Representing Motion
Why this matters
Imagine you're a scout for a major league baseball team, watching a high school phenom pitch. You don't just care if she throws a strike. You want to know everything about that pitch. How fast was it the moment it left her hand? Did it speed up or slow down on its way to the plate? Where exactly was the ball 0.2 seconds after the release?
Telling this full story—the "what, where, and when" of motion—is what physicists do. Simply saying "the ball was fast" isn't enough. We need a more precise language.
In this lesson, we'll learn how to be expert motion storytellers. We'll translate the narrative of an object's journey into the powerful languages of graphs and equations. You'll learn to look at a graph and instantly see the object's entire history, from its starting position to its final speed, just like that scout analyzing a perfect pitch.
Diagram
Concept map
flowchart TD
A[Position, x] -- "Slope is..." --> B[Velocity, v];
B -- "Slope is..." --> C[Acceleration, a];
C -- "Area is..." --> D[Change in Velocity, Δv];
B -- "Area is..." --> E[Displacement, Δx];
subgraph "Calculus Connections"
A
B
C
D
E
end
Core explanation
In physics, we need to be precise. We can't just say a car is "moving." We need to describe how it's moving. There are five main ways we do this, and they all connect to each other.
The Five Languages of Motion
Think of these as five different ways to tell the same story:
- 1Narrative DescriptionA plain English story. "A car starts from rest at a stoplight and speeds up steadily."
- 2Motion DiagramsA series of "snapshots" or dots showing the object's position at equal time intervals. If the dots get farther apart, the object is speeding up.
- 3GraphsThe most powerful visual tool. We use position-time (x-t), velocity-time (v-t), and acceleration-time (a-t) graphs.
- 4EquationsA mathematical summary of the motion.
- 5FiguresA simple drawing of the situation.
The AP exam expects you to be fluent in translating between these "languages," especially between graphs and equations.
The Big Three: Position, Velocity, and Acceleration
Let's get our key terms straight.
- Position (x)Where an object is located. It's measured in meters (m).
- Velocity (v)How fast position is changing, and in what direction. It's measured in meters per second (m/s).
- Acceleration (a)How fast velocity is changing. It's measured in meters per second squared (m/s²).
Motion Graphs: The Story at a Glance
Motion graphs are the heart of kinematics. If you understand how to read them, you can solve most problems. There are a few key relationships you must know.
1. From Position to Velocity The slope of a position-time (x-t) graph gives you the instantaneous velocity.
- Why? Slope is "rise over run," which is
Δx / Δt(change in position over change in time). That's the definition of velocity! - A steep slope means high velocity.
- A flat (zero slope) line means zero velocity (the object is stopped).
- A straight line means constant velocity.
- A curved line means the velocity is changing (there's acceleration).
2. From Velocity to Acceleration The slope of a velocity-time (v-t) graph gives you the acceleration.
- Why? Slope is
Δv / Δt(change in velocity over change in time). That's the definition of acceleration! - A steep slope means high acceleration.
- A flat (zero slope) line means zero acceleration (constant velocity).
This is where many students get confused. They see a flat horizontal line on a v-t graph and think the object is stopped. No! A flat line on a v-t graph means acceleration is zero, so the velocity is constant. The object is still moving, just not speeding up or slowing down.
3. Going Backwards: Using Area We can also go in the other direction using the area under the graph.
- The area under a velocity-time (v-t) graph gives you the displacement (Δx).
- The area under an acceleration-time (a-t) graph gives you the change in velocity (Δv).
Think about it: for a constant velocity, displacement is velocity × time. On a v-t graph, this is the height × width of the rectangle under the line—its area!
The Kinematic Equations: Your "Constant Acceleration" Toolkit
When an object's acceleration is constant, we can use three special equations to solve for unknown quantities. These are your best friends in this unit, but they come with a big warning label.
WARNING: Only use these equations when acceleration is constant!
Here they are, for motion in the x-direction:
-
vₓ = vₓ₀ + aₓt- (Final velocity = initial velocity + acceleration × time)
- This one doesn't involve position.
-
x = x₀ + vₓ₀t + (1/2)aₓt²- (Final position = initial position + initial velocity × time + one-half acceleration × time squared)
- This one doesn't involve final velocity.
-
vₓ² = vₓ₀² + 2aₓ(x − x₀)- (Final velocity squared = initial velocity squared + 2 × acceleration × displacement)
- This one doesn't involve time.
Here, the subscript ₀ means "initial" (at time t=0), and x just tells us the motion is along the horizontal axis. You can swap x for y for vertical motion.
A Special Case: Free Fall
One of the most common examples of constant acceleration is an object in free fall near Earth's surface (ignoring air resistance).
- Any object, regardless of its mass, will accelerate downwards at the same rate.
- This acceleration, called the acceleration due to gravity, is given the symbol
g. - On the AP exam, you can approximate
g ≈ 10 m/s².
This is a critical point: When we define the upward direction as positive, the acceleration a in our kinematic equations becomes a = -g ≈ -10 m/s². The negative sign is crucial because gravity pulls things down.
By mastering the interplay between graphs and these three equations, you can describe and predict the motion of almost anything in a straight line.
Worked examples
Let's put these ideas into practice. We'll use the same scenario and look at it through the lens of graphs and equations.
Using Kinematic Equations
Problem: A Tesla starting from rest accelerates uniformly at 5.0 m/s² for 3.0 seconds. How far does it travel in this time?
Step 1: Identify your knowns and unknowns. This is the most important first step. Let's list them out.
- "Starts from rest" means initial velocity,
v₀ = 0 m/s. - "Accelerates uniformly at 5.0 m/s²" means constant acceleration,
a = 5.0 m/s². - "for 3.0 seconds" means time,
t = 3.0 s. - "How far does it travel?" means we are looking for displacement,
Δx(which isx - x₀). We can set the initial positionx₀to 0 for simplicity. So we need to findx. - The unknown we don't care about is the final velocity,
v.
Step 2: Choose the right kinematic equation.
We need an equation that has x, v₀, a, and t, but not v. Looking at our list:
v = v₀ + at(No, hasv)x = x₀ + v₀t + (1/2)at²(Yes! This has everything we need.)v² = v₀² + 2a(x - x₀)(No, doesn't havet)
So, we'll use the second equation.
Step 3: Plug in the values and solve.
x = x₀ + v₀t + (1/2)at²
x = 0 + (0 m/s)(3.0 s) + (1/2)(5.0 m/s²)(3.0 s)²
This is where students often make a mistake: They forget to square the time. Remember the t²!
x = 0 + 0 + (1/2)(5.0 m/s²)(9.0 s²)
x = (1/2)(45 m)
x = 22.5 m
The car travels 22.5 meters.
Connecting to Graphs
Problem: For the same Tesla from Example 1, what is the area under its velocity-time graph from t=0 to t=3.0 s?
Step 1: Understand the question.
The question asks for the "area under the velocity-time graph." We know from our core concepts that this area is equal to the displacement, Δx. So, this is just another way of asking "how far did the car travel?" We should get the same answer: 22.5 m. Let's prove it.
Step 2: Find the final velocity to draw the graph.
First, we need to know what the v-t graph looks like. It starts at v₀ = 0. What's the final velocity? We can use the first kinematic equation:
v = v₀ + at
v = 0 + (5.0 m/s²)(3.0 s)
v = 15 m/s
Step 3: Sketch the graph and find the area. The v-t graph is a straight line starting at (0, 0) and going up to the point (3.0 s, 15 m/s). The "area under the curve" is the area of the triangle formed by this line, the time-axis, and the line t=3.0.
- The shape is a triangle.
- The formula for the area of a triangle is
(1/2) × base × height. - The base of our triangle is the time interval:
base = 3.0 s. - The height of our triangle is the final velocity:
height = 15 m/s.
Area = (1/2) * (3.0 s) * (15 m/s)
Area = (1/2) * 45 m
Area = 22.5 m
Try it yourself
Ready to try one on your own? Don't worry, I'll give you some hints.
Problem 1: Aaliyah is driving her car at a constant 20 m/s. She passes a sign for her exit and hits the brakes, slowing down with a constant acceleration of -4 m/s². a) How much time does it take for her to come to a complete stop? b) How far does her car travel while braking?
Hints:
- What does "come to a complete stop" tell you about her final velocity?
- For part (a), which kinematic equation connects initial velocity, final velocity, acceleration, and time?
- For part (b), you can now use your answer from part (a). Or, you could use a different kinematic equation that doesn't require time at all!
Problem 2: Sketch the position-time, velocity-time, and acceleration-time graphs for Aaliyah's motion while she is braking.
Hints:
- What does constant negative acceleration look like on an a-t graph?
- What does a constantly decreasing velocity look like on a v-t graph?
- If velocity is decreasing, what must be happening to the slope of the position-time graph?
Practice — 8 questions
In simple terms, representing motion is about using graphs, equations, and diagrams to tell the complete story of an object's movement, including its position, velocity, and acceleration.
- 1.3.A: Describe the position, velocity, and acceleration of an object using representations of that object's motion.
- 1.3.A.1
- Motion can be represented by motion diagrams, figures, graphs, equations, and narrative descriptions.
- 1.3.A.2
- For constant acceleration, three kinematic equations can be used to describe instantaneous linear motion in one dimension: vₓ = vₓ₀ + aₓt x = x₀ + vₓ₀t + (1/2)aₓt² vₓ² = vₓ₀² + 2aₓ(x − x₀) Note: The equations above are written to indicate motion in the x-direction, but these equations can be used in any single dimension as appropriate.
- 1.3.A.3
- Near the surface of Earth, the vertical acceleration caused by the force of gravity is downward, constant, and has a measured value approximately equal to a_g = g ≈ 10 m/s².
- 1.3.A.4
- Graphs of position, velocity, and acceleration as functions of time can be used to find the relationships between those quantities.
- 1.3.A.4.i
- An object's instantaneous velocity is the rate of change of the object's position, which is equal to the slope of a line tangent to a point on a graph of the object's position as a function of time.
- 1.3.A.4.ii
- An object's instantaneous acceleration is the rate of change of the object's velocity, which is equal to the slope of a line tangent to a point on a graph of the object's velocity as a function of time.
- 1.3.A.4.iii
- The displacement of an object during a time interval is equal to the area under the curve of a graph of the object's velocity as a function of time (i.e., the area bounded by the function and the horizontal axis for the appropriate interval).
- 1.3.A.4.iv
- The change in velocity of an object during a time interval is equal to the area under the curve of a graph of the acceleration of the object as a function of time.
flowchart TD
A[Position, x] -- "Slope is..." --> B[Velocity, v];
B -- "Slope is..." --> C[Acceleration, a];
C -- "Area is..." --> D[Change in Velocity, Δv];
B -- "Area is..." --> E[Displacement, Δx];
subgraph "Calculus Connections"
A
B
C
D
E
end
Read what Saavi narrates
Hey everyone, it's Saavi. Let's talk about motion.
Imagine you're a scout for a major league baseball team, watching a high school phenom pitch. You don't just care if she throws a strike. You want to know everything about that pitch. How fast was it the moment it left her hand? Did it slow down on its way to the plate? Where exactly was the ball zero-point-two seconds after release?
Telling this full story of motion is what we're learning to do today. We're going to learn how to translate the story of an object's journey into the powerful languages of graphs and equations.
Basically, we're learning to describe an object's position, velocity, and acceleration over time. We can tell this story using words, diagrams, graphs, and a set of powerful "shortcut" equations called the kinematic equations.
Let's walk through an example. A Tesla starting from rest accelerates uniformly at five meters per second squared for three seconds. How far does it travel?
First, we list what we know. "Starts from rest" means the initial velocity is zero. Acceleration is five. Time is three seconds. We're looking for how far it travels, which is displacement.
The kinematic equation that connects all these is: final position equals initial position plus initial velocity times time, plus one-half times acceleration times time squared. It's a mouthful, I know!
Let's plug in the numbers. Initial position is zero. Initial velocity is zero, so that whole term is zero. That leaves us with one-half times five meters per second squared, times three seconds squared.
Now, here's a common mistake I see all the time. Students forget to square the time. Don't be that student! Three squared is nine. So we have one-half times five times nine. That gives us twenty-two point five meters.
The key takeaway here is that these equations and the graphs we'll study are just different tools to tell the same story. Once you get comfortable translating between them, you'll feel like a physics superhero. Keep practicing, you've got this.
These equations are derived from the assumption that `a` is a constant number. If acceleration is changing (e.g., the problem gives `a(t) = 2t`), these equations will give a wrong answer.
If acceleration is not constant, you must use the graphical relationships. Use the area under the v-t graph for displacement, or the area under the a-t graph for change in velocity.
A flat line on an x-t graph means position isn't changing, so velocity is zero (object is stopped). A flat line on a v-t graph means velocity isn't changing, so acceleration is zero (object moves at a constant velocity).
Always look at the vertical axis label! Ask yourself, "What quantity is being held constant?" before interpreting the motion.
In problems where an object is thrown upwards, gravity acts to slow it down. If you define "up" as the positive direction, then the acceleration `a` must be negative (`a = -10 m/s²`). Using a positive value implies the object magically speeds up as it flies up.
Establish a coordinate system at the start of every problem. If up is positive, `a = -g`. If down is positive, `a = +g`. Stick with your choice.
At the very peak, the ball's instantaneous velocity is zero for a split second. However, gravity doesn't just switch off. The acceleration is *still* `-10 m/s²` downwards, pulling it back to Earth. This is what makes it change direction.
Remember that in free fall, the acceleration is *always* `g` downwards, from the moment it leaves the hand to the moment it's caught. The only exception is if it's on the ground or being pushed.
It's a simple algebraic error, but it's one of the most common on exams. It leads to a completely incorrect displacement.
When you write the equation, circle or underline the `²` exponent as a mental reminder. Double-check your calculator input before writing down the final answer.