FOCUSED REVIEW
Focused Review — Acceleration and Constant-Acceleration Motion — Calculus-Based
Reinforce the highest-leverage ideas and representative problem-solving tools for Acceleration and Constant-Acceleration Motion.
TIME
Approximately 15 minutes
BEST FOR
Targeted reinforcement
FINISH WITH
A readiness check
After this focused review, you'll be able to...
reinforce the key relationships, apply them to representative problems, and identify what still needs work.
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-T02 | TOPIC: Acceleration and Constant-Acceleration Motion | PARENT UNIT: MEC-U03 — Motion in One Dimension | COURSE LEVEL: Calculus-Based
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 15 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
Key Ideas
Connect the graph to the derivative.
What does the slope of a v-versus-t graph represent at a given time?
Reveal Answers
Instantaneous acceleration.
Why it works: The derivative dv/dt is the slope of v(t).
ACTIVITY 2
Common Mistakes
Distinguish an instantaneous value from a rate.
Can v = 0 while a ≠ 0? Give a one-dimensional example.
Reveal Answers
Yes. At the top of a vertical throw, v = 0 momentarily while acceleration remains downward.
Why it works: Velocity can pass through zero while its derivative remains finite.
ACTIVITY 3
Quick Application
Use an integral interpretation.
What does the signed area under an a-versus-t graph represent?
Reveal Answers
The change in velocity, Δv.
Why it works: Integrating acceleration over time accumulates velocity change.
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
Reinforce the two highest-leverage relationships, then use them in representative situations.
KEY CONCEPT 1
Differential Definition of Acceleration
Instantaneous acceleration is the derivative of velocity. Equivalently, it is the second derivative of position, so curvature in x(t) is tied directly to acceleration.
a = dv/dt = d²x/dt²
Example: For x(t) = 2 + 5t − t², v(t) = 5 − 2t and a(t) = −2 m/s².
Sensei note: A zero slope in x(t) means v = 0, not necessarily a = 0.
KEY CONCEPT 2
Integrating Constant Acceleration
When a is constant, integrate acceleration over time to obtain velocity, then integrate velocity to obtain position. Initial conditions determine the integration constants.
v(t) = v₀ + at
Example: Integrating constant a from 0 to t gives v − v₀ = at.
Sensei note: Initial conditions are essential; integration does not determine them automatically.
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
Apply the reinforced ideas to two representative situations, then use the strategy card to check your setup.
PRACTICE 1
Guided Example
Differentiate first, then evaluate.
A particle has x(t) = 1 + 4t + 1.5t² in SI units. Find v(2 s) and a.
Reveal Answers
v(t) = 4 + 3t, so v(2) = 10 m/s; a = 3 m/s².
Why it works: Differentiation converts position to velocity and then to acceleration.
PRACTICE 2
Independent Check
Use the constant-acceleration result with signed quantities.
A particle has v₀ = −3 m/s and constant a = +2 m/s². Find the time when it reverses direction and the displacement up to that instant.
Reveal Answers
Set 0 = −3 + 2t, so t = 1.5 s. Then Δx = (−3)(1.5) + ½(2)(1.5²) = −2.25 m.
Why it works: The reversal occurs at v = 0; the signed displacement follows from the integrated constant-acceleration motion.
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
Second Derivative
Differentiate twice.
For x(t) = 6 − 2t + 4t², what are v(t) and a(t)?
Reveal Answers
v(t) = −2 + 8t and a(t) = +8 m/s².
Why it works: The first derivative gives velocity; the second derivative gives acceleration.
QUICK CHECK 2
Area under Acceleration
Use Δv = ∫ a dt for constant a.
If a = −5 m/s² for 0.60 s, what is the change in velocity?
Reveal Answers
Δv = (−5)(0.60) = −3.0 m/s.
Why it works: For constant acceleration, the acceleration–time area is simply aΔt.
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 essential ideas you’ll want to remember.
KEY TAKEAWAY 1
Derivative View
Acceleration is the local derivative dv/dt and also d²x/dt². Graph slopes encode these derivative relationships.
KEY TAKEAWAY 2
Integral View
Integrating acceleration gives velocity change; integrating velocity gives displacement. Constant acceleration yields linear v(t) and quadratic x(t).
Next Step
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