Kinematics — AP Physics 1 Phys 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: 1D position/velocity/acceleration, constant acceleration kinematic equations, free fall/projectile motion, velocity-time graph analysis, and reference frames/relative velocity for AP Physics 1 Kinematics.

You should already know: Algebra 2, basic trig, no calculus required.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Kinematics?

Kinematics is the branch of classical mechanics that describes the motion of objects without considering the forces that cause that motion. Its core goal is to answer three key questions about moving objects: where is it, how fast is it moving, and how is its speed changing? Standard SI units are used for all kinematic quantities: meters (m) for position, meters per second (m/s) for velocity, and meters per second squared (m/s²) for acceleration. Sometimes called "descriptive motion", kinematics makes up 10-16% of the AP Physics 1 exam per the 2024-25 CED, and is the foundational framework for all subsequent mechanics topics including dynamics, work and energy, and rotational motion.

2. Position, velocity, acceleration in 1D

One-dimensional (1D) motion describes objects moving along a straight line, and uses three core vector quantities (values with both magnitude and direction):

1. Position ($x$ or $y$): The location of an object relative to a defined origin. Displacement ($\Delta x$) is the change in position: $\Delta x=x_{\text{final}}-x_{\text{initial}}$, and is distinct from distance (the total length of the path traveled, a scalar quantity with no direction).
2. Velocity ($v$): The rate of change of displacement. Average velocity is $v_{\text{avg}}=\frac{\Delta x}{\Delta t}$, while instantaneous velocity is the velocity of the object at a single point in time. Speed is the scalar magnitude of velocity, with no direction.
3. Acceleration ($a$): The rate of change of velocity, calculated as $a_{\text{avg}}=\frac{\Delta v}{\Delta t}$. Positive acceleration does not always mean an object is speeding up: if velocity is negative, positive acceleration means the object is slowing down.

Worked Example

A delivery truck travels from $x=0\text{m}$ to $x=120\text{m}$ in 12 s, then reverses to $x=30\text{m}$ in 6 s. Calculate its average speed, average velocity, and average acceleration if it starts and ends at rest.

• Total distance traveled: $120\text{m}+90\text{m}=210\text{m}$, so average speed = $\frac{210}{18}=11.7\text{m/s}$
• Total displacement: $\Delta x=30-0=30\text{m}$, so average velocity = $\frac{30}{18}=1.67\text{m/s}$
• Total change in velocity: $\Delta v=0-0=0$, so average acceleration = $\frac{0}{18}=0{\text{m/s}}^2$

Exam tip: Examiners frequently test the difference between scalars and vectors here, so always confirm if a question asks for a signed value (vector) or magnitude only (scalar).

3. Kinematic equations (constant acceleration)

The four core kinematic equations only apply when acceleration is constant, the case for nearly all AP Physics 1 kinematics problems. They are derived directly from the definitions of velocity and acceleration:

1. Rearranging the acceleration definition gives: $$v=v_0+at$$ (no displacement term)
2. Average velocity for constant acceleration is the midpoint of initial and final velocity, so: $$\Delta x=\frac{(v_0+v)}{2}t$$ (no acceleration term)
3. Substitute $v$ from equation 1 into equation 2 to eliminate final velocity: $$\Delta x=v_{0}t+\frac{1}{2}a{t}^2$$ (no final velocity term)
4. Eliminate time from equations 1 and 2 to get the time-independent form: $$v^2=v_0^2+2a\Delta x$$ (no time term)

Worked Example

A motorcycle accelerates from rest at $3{\text{m/s}}^2$ for 7 s. What is its final speed and total distance traveled?

• Known values: $v_0=0$, $a=3{\text{m/s}}^2$, $t=7\text{s}$
• Final speed: $v=0+(3)(7)=21\text{m/s}$
• Distance traveled: $\Delta x=0+\frac{1}{2}(3)(7{)}^2=73.5\text{m}$

Exam tip: Always select the equation that avoids calculating intermediate variables you do not need, to reduce arithmetic error and save time on the exam.

4. Free fall and projectile motion

Free fall describes motion where the only acceleration acting on an object is gravity, with air resistance ignored for all AP Physics 1 problems unless explicitly stated. The acceleration due to gravity $g=9.8{\text{m/s}}^2$, or $10{\text{m/s}}^2$ for approximate calculations if allowed. If you define upward as the positive direction, acceleration in free fall is always $a=-g$.

1D Free Fall Worked Example

A ball is thrown upward at $18\text{m/s}$ from ground level. What maximum height does it reach, and how long does it take to hit the ground?

• At maximum height, vertical velocity $v=0$. Use $v^2=v_0^2+2a\Delta y$: $$0=(18{)}^2+2(-9.8)\Delta y⟹\Delta y=\frac{324}{19.6}=16.5\text{m}$$
• When it hits the ground, $\Delta y=0$. Use $\Delta y=v_{0}t+\frac{1}{2}a{t}^2$: $$0=18t-4.9t^2⟹t=\frac{18}{4.9}=3.67\text{s}$$

Projectile motion is 2D motion where the only acceleration is gravity, and horizontal and vertical motion are completely independent. Split all quantities into x (horizontal) and y (vertical) components immediately:

• Horizontal axis: No air resistance, so $a_x=0$, horizontal velocity is constant: $\Delta x=v_{0x}t$, $v_{0x}=v_0\cos \theta$ where $\theta$ is the launch angle above horizontal.
• Vertical axis: Identical to 1D free fall, $v_{0y}=v_0\sin \theta$, $a_y=-g$.

Projectile Motion Worked Example

A soccer ball is kicked at $16\text{m/s}$ at a $45^{\circ }$ angle above horizontal, from ground level. Calculate its range (horizontal distance when it hits the ground).

• Initial components: $v_{0x}=16\cos 45=11.31\text{m/s}$, $v_{0y}=16\sin 45=11.31\text{m/s}$
• Time of flight from vertical motion: $\Delta y=0$, so $0=11.31t-4.9t^2⟹t=2.31\text{s}$
• Range: $\Delta x=v_{0x}t=11.31\ast 2.31=26.1\text{m}$

5. Velocity-time graphs and area = displacement

Velocity-time (v-t) plots show velocity on the y-axis and time on the x-axis, and have two core interpretation rules that appear on nearly every AP Physics 1 exam:

1. Slope of the v-t graph = acceleration: A horizontal line means constant velocity (0 acceleration), an upward slope means positive acceleration, and a downward slope means negative acceleration.
2. Area under the v-t graph between two times = net displacement between those times: Area above the t-axis counts as positive displacement, while area below the t-axis counts as negative displacement. The sum of the absolute values of all areas gives total distance traveled.

Worked Example

A v-t graph has a line from (0,0) to (3,6), stays horizontal until t=7 s, then decreases linearly to (10,0). Calculate total displacement and average acceleration over 10 s.

• Area 0-3 s: Triangle, $\frac{1}{2}\ast 3\ast 6=9\text{m}$
• Area 3-7 s: Rectangle, $4\ast 6=24\text{m}$
• Area 7-10 s: Triangle, $\frac{1}{2}\ast 3\ast 6=9\text{m}$
• Total displacement: $9+24+9=42\text{m}$
• Average acceleration: $\Delta v=0-0=0$, so $a_{\text{avg}}=0{\text{m/s}}^2$

6. Reference frames and relative velocity

All motion is measured relative to a reference frame, a coordinate system used to define position and velocity. AP Physics 1 exclusively uses inertial reference frames (frames with no acceleration, where Newton's first law holds). There is no "absolute" rest frame, so velocity values depend entirely on the frame they are measured in.

For 1D relative velocity, if object A has velocity $v_A$ relative to the ground, and object B has velocity $v_B$ relative to the ground, the velocity of A relative to B is: $$v_{AB}=v_A-v_B$$ For 2D relative velocity, apply the same rule as vector subtraction, calculating x and y components separately.

Worked Example

Train A travels east at $30\text{m/s}$ relative to the ground, while Train B travels west at $25\text{m/s}$ relative to the ground. What is the velocity of Train A relative to passengers on Train B?

• Define east as positive: $v_A=30\text{m/s}$, $v_B=-25\text{m/s}$
• $v_{AB}=30-(-25)=55\text{m/s}$ east, meaning passengers on Train B see Train A approaching at 55 m/s.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Mixing up scalar and vector quantities, e.g., reporting displacement as a positive value when the object moves left of the origin. Why: Students use distance/displacement and speed/velocity interchangeably. Correct move: Always check if a question asks for a scalar (magnitude only) or vector (sign + magnitude, or magnitude + direction) before answering.
• Wrong move: Applying kinematic equations to non-constant acceleration motion, e.g., a car with changing engine thrust. Why: Students memorize the equations but forget their use case. Correct move: Confirm acceleration is constant before using the 4 kinematic equations; if not, use v-t graph area for displacement or slope for acceleration.
• Wrong move: Using $a=+g$ for upward free fall motion. Why: Students associate $g$ with a positive value, ignoring direction. Correct move: Define your coordinate system explicitly first (almost always upward = positive), so $a_y=-9.8{\text{m/s}}^2$ for all free fall motion.
• Wrong move: Combining horizontal and vertical components in projectile problems. Why: Students use total initial velocity for both axes instead of splitting it. Correct move: Split all velocity, acceleration, and displacement values into x and y components immediately, solve each axis separately, and only combine values if asked for total velocity or displacement.
• Wrong move: Counting area below the t-axis on a v-t graph as positive displacement. Why: Students memorize "area = displacement" but forget sign conventions. Correct move: Sum signed areas for net displacement, sum absolute values of areas for total distance traveled.

8. Practice Questions (AP Physics 1 Style)

Question 1

A student drops a stone from the top of a 50 m tall building. At the same moment, a second student throws a stone upward from the ground at $22\text{m/s}$. At what height above the ground do the two stones pass each other? Ignore air resistance.

Solution

Define upward as positive, ground = 0, $a=-9.8{\text{m/s}}^2$.

• Position of dropped stone: $y_1(t)=50-4.9t^2$
• Position of thrown stone: $y_2(t)=22t-4.9t^2$
• Set equal: $50-4.9t^2=22t-4.9t^2$. The $t^2$ terms cancel, so $t=\frac{50}{22}=2.27\text{s}$
• Substitute back to find height: $y=22(2.27)-4.9(2.27{)}^2=24.6\text{m}$

Question 2

A v-t graph for a skateboarder follows these points: (0, -3) to (2,3), horizontal to (6,3), then linear to (8, -3). Calculate (a) net displacement over 8 s, (b) total distance traveled over 8 s.

Solution

(a) Split into segments:

• 0-2 s: Crosses t-axis at t=1 s. Area below axis: $\frac{1}{2}\ast 1\ast 3=-1.5\text{m}$, area above axis: $\frac{1}{2}\ast 1\ast 3=1.5\text{m}$, net 0 m.
• 2-6 s: Rectangle, $4\ast 3=12\text{m}$
• 6-8 s: Crosses t-axis at t=7 s. Area above axis: $\frac{1}{2}\ast 1\ast 3=1.5\text{m}$, area below axis: $\frac{1}{2}\ast 1\ast 3=-1.5\text{m}$, net 0 m.
• Total displacement: $0+12+0=12\text{m}$ (b) Total distance = sum of absolute areas: $1.5+1.5+12+1.5+1.5=18\text{m}$

Question 3

A ferry is moving north at $7\text{m/s}$ relative to the water. The river current flows east at $2\text{m/s}$ relative to the shore. At what angle west of north should the ferry point to travel directly north relative to the shore?

Solution

To travel directly north, the westward component of the ferry's velocity relative to the water must cancel the eastward current.

• Let $\theta$ = angle west of north. The westward component of the ferry's velocity is $7\sin \theta$, which must equal $2\text{m/s}$ to cancel the current.
• $\sin \theta =\frac{2}{7}⟹\theta =\arcsin \left(\frac{2}{7}\right)=16.6^{\circ }$ west of north.

9. Quick Reference Cheatsheet

Core Kinematic Formulas (Constant Acceleration Only)

Key Rules

1. Slope of v-t graph = acceleration
2. Net area under v-t graph = displacement; sum of absolute areas = distance
3. Horizontal and vertical projectile motion are independent, solve separately
4. Always define your positive direction and coordinate system before solving problems

10. What's Next

Kinematics is the foundational building block for all remaining mechanics content in AP Physics 1. You will use the constant acceleration equations, vector splitting rules, and sign conventions you learned here for dynamics (where you relate acceleration to net force), uniform circular motion, work and energy, impulse and momentum, and rotational kinematics later in the course. Mastering these core rules now will eliminate 80% of common errors in these later units, which make up more than 60% of the total AP Physics 1 exam score.

If you have questions about any of the worked examples, pitfalls, or formula applications covered in this guide, you can ask Ollie, our AI tutor, for customized explanations or extra practice problems at any time on the homepage. You can also move on to our next study guide covering dynamics, the next unit in the 2024-25 AP Physics 1 CED, to learn how forces cause the motion you just learned to describe.