FULL REVIEW
Full Review: Motion in One Dimension
Review one-dimensional kinematics through derivatives, integrals, motion graphs, constant acceleration, and variable-acceleration models.
TIME
45–60 minutes
BEST FOR
A complete unit review
FINISH WITH
A readiness check
After this full review, you'll be able to...
move fluently between x(t), v(t), and a(t), use definite integrals to reconstruct motion, and recognize constant-acceleration formulas as a special case.
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: Calculus-Based
BEST USED
✓ Before calculus-based kinematics homework
✓ Before a quiz or exam
✓ When derivative, integral, or variable-acceleration models need reinforcement
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
Differentiate position once for velocity and twice for acceleration.
If x(t) = 2t³ − 5t² + 4, write v(t) and a(t).
Reveal Answers
v(t) = 6t² − 10t; a(t) = 12t − 10.
Why it works: Instantaneous velocity is dx/dt and acceleration is d²x/dt².
ACTIVITY 2
Recall Activity
Use a definite integral to accumulate signed displacement.
If v(t) is known, what definite integral gives the displacement from t₁ to t₂?
Reveal Answers
Δx = ∫[t₁,t₂] v(t) dt.
Why it works: Integrating the instantaneous velocity over time accumulates the signed changes in position.
ACTIVITY 3
Recall Activity
Use the derivative condition, then check whether the sign changes.
For a differentiable x(t), what condition identifies a possible turning point?
Reveal Answers
v(t) = dx/dt = 0, followed by a sign change in v for a true reversal.
Why it works: A zero derivative marks an instantaneous horizontal tangent, but the direction reverses only if velocity changes sign.
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
Instantaneous velocity is the first derivative of position. Acceleration is the derivative of velocity and the second derivative of position.
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
Integrating velocity gives displacement; integrating acceleration gives velocity change. Definite integrals naturally handle motion whose rate changes continuously.
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 familiar algebraic kinematics equations follow by integrating a constant acceleration. When acceleration varies, use the underlying derivative/integral relations instead of constant-a shortcuts.
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
Differentiate a position function and solve v(t)=0 to identify rest times.
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
Integrate acceleration, use the initial condition, then integrate velocity.
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
Use the definite integral of velocity and check whether v changes sign.
For 0≤t≤4 s, v(t)=5−t m/s. Find the displacement and total distance.
Reveal Answers
Displacement = 12 m; total distance = 12 m.
Why it works: ∫₀⁴(5−t)dt = 12 m. Because v(t)>0 throughout 0≤t≤4, speed equals velocity and distance equals displacement.
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
Find the instantaneous velocity from x(t).
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 initial velocity as the constant of integration.
If a(t)=2t and v(0)=−1 m/s, find v(3 s).
Reveal Answers
8 m/s.
Why it works: v(3)=−1+∫₀³2t dt=−1+9=8 m/s.
QUICK CHECK 3
Apply the Free-Fall Sign Convention
Recognize the assumption hidden in a shortcut equation.
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
Local slopes are calculus statements: differentiate position for velocity and velocity for acceleration.
KEY TAKEAWAY 2
Match the Equation to the Model
Signed areas under v(t) and a(t) accumulate displacement and velocity change.
KEY TAKEAWAY 3
Use Graphs and Physical Checks
Constant acceleration produces the familiar shortcut equations; variable acceleration requires the underlying calculus model.
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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