ChemSTEM Kinematic Equations Study Guide

Kinematic Equations

Study Guide and Crash Course

A clear, exam-ready guide to constant-acceleration motion using words, equations, graphs, tables, diagrams, and units.

How this guide is organized

This guide teaches the kinematic equations as more than formulas to memorize. You will connect position, displacement, velocity, acceleration, time, motion graphs, and free fall. The guide starts with meaning and vocabulary, then builds toward equation choice, graph interpretation, projectile motion, worked examples, mixed practice, answers, and a final review.

WordsWhat is changing?

GraphsSlope and area

EquationsChoose by variables

Units & SignsCheck meaning

Tutor note: Mindset for kinematics

If kinematics feels intimidating, slow the problem down. First ask: What object is moving? Then ask: What interval of time are we studying? Most mistakes happen before the equation is chosen. A careful sketch, a sign convention, and a known/unknown list often matter more than algebra speed.

Quick Start: The Big Picture

Key idea: Kinematics describes motion without asking why

Kinematics is the part of mechanics that describes motion: where an object is, how fast it moves, and how its motion changes. It does not focus on forces yet. The main question is: Can we describe and predict the motion using position, velocity, acceleration, and time?

Kinematic equations are most commonly used when acceleration is constant. That means velocity changes by the same amount during each equal time interval. For example, an object in ideal free fall near Earth’s surface has approximately constant acceleration \(a=-9.8\,\mathrm{m/s^2}\) when upward is positive.

Increasing spacing between dots means the object is speeding up to the right.

Formula focus: The constant-acceleration family

\[v=v_0+at\]

\[\Delta x=v_0t+\frac{1}{2}at^2\]

\[v^2=v_0^2+2a\Delta x\]

\[\Delta x=\frac{v_0+v}{2}t\]

Here, \(v_0\) is initial velocity, \(v\) is final velocity, \(a\) is acceleration, \(t\) is time, and \(\Delta x=x-x_0\) is displacement.

Tutor note: What learners should be able to do

Define each motion variable, choose a sign convention, recognize constant acceleration, select an equation based on the missing variable, solve one-dimensional and basic projectile problems, and connect slope and area on motion graphs to physical meaning.

Core Vocabulary and Reference

Key idea: Vocabulary is part of the math

The equations are compact, but every symbol carries meaning. Before substituting numbers, translate the situation into motion vocabulary. This prevents the most common errors: confusing distance with displacement, speed with velocity, or velocity with acceleration.

Term	Meaning	Common unit
Position, \(x\)	Location measured from a chosen origin. In one dimension, position can be positive, negative, or zero.	\(\mathrm{m}\)
Displacement, \(\Delta x\)	Change in position: \(\Delta x=x_f-x_i\). It includes direction through its sign.	\(\mathrm{m}\)
Distance	Total path length traveled. It is never negative and may be larger than displacement.	\(\mathrm{m}\)
Speed	How fast distance is covered. Speed has no direction.	\(\mathrm{m/s}\)
Velocity, \(v\)	Rate of change of position. Velocity includes direction through sign or vector direction.	\(\mathrm{m/s}\)
Acceleration, \(a\)	Rate of change of velocity. Acceleration tells how velocity changes, not directly how position changes.	\(\mathrm{m/s^2}\)
Time interval, \(t\)	Duration of the motion being modeled. In basic kinematics, \(t\ge 0\).	\(\mathrm{s}\)

Displacement depends on final position minus initial position, not just the path drawn.

Common trap: Distance and displacement are not the same

If a runner goes 100 m east and then 100 m west, the distance is 200 m, but the displacement is 0 m. Kinematic equations use displacement, not total path length.

Formula focus: Reference equations and meanings

Average velocity	\(v_{\mathrm{avg}}=\dfrac{\Delta x}{\Delta t}\); slope of a position-time graph over an interval.
Average acceleration	\(a_{\mathrm{avg}}=\dfrac{\Delta v}{\Delta t}\); slope of a velocity-time graph over an interval.
Constant acceleration	\(a\) is the same at every moment in the interval. The velocity-time graph is a straight line.
Displacement from velocity	\(\Delta x=\text{area under the velocity-versus-time graph}\).
Change in velocity	\(\Delta v=\text{area under the acceleration-versus-time graph}\).

Signs, Direction, and Setup

Key idea: Choose a positive direction and stay consistent

In one-dimensional kinematics, signs are not decorations. They represent direction. You may choose right as positive, upward as positive, or downhill as positive. The choice is flexible, but once chosen, every velocity, acceleration, and displacement must follow it.

Velocity and acceleration can point in opposite directions. Then the object slows down.

Tutor note: Separate “moving” from “accelerating”

Point in the direction of velocity, then point in the direction of acceleration. If the arrows point the same way, the object speeds up. If they point opposite ways, the object slows down.

Worked example: A car slowing down

A car moves to the right at 20 m/s and brakes with acceleration −5 m/s². Let right be positive.

\(v_0=+20\,\mathrm{m/s}\)\(a=-5\,\mathrm{m/s^2}\)\(t=3.0\,\mathrm{s}\)

\[v=v_0+at=20+(-5)(3)=5\,\mathrm{m/s}\]

The final velocity is still positive, so the car is still moving right, but more slowly.

Common trap: Negative acceleration does not always mean slowing down

A negative acceleration means acceleration points in the negative direction. If velocity is also negative, the object speeds up in the negative direction. Slowing down happens when velocity and acceleration have opposite signs.

Practice: Sign check

A ball is thrown upward. Choose upward as positive. While the ball is moving upward, what are the signs of \(v\) and \(a\)? What about on the way down?

Answer. Upward trip: \(v>0\) and \(a<0\). Downward trip: \(v<0\) and \(a<0\). Gravity points downward the whole time.

The Constant-Acceleration Equations

Key idea: Each equation leaves out one variable

The four main kinematic equations are connected. List what you know, list what you need, and notice which variable is missing from the problem.

Formula focus: The big four

\[\boxed{v=v_0+at}\]

\[\boxed{\Delta x=v_0t+\frac12at^2}\]

\[\boxed{v^2=v_0^2+2a\Delta x}\]

\[\boxed{\Delta x=\frac{v_0+v}{2}t}\]

Equation	Best when you know or need	Variable not included
\(v=v_0+at\)	Velocity, acceleration, and time	\(\Delta x\)
\(\Delta x=v_0t+\frac12at^2\)	Displacement with time	\(v\)
\(v^2=v_0^2+2a\Delta x\)	Displacement but no time	\(t\)
\(\Delta x=\dfrac{v_0+v}{2}t\)	Average of initial and final velocities	\(a\)

Sketch and choose signs

List knowns and unknowns

Choose by missing variable

Substitute, solve, check units

Interpret sign and size

Worked example: Choosing the no-time equation

A cyclist increases speed from 4.0 m/s to 10.0 m/s while accelerating at 2.0 m/s². How far does the cyclist travel?

\(v_0=4.0\,\mathrm{m/s}\)\(v=10.0\,\mathrm{m/s}\)\(a=2.0\,\mathrm{m/s^2}\)\(\Delta x=?\)

Time is not given and not asked for, so use \(v^2=v_0^2+2a\Delta x\).

\[10^2=4^2+2(2)\Delta x\]

\[100=16+4\Delta x\Rightarrow \Delta x=21\,\mathrm{m}\]

Answer. The cyclist travels 21 m.

Common trap: Using the wrong velocity

Do not use \(\Delta x=vt\) unless \(v\) is a constant velocity or an average velocity. If an object accelerates, the final velocity is usually not the same as the average velocity.

Reading Motion Graphs

Key idea: Graphs are equations in visual form

On a position-time graph, slope is velocity. On a velocity-time graph, slope is acceleration and area is displacement.

Formula focus: Graph meanings

Graph	Slope means	Area means
Position vs. time	Velocity	Usually not a basic kinematics quantity
Velocity vs. time	Acceleration	Displacement
Acceleration vs. time	Usually not needed in basic problems	Change in velocity

Worked example: Area under a velocity-time graph

A cart starts at \(v_0=2\,\mathrm{m/s}\) and reaches \(v=10\,\mathrm{m/s}\) after \(4\,\mathrm{s}\) with constant acceleration.

\[\Delta x=\frac{v_0+v}{2}t=\frac{2+10}{2}(4)=24\,\mathrm{m}\]

Answer. The displacement is 24 m.

Common trap: Slope versus height

On a position-time graph, a high point does not mean high speed. Speed is related to the steepness of the graph. A horizontal position-time graph means the object is at rest, even if it is far from the origin.

Practice: Quick graph check

A velocity-time graph is a horizontal line at \(v=6\,\mathrm{m/s}\) from \(t=0\) to \(t=5\,\mathrm{s}\). What is the acceleration? What is the displacement?

Answer. Acceleration is \(0\,\mathrm{m/s^2}\) because the slope is zero. Displacement is area: \((6)(5)=30\,\mathrm{m}\).

Free Fall and Vertical Motion

Key idea: Free fall is kinematics with gravity as the acceleration

Free fall means gravity is the only important force affecting the motion. Near Earth’s surface, acceleration is approximately 9.8 m/s² downward. If upward is positive, \(a=-9.8\,\mathrm{m/s^2}\). If downward is positive, \(a=+9.8\,\mathrm{m/s^2}\).

Spacing grows because the object speeds up downward.

Formula focus: Vertical free-fall equations

\[v_y=v_{0y}+a_yt\]

\[\Delta y=v_{0y}t+\frac12a_yt^2\]

\[v_y^2=v_{0y}^2+2a_y\Delta y\]

With upward positive near Earth, \(a_y=-g\approx-9.8\,\mathrm{m/s^2}\).

Worked example: Dropped object

A stone is dropped from rest from a bridge 45 m above the water. Choose upward as positive and set the release point as \(y=0\).

\(v_{0y}=0\)\(\Delta y=-45\,\mathrm{m}\)\(a_y=-9.8\,\mathrm{m/s^2}\)\(t=?\)

\[-45=\frac12(-9.8)t^2=-4.9t^2\]

\[t^2=\frac{45}{4.9}=9.18\Rightarrow t=3.03\,\mathrm{s}\]

Answer. About 3.0 s.

Common trap: Acceleration at the top is not zero

For a ball thrown upward, velocity is momentarily zero at the top. Acceleration is still downward, about \(9.8\,\mathrm{m/s^2}\).

Practice: Top of the path

A ball is thrown upward at \(18\,\mathrm{m/s}\). With upward positive, how long does it take to reach its highest point?

Answer. At the top, \(v_y=0\): \(0=18-9.8t\), so \(t=1.84\,\mathrm{s}\).

Equation Choice: A Tutor-Friendly Procedure

Key idea: Match the equation to the information

Do not start by asking, “Which equation do I remember?” Start by asking, “What quantities are involved, and which quantity is missing?”

Formula focus: Known/unknown checklist

For constant acceleration in one dimension, check the five major variables:

\[\boxed{v_0,\quad v,\quad a,\quad t,\quad \Delta x}\]

A standard problem usually gives three and asks for one. The remaining unused variable tells you which equation to avoid or choose.

Missing	Good equation	Why it helps
\(\Delta x\)	\(v=v_0+at\)	Connects velocity change directly to acceleration and time.
\(v\)	\(\Delta x=v_0t+\frac12at^2\)	Finds displacement without final velocity.
\(t\)	\(v^2=v_0^2+2a\Delta x\)	Avoids time completely.
\(a\)	\(\Delta x=\frac{v_0+v}{2}t\)	Uses average velocity when acceleration is constant.
Another variable	Rearrange the equation containing the target and knowns.	Algebra isolates the target.

Worked example: Stopping distance

A car traveling 24 m/s brakes uniformly to rest with acceleration −6.0 m/s².

\(v_0=24\,\mathrm{m/s}\)\(v=0\)\(a=-6.0\,\mathrm{m/s^2}\)\(\Delta x=?\)

\[0^2=24^2+2(-6.0)\Delta x\]

\[0=576-12\Delta x\Rightarrow \Delta x=48\,\mathrm{m}\]

Answer. The stopping distance is 48 m.

Tutor note: A strong prompt

Complete this sentence before substituting: “I am using ______ because the problem gives ______, asks for ______, and does not mention ______.”

Two-Dimensional Motion and Projectiles

Key idea: Split projectile motion into horizontal and vertical parts

Horizontal and vertical motion happen at the same time, but they are analyzed separately. Usually \(a_x=0\) and \(a_y=-g\) when upward is positive.

Horizontal: constant velocity. Vertical: constant acceleration.

Formula focus: Component setup

For launch speed \(v_0\) at angle \(\theta\):

\[v_{0x}=v_0\cos\theta,\qquad v_{0y}=v_0\sin\theta\]

\[\Delta x=v_{0x}t,\qquad a_x=0\]

\[\Delta y=v_{0y}t-\frac12gt^2,\qquad v_y=v_{0y}-gt\]

The same time \(t\) applies to both directions.

Worked example: Horizontal launch from a table

A ball rolls horizontally off a 1.25 m-high table at 3.0 m/s.

Vertical motion determines time:

\[-1.25=-4.9t^2\Rightarrow t=0.505\,\mathrm{s}\]

Then horizontal displacement is

\[\Delta x=v_{0x}t=(3.0)(0.505)=1.52\,\mathrm{m}\]

Answer. About 1.5 m from the base.

Common trap: Horizontal velocity does not keep a projectile in the air

Gravity immediately accelerates the object downward. Horizontal velocity affects range; vertical motion controls time in the air when launch height is known.

Multiple Representations

Key idea: Move between words, tables, graphs, and equations

The same motion can be described with a sentence, a data table, a graph, and an equation. Exam problems often test whether you can translate between these forms.

\(t\) (s)	0	1	2	3	4
\(x\) (m)	2	6.5	14	24.5	38

Worked example: Extracting motion from an equation

Given \(x=2+3t+1.5t^2\), compare with \(x=x_0+v_0t+\frac12at^2\).

\[x_0=2\,\mathrm{m},\qquad v_0=3\,\mathrm{m/s},\qquad \frac12a=1.5\]

So \(a=3.0\,\mathrm{m/s^2}\).

Common trap: Assuming every position-time graph is linear

Constant velocity gives a straight position-time graph. Constant acceleration usually gives a curved position-time graph and a straight velocity-time graph.

Common Problem Types

Key idea: Most exam problems are variations on a few patterns

Common patterns include speeding up, slowing down, stopping distance, dropped objects, thrown objects, graph interpretation, and projectile motion.

Worked example: Accelerating from rest

A toy car starts from rest and accelerates at \(1.8\,\mathrm{m/s^2}\) for \(5.0\,\mathrm{s}\).

\[v=0+(1.8)(5.0)=9.0\,\mathrm{m/s}\]

\[\Delta x=\frac12(1.8)(5.0)^2=22.5\,\mathrm{m}\]

Answer. Final speed: 9.0 m/s; displacement: 22.5 m.

Worked example: Finding acceleration from distance and time

A cart starts from rest and rolls \(12.0\,\mathrm{m}\) in \(4.0\,\mathrm{s}\).

\[12.0=\frac12a(4.0)^2=8a\Rightarrow a=1.5\,\mathrm{m/s^2}\]

Worked example: Thrown upward and returning to the same height

A ball is thrown upward from ground level at \(12\,\mathrm{m/s}\).

\[0=12t-4.9t^2=t(12-4.9t)\]

The launch root is \(t=0\). The return root is \(t=2.45\,\mathrm{s}\).

Common trap: Throwing away the wrong root

Quadratic kinematics problems can produce two times. One may be the starting instant, one may be landing, or both may represent times the object is at the same height. Interpret both roots physically before rejecting one.

Practice Set: Mixed Kinematic Equations

How to use the practice set

For each computational problem, write a known/unknown list, choose a positive direction, and state which equation you used. For graph problems, identify slope or area before calculating.

Part A: Concepts and setup

Define displacement in one sentence. How is it different from distance?
A car moves left while slowing down. If right is positive, what are the signs of velocity and acceleration?
A ball thrown upward reaches its highest point. What is its velocity at that instant? What is its acceleration?
A position-time graph is a straight line with positive slope. What does that tell you about velocity and acceleration?
A velocity-time graph crosses from positive velocity to negative velocity while the slope stays negative. What does the crossing point represent?

Part B: One-dimensional calculations

A runner accelerates from rest at 2.4 m/s² for 3.0 s. Find final speed.
A skateboarder moving at 5.0 m/s accelerates at 1.2 m/s² for 6.0 s. Find displacement.
A car slows from 30 m/s to 12 m/s in 4.5 s. Find acceleration.
A train starts from rest and travels 180 m in 12 s. Find acceleration.
A bike increases speed from 3.0 m/s to 9.0 m/s over 24 m. Find acceleration.
A car traveling 28 m/s brakes at −7.0 m/s². How far before stopping?
A cart has \(v_0=2.0\,\mathrm{m/s}\), \(a=0.80\,\mathrm{m/s^2}\), and \(t=5.0\,\mathrm{s}\). Find \(v\) and \(\Delta x\).
A motorcycle accelerates from 10 m/s to 26 m/s while traveling 144 m. Find time.

Part C: Free fall and vertical motion

A rock is dropped from rest from a height of 20 m. Find the time to hit the ground.
A ball is thrown upward at 15 m/s. How long to reach the top?
For the ball in problem 15, what maximum height above the release point does it reach?
A stone is thrown downward at 4.0 m/s and falls 50 m. Choose downward as positive. Find impact speed.

Part D: Graphs and projectiles

A velocity-time graph goes from \((0\,\mathrm{s},2\,\mathrm{m/s})\) to \((6\,\mathrm{s},14\,\mathrm{m/s})\). Find acceleration.
For problem 18, find displacement from \(0\) to \(6\,\mathrm{s}\).
A projectile is launched horizontally at 8.0 m/s from a height of 5.0 m. How long is it in the air?
For problem 20, how far horizontally does it travel?
A projectile is launched at 20 m/s at 30°. Find \(v_{0x}\) and \(v_{0y}\).
Position is modeled by \(x=4+6t-2t^2\). Identify \(x_0\), \(v_0\), and \(a\).

Answers and Worked Checks

Use answers to diagnose, not just score

Identify the error type: setup, sign, equation choice, algebra, units, or interpretation. Then redo the problem from the step before the error.

Displacement is final position minus initial position, \(\Delta x=x_f-x_i\). Distance is total path length.
Moving left: \(v<0\). Slowing down means acceleration points right: \(a>0\).
At the top, \(v=0\) momentarily. Acceleration is still downward: \(a=-9.8\,\mathrm{m/s^2}\) if upward is positive.
Constant positive velocity; acceleration is zero.
The object changes direction at the crossing point because \(v=0\) there.
\(v=0+(2.4)(3.0)=7.2\,\mathrm{m/s}\).
\(\Delta x=(5.0)(6.0)+\frac12(1.2)(6.0)^2=51.6\,\mathrm{m}\).
\(a=(12-30)/4.5=-4.0\,\mathrm{m/s^2}\).
\(180=\frac12a(12)^2\Rightarrow a=2.5\,\mathrm{m/s^2}\).
\(9^2=3^2+2a(24)\Rightarrow a=1.5\,\mathrm{m/s^2}\).
\(0=28^2+2(-7.0)\Delta x\Rightarrow \Delta x=56\,\mathrm{m}\).
\(v=6.0\,\mathrm{m/s}\); \(\Delta x=20\,\mathrm{m}\).
\(144=\frac{10+26}{2}t=18t\Rightarrow t=8.0\,\mathrm{s}\).
\(20=4.9t^2\Rightarrow t=2.02\,\mathrm{s}\).
\(0=15-9.8t\Rightarrow t=1.53\,\mathrm{s}\).
\(0^2=15^2+2(-9.8)\Delta y\Rightarrow \Delta y=11.5\,\mathrm{m}\).
\(v^2=4.0^2+2(9.8)(50)\Rightarrow v=31.6\,\mathrm{m/s}\) downward.
\(a=(14-2)/(6-0)=2.0\,\mathrm{m/s^2}\).
\(\Delta x=\frac{2+14}{2}(6)=48\,\mathrm{m}\).
\(5.0=4.9t^2\Rightarrow t=1.01\,\mathrm{s}\).
\(\Delta x=(8.0)(1.01)=8.1\,\mathrm{m}\).
\(v_{0x}=20\cos30^\circ=17.3\,\mathrm{m/s}\); \(v_{0y}=20\sin30^\circ=10.0\,\mathrm{m/s}\).
\(x_0=4\,\mathrm{m}\), \(v_0=6\,\mathrm{m/s}\), and \(a=-4\,\mathrm{m/s^2}\).

Final Review: One-Page Summary

The kinematics habit

For every problem: sketch the motion, choose a positive direction, list knowns and unknowns, choose an equation based on the missing variable, solve with units, and interpret the sign.

Average quantities

\[v_{\mathrm{avg}}=\frac{\Delta x}{\Delta t}\]

\[a_{\mathrm{avg}}=\frac{\Delta v}{\Delta t}\]

Constant acceleration

\[v=v_0+at\]

\[\Delta x=v_0t+\frac12at^2\]

\[v^2=v_0^2+2a\Delta x\]

\[\Delta x=\frac{v_0+v}{2}t\]

Formula focus: Graph reminders

Graph	Slope	Area
\(x\) versus \(t\)	Velocity	Not usually used
\(v\) versus \(t\)	Acceleration	Displacement
\(a\) versus \(t\)	Not usually used	Change in velocity

Tutor note: Final exam checklist

Use displacement, not distance, in the equation.
Keep the sign convention consistent.
Use \(g=9.8\,\mathrm{m/s^2}\) in the correct direction.
Do not use final velocity as average velocity.
Match the answer unit to the quantity.
Check that the sign and size make physical sense.

Common trap: High-yield reminders

Velocity zero at the top does not mean acceleration zero.
Negative acceleration does not automatically mean slowing down.
Distance and displacement are not interchangeable.
Horizontal and vertical projectile motion share the same time, but not the same acceleration.
Slope and area have different meanings on different motion graphs.

Last self-check

Can you explain why \(v^2=v_0^2+2a\Delta x\) is useful when time is not given? Can you explain why the area under a velocity-time graph gives displacement? If yes, you are thinking like a physics problem-solver, not just a formula memorizer.