FULL REVIEW
Full Review: Motion in One Dimension
Review straight-line motion through physical meaning, direction, signs, motion graphs, and qualitative reasoning before relying on heavy calculation.
TIME
45–60 minutes
BEST FOR
A complete unit review
FINISH WITH
A readiness check
After this full review, you'll be able to...
explain what position, velocity, acceleration, and motion graphs mean physically and use those ideas to predict one-dimensional motion.
Choose how you want to review
Unit Alignment
This bundle is aligned to the approved Physics Sensei unit specification below. Use it to recover the unit structure, reinforce key decisions, and confirm readiness for the next study task.
ARCHITECTURE: Physics Sensei Independent Mechanics
UNIT: MEC-U02 — Motion in One Dimension
SCOPE: Unit Review
PHYSICS LEVEL: Foundational
BEST USED
✓ Before algebra-heavy kinematics homework
✓ When graph meaning or signs feel confusing
✓ When you want to understand the physics before calculating
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
Separate path length from net change in position.
A student walks from x = 2 m to x = 8 m, then back to x = 5 m. What are the displacement and the total distance traveled?
Reveal Answers
+3 m displacement; 9 m distance.
Why it works: Displacement uses final minus initial position: 5 − 2 = +3 m. Distance adds the path lengths: 6 m + 3 m = 9 m.
ACTIVITY 2
Recall Activity
Compare the signs of velocity and acceleration.
A car is moving to the right but slowing down. If +x points right, what are the signs of velocity and acceleration?
Reveal Answers
v > 0 and a < 0.
Why it works: The car moves in +x, so velocity is positive. Because its positive velocity is decreasing, acceleration points in the opposite direction.
ACTIVITY 3
Recall Activity
Translate graph slope into a physical statement.
What does a horizontal tangent on a position–time graph mean at that instant?
Reveal Answers
The instantaneous velocity is zero.
Why it works: Velocity is represented by the slope of the position–time graph. A horizontal tangent has zero slope.
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, Displacement, Velocity, and Signs
Position is measured relative to a chosen origin. Displacement is the signed change from initial to final position. Distance is the total path length and is never negative.
Δx = xf − xi; vavg = Δx/Δt.
Example: From x = +3 m to x = −9 m in 4 s, Δx = −12 m and vavg = −3 m/s.
Sensei Note: A negative velocity means motion in the negative coordinate direction. It does not automatically mean the object is slowing down.
KEY CONCEPT 2
Constant Acceleration Gives a Compact Kinematics Toolkit
Velocity tells how position changes and includes direction. Acceleration tells how velocity changes. An object speeds up when velocity and acceleration have the same sign and slows down when they have opposite signs.
v = v0 + at; Δx = v0t + ½at2; v2 = v02 + 2aΔx; Δx = ½(v0 + v)t.
Example: A car with v0 = 5 m/s and a = 2 m/s2 for 4 s reaches v = 13 m/s and moves Δx = 36 m.
Sensei Note: These shortcut equations require constant acceleration over the interval. If the motion changes stages, reset the initial conditions for each stage.
KEY CONCEPT 3
Motion Graphs and Vertical Free Fall Tell the Same Story
The slope of a position–time graph tells velocity. The slope of a velocity–time graph tells acceleration. The signed area under a velocity–time graph tells displacement.
slope(x–t) → v; slope(v–t) → a; area under v–t → Δx. If +y is upward, ay = −g.
Example: A ball thrown straight upward has v = 0 for an instant at the top, but its acceleration remains −9.80 m/s2 when +y is upward.
Sensei Note: Zero velocity at an instant does not imply zero acceleration. Keep the chosen vertical sign convention from start to finish.
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
Read a motion story in pieces and combine the signed displacements.
A car travels at 12.0 m/s for 5.0 s, then accelerates uniformly at 3.00 m/s2 for 4.00 s. Find the total displacement and final speed.
Reveal Answers
Total displacement = 132 m; final speed = 24.0 m/s.
Why it works: First interval: Δx1 = (12.0)(5.0) = 60.0 m. Second interval: Δx2 = (12.0)(4.0) + ½(3.00)(4.002) = 72.0 m. Total = 132 m. Final speed is v = 12.0 + (3.00)(4.00) = 24.0 m/s.
PRACTICE 2
Guided Problem
Use the turning point to distinguish zero velocity from zero acceleration.
A bicycle moving at 10.0 m/s brakes with constant acceleration −2.50 m/s2. Find the stopping time and stopping distance.
Reveal Answers
Stopping time = 4.00 s; stopping distance = 20.0 m.
Why it works: Use 0 = 10.0 − 2.50t to obtain t = 4.00 s. Then 0 = (10.0)2 + 2(−2.50)Δx gives Δx = 20.0 m.
PRACTICE 3
Independent Problem
Connect constant acceleration to the shape of a velocity–time graph.
A cyclist moves at 4 m/s and speeds up uniformly at 2 m/s² for 3 s. What is the final velocity, and what does the v–t graph look like?
Reveal Answers
Final velocity = 10 m/s; the v–t graph is a straight rising line.
Why it works: Constant positive acceleration means velocity increases by the same amount each second: 4 → 6 → 8 → 10 m/s.
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
Choose the Equation That Removes Time
Compare the signs of v and a.
An object has v0 = 4.0 m/s, a = 3.0 m/s2, and Δx = 10 m. Which constant-acceleration equation finds the final velocity without first finding time?
Reveal Answers
v2 = v02 + 2aΔx.
Why it works: It contains the desired final velocity and the given v0, a, and Δx, but no time.
QUICK CHECK 2
Use Area on a Velocity–Time Graph
Use the sign of the slope and how the slope changes.
An x–t graph rises and becomes progressively steeper. What are the signs of velocity and acceleration?
Reveal Answers
v > 0 and a > 0.
Why it works: The positive slope gives positive velocity. Because the slope becomes more positive, velocity is increasing, so acceleration is positive.
QUICK CHECK 3
Apply the Free-Fall Sign Convention
Multiply the constant velocity by the elapsed time.
A rock is dropped from rest and falls for 2.0 s. With +y upward and g = 9.80 m/s2, find its velocity and displacement.
Reveal Answers
v = −19.6 m/s; Δy = −19.6 m.
Why it works: With v0 = 0 and a = −9.80 m/s2, v = at = −19.6 m/s and Δy = ½at2 = −19.6 m.
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
Define the Axis Before the Algebra
Define the positive axis first; then let the signs of position, velocity, and acceleration carry physical meaning.
KEY TAKEAWAY 2
Match the Equation to the Model
Velocity describes how position changes; acceleration describes how velocity changes.
KEY TAKEAWAY 3
Use Graphs and Physical Checks
Use slope and signed area to connect position, velocity, and acceleration.
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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