FULL REVIEW
Full Review — Position, Displacement, and Velocity — Algebra-Based
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: Algebra-Based 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
Write the core relationships.
State Δx and v̄ for an interval from (t₁, x₁) to (t₂, x₂).
Reveal Answers
Δx = x₂ − x₁; Δt = t₂ − t₁; v̄ = Δx/Δt.
Why it works: Using matching final-minus-initial differences gives the correct signed rate of change.
ACTIVITY 2
Recall Activity 2
Check a return trip.
A particle leaves x = 0, moves 6 m right, then 6 m left. What are distance and displacement?
Reveal Answers
Distance = 12 m; displacement = 0 m.
Why it works: Distance accumulates path length; displacement compares endpoints.
ACTIVITY 3
Recall Activity 3
Interpret a negative slope.
What does a negative slope on an x–t graph mean?
Reveal Answers
Velocity is negative.
Why it works: Position decreases as time increases, so motion is in the negative coordinate 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
Position and displacement are algebraic quantities
Position x is a coordinate relative to a chosen origin. Displacement is the signed difference between final and initial position. A multi-leg path can have large distance while its displacement is small or zero.
Δx = x₂ − x₁
Example: x₁ = −3 m and x₂ = +7 m gives Δx = +10 m.
Sensei note: Never erase coordinate signs before evaluating final minus initial.
KEY CONCEPT 2
Average velocity is net rate of position change
Average velocity over a finite interval divides displacement by elapsed time. On an x–t graph, it is the secant slope through the two interval endpoints.
v̄ = (x₂ − x₁)/(t₂ − t₁)
Example: From (2 s, 5 m) to (8 s, −7 m), v̄ = −2 m/s.
Sensei note: Average velocity can be zero even when the object moves throughout the interval.
KEY CONCEPT 3
Instantaneous velocity is local x–t slope
When an x–t graph is curved, velocity can change during the interval. Instantaneous velocity is represented by the slope of the tangent line at a particular time; for a straight segment, instantaneous and average velocity are the same.
instantaneous velocity = tangent slope of the x–t graph
Example: If the tangent rises 6 m for a 2 s run, the instantaneous velocity represented by that tangent is +3 m/s.
Sensei note: Do not calculate a long-interval secant slope when the question asks for velocity at one instant.
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
Compute interval quantities.
At t₁ = 1 s a particle is at x₁ = +4 m; at t₂ = 5 s it is at x₂ = −8 m. Find Δx, Δt, and v̄.
Reveal Answers
Δx = −12 m; Δt = 4 s; v̄ = −3 m/s.
Why it works: Use final minus initial for both x and t, then divide.
PRACTICE 2
Guided Problem
Separate distance and displacement.
A cart moves 10 m right, 4 m left, then 2 m right. It starts at x = −1 m. Find final position, distance, and displacement.
Reveal Answers
Final position = +7 m; distance = 16 m; displacement = +8 m.
Why it works: The signed moves sum to +8 m, while distance adds 10 + 4 + 2 = 16 m.
PRACTICE 3
Independent Problem
Analyze an x–t graph segment.
Position changes linearly from −6 m at 2 s to +14 m at 7 s. Find the velocity and predict the position 1 s later if the same velocity continues.
Reveal Answers
Velocity = +4 m/s; at 8 s, x = +18 m.
Why it works: The straight-line slope is 20 m/5 s = 4 m/s; one more second adds 4 m.
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
Signed average velocity
Calculate carefully.
A particle moves from +9 m at 3 s to −3 m at 9 s. Find v̄.
Reveal Answers
−2 m/s.
Why it works: Δx = −12 m and Δt = 6 s.
QUICK CHECK 2
Return trip
Compare distance and displacement.
A particle travels 20 m but finishes exactly where it started after 8 s. What is its average velocity?
Reveal Answers
0 m/s.
Why it works: Average velocity uses zero displacement, not the 20 m distance.
QUICK CHECK 3
Graph reading
Use the slope.
An x–t tangent at one instant has slope −5 m/s. What does that say about the instantaneous velocity?
Reveal Answers
It is −5 m/s.
Why it works: The tangent slope at that instant is the instantaneous velocity.
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
Displacement is final minus initial
Signed coordinates determine Δx; path length determines distance.
KEY TAKEAWAY 2
Average velocity is a secant slope
Use v̄ = Δx/Δt for a finite time interval.
KEY TAKEAWAY 3
Instantaneous velocity is a tangent slope
On an x–t graph, velocity at one instant is represented by the local slope.
Ready for your next step?
You've reviewed the essential ideas one last time. Now choose the resource that best matches how confident you feel. Need a quick reminder?
Next Step
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