. Compare the starting point at $t = 0$ and the ending point at $t = 4$ . The vertical distance between these points is the displacement.

The slope of the tangent line at a point gives the instantaneous velocity at that time.
The slope of the secant line between two points gives the average velocity over that time interval.

Example problem 1

An object moves according to the function

$s (t) = - t^{2} + 3 t + 4, t \in [0, 4] .$

a) Using the graph, determine the displacement from $t = 0$ s to $t = 4$ s.

b) Explain how you would estimate the average velocity over this interval.

Solution:

(spoiler)

Write down the position function.

$s (t) = - t^{2} + 3 t + 4.$
Evaluate the position at $t = 0$ and $t = 4$ .

$s (0) = - 0^{2} + 3 \cdot 0 + 4 = 4, s (4) = - 4^{2} + 3 \cdot 4 + 4 = - 16 + 12 + 4 = 0.$
Find the displacement from $t = 0$ to $t = 4$ .

$Displacement = s (4) - s (0) = 0 - 4 = - 4 (units) .$

The negative sign means the final position is 4 units below the initial position on the $s$ -axis.
Estimate the average velocity over the interval $[0, 4]$ .

Use displacement over elapsed time:

$v_{avg} = \frac{Δ s}{Δ t} = \frac{s ( 4 ) - s ( 0 )}{4 - 0} = \frac{- 4}{4} = - 1 (units/s) .$

On the graph, this is the slope of the secant line connecting $(0, s (0))$ and $(4, s (4))$ .

Velocity-time graphs

A velocity-time graph shows velocity on the vertical axis and time on the horizontal axis. Two key ideas come directly from the geometry of the graph:

Definitions

Acceleration: The slope of the graph (change in velocity over time).
Displacement: The area under the curve represents the displacement over the given time interval. To approximate this area, you can break it into basic shapes such as rectangles and trapezoids.

For example, if the graph is a straight line, the area under the line between two time points forms a trapezoid.

In the graph below, the slope $Δ v /Δ t$ corresponds to acceleration, and the shaded area under the curve represents the displacement over that time interval.

Example problem 2

An object moves along a straight line with the following velocity-time profile:

$v (t) = 2, 6 - t, if 0 \leq t < 4, if 4 \leq t \leq 8.$

Use the velocity-time graph to:

Determine the displacement from $t = 0$ to $t = 4$ seconds.

Determine the displacement from $t = 4$ to $t = 8$ seconds.

Find the net displacement from $t = 0$ to $t = 8$ seconds.

Find the total distance traveled over the entire 8-second interval.

Solution:

(spoiler)

Velocity-time graph setup:
- From $t = 0$ to $t = 4$ s, the velocity is constant at $+ 2$ m/s.
- From $t = 4$ to $t = 8$ s, the velocity decreases linearly from $+ 2$ m/s to $- 2$ m/s.
The velocity crosses zero between 4 and 8 seconds, which indicates a change in direction.

Displacement from $t = 0$ to $t = 4$ s:

$Δ s_{0 \to 4} = 2 m/s \times 4 s = 8 m .$
Displacement from $t = 4$ to $t = 8$ s:

The area under the velocity curve from $t = 4$ to $t = 8$ is a trapezoid with bases $+ 2$ and $- 2$ and a width of 4 s:

$Δ s_{4 \to 8} = \frac{( 2 + ( - 2 ))}{2} \times 4 = 0 m .$

Thus, the net displacement from $(t = 4)$ to $(t = 8)$ is zero (the object returns to the same position it had at $(t = 4)$ , but not necessarily to the same position it started at $(t = 0)$ ).

Net displacement from $t = 0$ to $t = 8$ s:

$Δ s_{0 \to 8} = 8 + 0 = 8 m .$
Total distance traveled from $t = 0$ to $t = 8$ s:

Net displacement can be zero even when the object moves, because motion in opposite directions cancels in the signed area. To find total distance, split the interval $[4, 8]$ at the instant where $v = 0$ .
- $t_{turn}$ is the time when velocity crosses zero.
- From $t = 4$ to $t_{turn}$ , the velocity is positive, so the object moves forward.
- From $t_{turn}$ to $t = 8$ , the velocity is negative, so the object moves backward.
- From $t = 0$ to $t = 4$ : Distance = 8 m.
- From $t = 4$ to $t = 8$ : Distance = $(2 + 2)$ m.
- Total distance $= 8 + 2 (2) = 12$ m.

Acceleration-time graphs

An acceleration-time graph plots acceleration on the vertical axis versus time on the horizontal axis.

For a constant acceleration, the graph is a horizontal line.
The change in velocity over a time interval is determined by the area under the curve. When acceleration varies, approximate this area using basic geometric shapes (rectangles or trapezoids).

Below is an acceleration-time graph showing a constant acceleration of $a (t) = 3 m/s^{2}$ . The shaded area between $t = 1$ and $t = 3$ s represents the change in velocity over that interval.

Example problem 3

Consider the acceleration function:

$a (t) = 4 - t, t \in [0, 4] .$

a) Estimate the change in velocity between $t = 1$ s and $t = 3$ s by approximating the area under the acceleration-time graph.

b) Explain what the sign of the change in velocity indicates about the object’s motion.

Solution:

(spoiler)

At $t = 1$ , $a (1) = 4 - 1 = 3 m/s^{2}$ and at $t = 3$ , $a (3) = 4 - 3 = 1 m/s^{2}$ .

Approximating the area under the graph as a trapezoid:

$Δ v \approx \frac{3 + 1}{2} \times (3 - 1) = 2 \times 2 = 4 m/s .$
A positive change in velocity means the object’s velocity increases by $4 m/s$ over the interval.

Example problem 4

An object moves along a straight line for the time interval $0 \leq t \leq 6$ seconds. The following two graphs describe the motion:

a) From the position-time graph, determine the displacement of the object between $t = 0$ s and $t = 3$ s.

b) From the velocity-time graph, calculate the net displacement of the object over the entire interval $t = 0$ s to $t = 6$ s.

c) Using the velocity-time graph, determine the total distance traveled by the object from $t = 0$ s to $t = 6$ s.

d) At $t = 2$ s, use the slope of the position-time graph to find the instantaneous velocity.

e) From the velocity-time graph, find the average acceleration of the object during the first 3 seconds.

Solution:

(spoiler)

(a) Displacement from $t = 0$ to $t = 3$ s (Position-time graph):

At $t = 0$ , the position is $s (0) = 0$ m.
At $t = 3$ , the position is $s (3) = 6$ m.
Displacement: $6 - 0 = 6$ m.

(b) Net displacement from $t = 0$ to $t = 6$ s (Velocity-time graph):

From $t = 0$ to $t = 3$ : $v = + 2$ m/s, area = $2 \times 3 = 6$ m.
From $t = 3$ to $t = 4$ : $v = 0$ m/s, area = 0 m.
From $t = 4$ to $t = 6$ : $v = - 2$ m/s, area = $- 2 \times 2 = - 4$ m.
Net displacement: $6 + 0 - 4 = 2$ m.

Distance from $0$ to $t = 3$ : 6 m.
Distance from $t = 4$ to $t = 6$ : $∣ - 4 ∣ = 4$ m.
Total distance: $6 + 0 + 4 = 10$ m.

(d) Instantaneous velocity at $t = 2$ s (Position-time graph):

The slope between $t = 0$ and $t = 3$ is $\frac{6 - 0}{3 - 0} = 2$ m/s.
Thus, $v (2) = 2$ m/s.

(e) Average acceleration from $t = 0$ to $t = 3$ s (Velocity-time graph):

Since the velocity is constant at $2$ m/s from $t = 0$ to $t = 3$ , $Δ v = 2 - 2 = 0$ m/s.
Therefore, $a_{avg} = \frac{0}{3} = 0 m/s^{2}$ .

Achievable AP Physics 1

3. Kinematics

Our AP Physics 1 course is currently in development and is a work-in-progress.