Kinematics
A calculus-based introduction to motion using vectors, graphs, uncertainty, and model-building. This page emphasizes how to describe motion before asking what force caused it.
How to Use This Page
Core Idea
Kinematics is not a list of equations. It is the process of building a mathematical model of position, velocity, and acceleration.
- Use vectors to describe direction.
- Use derivatives to describe rates of change.
- Use integrals to accumulate motion.
- Use data to test the model.
What students should be able to do
By the end of this unit, students should be able to represent motion with words, diagrams, vectors, graphs, equations, and data. The goal is to move between those representations without losing physical meaning.
AP Physics C focus
Every equation in this unit can be understood through derivatives or integrals. Algebraic kinematics is useful, but AP Physics C requires students to understand where the relationships come from.
General Problem-Solving Method
- Draw the situation before doing algebra.
- Choose a coordinate system and define positive direction.
- Write position, velocity, and acceleration as components when needed.
- Separate motion into independent directions.
- Choose equations based on missing and known variables.
- Check whether the sign, units, and magnitude of the answer make sense.
1. Lab Skills and Experimental Language
Even when the unit is not centered on a formal lab, kinematics should be taught as experimental modeling. Motion is observed, measured, graphed, fit with functions, and compared to theory.
How close a measured or calculated value is to the accepted or theoretical value.
How repeatable measurements are. A measurement can be precise but not accurate if the method has systematic bias.
Scatter caused by timing variation, reading uncertainty, or small uncontrolled differences from trial to trial.
A consistent shift in one direction, often caused by calibration, an incorrect starting point, or a biased procedure.
A measure of spread in repeated data. In kinematics, it is useful when comparing many timing or distance trials.
Check Your Understanding: Measurement
A student predicts that a cart will take 1.42 s to travel down a ramp. Five measured trials are 1.49 s, 1.46 s, 1.51 s, 1.47 s, and 1.50 s. Is the main issue accuracy, precision, or both?
Show solution
The trials are close to one another, so the measurements are reasonably precise. The average is about 1.49 s, which is noticeably above the predicted 1.42 s. That suggests an accuracy issue or a model mismatch. A strong answer would investigate whether the prediction ignored friction, starting delay, ramp angle uncertainty, or release method.
2. Vectors in Kinematics
Kinematics uses vectors because motion has direction. A standalone vector unit may go deeper into vector algebra, dot products, and cross products, but this unit uses vector notation as the language of motion.
Position
Position tells where an object is relative to an origin.
Velocity
Velocity tells how position changes in time.
Acceleration
Acceleration tells how velocity changes in time.
Component Thinking
In two-dimensional motion, treat the x-direction and y-direction as separate one-dimensional problems that share the same time variable.
Core skill
Write the vector first, then decide what each component does.
If the projectile is launched at speed \(v\) and angle \(\theta\), the horizontal and vertical components are different parts of one velocity vector.
Problem: Velocity Vector
A ball is launched at 20.0 m/s at 30.0° above the horizontal. Write its initial velocity vector.
Show solution
Use \(v_x=v\cos\theta\) and \(v_y=v\sin\theta\).
So the vector is \(\vec v_0=(17.3\hat i+10.0\hat j)\text{ m/s}\).
3. One-Dimensional Motion
One-dimensional kinematics is the foundation for more complicated motion. The key is choosing a direction convention and sticking with it.
Constant Acceleration Relationships
Useful when displacement is not needed.
Useful when final velocity is not needed.
Useful when time is not needed.
Worked Example: Braking Car
A car travels at 20.0 m/s. The driver applies the brakes, producing a constant acceleration of −2.00 m/s² until the car stops.
- How long does it take to stop?
- How far does the car travel while braking?
- What is the car’s speed halfway through the stopping time?
- What is the car’s speed halfway through the stopping distance?
Show solution
1. Use \(v_f=v_i+at\).
2. Use \(v_f^2-v_i^2=2a\Delta x\).
3. Halfway through the time is 5.00 s.
4. Halfway through the distance is 50.0 m.
The halfway-time speed and halfway-distance speed are not the same because distance does not accumulate evenly during accelerated motion.
4. Deriving the Kinematic Equations with Separable Differential Equations
The constant-acceleration equations should not feel like magic. In AP Physics C, they come directly from the definitions of velocity and acceleration. When acceleration is constant, the differential equations are separable and can be integrated.
Derivation 1: Velocity as a Function of Time
Start with the definition of acceleration in one dimension:
Separate the variables:
Integrate from the initial state to the final state:
For constant acceleration, \(a\) comes out of the integral:
Derivation 2: Position as a Function of Time
Use velocity as the derivative of position:
Since \(v(t)=v_i+at\), substitute that expression for velocity:
Separate and integrate:
Evaluate the integral:
Derivation 3: The Time-Independent Equation
Sometimes time is not given. Start with acceleration and use the chain rule:
Since \(\frac{dx}{dt}=v\), this becomes:
Separate variables:
Integrate from the initial state to the final state:
For constant acceleration:
Check Your Understanding
A particle starts from rest and accelerates at a constant \(3.0\text{ m/s}^2\). Use calculus, not memorized kinematics, to find its displacement after \(4.0\text{ s}\).
Show solution
Start with acceleration:
Integrate to find velocity:
Since the particle starts from rest, \(v(0)=0\), so \(C=0\). Then:
Now use \(v=\frac{dx}{dt}\):
5. Graphs and Calculus Interpretation
AP Physics C expects students to connect kinematics to calculus. Derivatives describe slopes. Integrals describe accumulated change.
Velocity from Position
Velocity is the derivative of position with respect to time.
Acceleration from Velocity
Acceleration is the derivative of velocity and the second derivative of position.
Displacement from Velocity
Displacement is the area under the velocity-time graph.
| Graph | Slope Means | Area Means |
|---|---|---|
| Position vs. time | Velocity | Usually not physically central in this course |
| Velocity vs. time | Acceleration | Displacement |
| Acceleration vs. time | Rate of change of acceleration (jerk) | Change in velocity |
Problem: Calculus Kinematics
A particle moves along the x-axis with position \(x(t)=3t^3-4t^2+2\), where x is in meters and t is in seconds.
- Find \(v(t)\).
- Find \(a(t)\).
- Find the velocity and acceleration at \(t=2.00\text{ s}\).
Show solution
Differentiate position to get velocity:
Differentiate velocity to get acceleration:
At \(t=2.00\text{ s}\):
6. Projectile Motion
Projectile motion is two-dimensional kinematics with constant vertical acceleration and, if air resistance is ignored, constant horizontal velocity.
Projectile Model
Horizontal Motion
With no air resistance, \(a_x=0\).
Vertical Motion
Near Earth, \(a_y=-g\).
Velocity Components
The velocity vector changes because the vertical component changes.
Case A: Launch and Land at Same Height
Problem
A projectile is launched from level ground at 20.0 m/s and 30.0° above the horizontal. Ignore air resistance. Find the total time of flight, maximum height, and range.
Show solution
Resolve the initial velocity:
At the top, \(v_y=0\):
Total time is twice the time to the top:
Maximum height:
Range:
Case B: Horizontal Launch
Problem
A package is released from an aircraft moving horizontally at 686 m/s at an altitude of 21,000 m. Ignore air resistance. How far horizontally does the package travel before hitting the ground?
Show solution
Vertical motion determines the time. The initial vertical velocity is zero.
Use horizontal motion to find range:
The package lands about \(44.9\text{ km}\) horizontally from the release point.
Case C: Symbolic Range Derivation
Problem
Derive the range of a projectile launched and landing at the same height with launch speed \(v_0\) and angle \(\theta\).
Show solution
The vertical position returns to zero:
Ignoring the starting time \(t=0\), solve for total flight time:
Horizontal range is horizontal velocity times time:
Use \(2\sin\theta\cos\theta=\sin(2\theta)\):
7. Relative Motion and Circular Motion Preview
Relative motion and circular motion both require careful vector thinking. This page introduces the ideas briefly; later units can treat them in more depth.
Relative Velocity
Velocity depends on the observer. Riverboat and airplane-wind problems are applications of vector addition.
Circular Motion Preview
Even when speed is constant, velocity changes direction. That change in direction requires acceleration toward the center.
Constant speed does not mean zero acceleration. Velocity is a vector, so changing direction counts as acceleration.
Problem: River Crossing
A boat points directly across a river and moves at 4.0 m/s relative to the water. The river current flows downstream at 3.0 m/s. Find the boat’s velocity relative to the shore.
Show solution
The boat’s across-river velocity and the river’s downstream velocity are perpendicular components.
The boat moves at 5.0 m/s relative to the shore, angled 36.9° downstream from straight across.
7. Additional Practice Problems
Use these problems after attempting the main examples. Solutions are hidden so students can check their work after making a serious attempt.
Conceptual Question 1
In ideal projectile motion, why does the horizontal velocity remain constant while the vertical velocity changes?
Show solution
With air resistance ignored, the only acceleration is gravity, which points vertically downward. Therefore \(a_x=0\), so \(v_x\) remains constant. The vertical component changes because \(a_y=-g\).
Conceptual Question 2
A car travels around a circular track at constant speed. Is its acceleration zero? Explain.
Show solution
No. The speed is constant, but the velocity vector changes direction continuously. Since acceleration is the rate of change of velocity, the car has centripetal acceleration toward the center of the circle.
Algebraic Problem
A cart starts from rest and accelerates at 1.60 m/s² for 5.00 s. Find its final speed and displacement.
Show solution
AP-Style Multi-Part Problem
A particle moves in the xy-plane with position vector \(\vec r(t)=(2t^2+1)\hat i+(6t-t^2)\hat j\), where position is in meters and time is in seconds.
- Find \(\vec v(t)\).
- Find \(\vec a(t)\).
- Find the speed at \(t=2.00\text{ s}\).
- Determine whether the particle’s acceleration is constant.
Show solution
Differentiate the position vector component by component:
Differentiate velocity:
At \(t=2.00\text{ s}\):
The acceleration is constant because \(\vec a(t)=4\hat i-2\hat j\) does not depend on time.