FULL REVIEW
Full Review — Position, Displacement, and Velocity — Foundational
Review the essential ideas, relationships, and problem-solving tools for Position, Displacement, and Velocity.
TIME
45–60 minutes
BEST FOR
A complete topic review
FINISH WITH
A readiness check
After this full review, you'll be able to...
recall the essential ideas, apply them to representative problems, and determine what to study next.
Choose how you want to review
Course Alignment
This Physics Sensei Topic Review is an independent learning resource. Use it to reinforce key concepts, prepare for homework, or review before a quiz or exam.
Related Unit Review: If you need to review the complete unit material, review Motion in One Dimension here →
RESOURCE: Physics Sensei Topic Review | TOPIC ID: MEC-U03-T01 | TOPIC: Position, Displacement, and Velocity | PARENT UNIT: MEC-U03 — Motion in One Dimension | COURSE LEVEL: Foundational introductory physics
BEST USED ✓ After learning the topic ✓ Before starting homework ✓ Before a quiz or exam
Your Review Plan
Complete these six stages in order. Each stage builds on the previous one and prepares you for the final readiness check.
6 Stages • Approximately 45–60 minutes.
Warm-Up Check
Before you begin, take a moment to see what you already remember. Do not worry about getting everything right. This is only a starting point.
ACTIVITY 1
Recall Activity 1
Define the coordinate system.
What must be specified before a position value such as x = −4 m can be interpreted?
Reveal Answers
An origin and a positive direction.
Why it works: Position is coordinate-dependent, so its sign is meaningful only after the axis is defined.
ACTIVITY 2
Recall Activity 2
Compare position and displacement.
A car starts at x = −6 m and ends at x = +2 m. Which number is the final position and which number is the displacement?
Reveal Answers
Final position = +2 m; displacement = +8 m.
Why it works: Position is a coordinate; displacement is the difference between final and initial coordinates.
ACTIVITY 3
Recall Activity 3
Read velocity direction.
On a position-versus-time graph, what does an upward trend mean about velocity?
Reveal Answers
Velocity is positive during an upward trend.
Why it works: Increasing position with increasing time corresponds to motion in the positive direction.
Ready to strengthen your understanding?
You've refreshed what you already know. Next, you'll reinforce the essential concepts that will help you solve problems with confidence. Need to see the learning path again?
Core Concepts
Let's rebuild the key ideas one step at a time. Focus on understanding the relationships before worrying about solving problems.
KEY CONCEPT 1
Reference frame, position, and displacement
One-dimensional motion begins with a coordinate axis. Position x gives location relative to the origin. Displacement compares two positions and is signed, so it tells both the magnitude and direction of the net change.
Δx = x₂ − x₁
Example: From −6 m to +2 m, Δx = +8 m.
Sensei note: Write final minus initial every time; reversing the order reverses the physical direction.
KEY CONCEPT 2
Average velocity describes net position change
Average velocity is displacement divided by elapsed time. Because elapsed time is positive, the sign of average velocity matches the sign of displacement. A return to the starting position gives zero average velocity even if the object traveled a long distance.
average velocity = Δx/Δt
Example: A +15 m displacement during 5 s gives +3 m/s.
Sensei note: Do not use total distance in the numerator when the question asks for average velocity.
KEY CONCEPT 3
Position-time graphs encode velocity
On an x–t graph, the slope tells how position changes with time. Upward slope means positive velocity, downward slope means negative velocity, and a horizontal segment means zero velocity. A steeper slope corresponds to a larger speed.
x–t slope = change in position ÷ change in time
Example: A straight line from 0 m at 0 s to 12 m at 4 s has slope +3 m/s.
Sensei note: A high position on the graph does not mean high velocity; slope, not height, describes velocity.
Ready to apply these ideas?
You've reinforced the essential concepts. Now it's time to put them into practice by working through guided examples and building your problem-solving confidence. Need a quick reminder?
Guided Practice
Now it's time to apply what you've reviewed.
Work through each activity in order. The examples become gradually more challenging, and each one prepares you for the final readiness check.
PRACTICE 1
Worked Example
Track endpoints first.
A student walks from x = −2 m to x = +6 m in 4 s. Find displacement and average velocity.
Reveal Answers
Δx = +8 m; average velocity = +2 m/s.
Why it works: Endpoints give the signed displacement; dividing by the interval gives the average velocity.
PRACTICE 2
Guided Problem
Separate path length from net change.
A robot moves from x = 0 m to x = +5 m, then to x = +1 m. Find the total distance and displacement.
Reveal Answers
Distance = 9 m; displacement = +1 m.
Why it works: Distance adds path segments: 5 m + 4 m. Displacement compares only the final position x₂ = +1 m with the initial position x₁ = 0 m.
PRACTICE 3
Independent Problem
Use a graph slope.
A position-time graph rises linearly from x = −4 m at t = 1 s to x = +8 m at t = 5 s. Find the average velocity over that interval.
Reveal Answers
Displacement = +12 m; elapsed time = 4 s; average velocity = +3 m/s.
Why it works: For a straight x–t segment, the slope equals the velocity over the interval.
Ready to check your understanding?
You've practiced the essential skills with guidance. Now it's time to solve a few short problems on your own and confirm you're ready to move forward. Need a quick reminder?
Confidence Check
You've rebuilt the key ideas and practiced them with guidance. Now try these short questions on your own to check your understanding before moving on.
QUICK CHECK 1
Zero net change
Answer using endpoints.
A runner starts and ends at x = 10 m after 20 s. What is the average velocity?
Reveal Answers
0 m/s.
Why it works: The displacement is zero because the endpoints are the same.
QUICK CHECK 2
Sign check
Calculate final minus initial.
A cart moves from x = +3 m to x = −9 m in 6 s. What is its average velocity?
Reveal Answers
Δx = −12 m, so average velocity = −2 m/s.
Why it works: The negative sign records motion in the negative coordinate direction.
QUICK CHECK 3
Graph interpretation
Use slope, not graph height.
An x–t graph is horizontal for 3 s. What is the velocity during that interval?
Reveal Answers
0 m/s.
Why it works: A horizontal x–t segment has zero slope, so position is not changing with time.
How did it go?
You've checked your understanding. Take one final look at the essential ideas before deciding what to do next. Need a quick reminder?
Summary
Before moving on, take one final look at the most important ideas from this review.
KEY TAKEAWAY 1
Position needs a reference
Position is a coordinate measured from an origin along a chosen positive direction.
KEY TAKEAWAY 2
Displacement is signed endpoint change
Use Δx = x₂ − x₁; the path length does not determine displacement.
KEY TAKEAWAY 3
Velocity comes from position change
Average velocity is Δx/Δt, and an x–t graph shows velocity through its slope.
Ready for your next step?
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