FULL REVIEW
Full Review — Acceleration and Constant-Acceleration Motion — Calculus-Based
Review the essential ideas, relationships, and problem-solving tools for Acceleration and Constant-Acceleration Motion.
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-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 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
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
Recall Activity 2
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
Recall Activity 3
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
Let's rebuild the key ideas one step at a time. Focus on understanding the relationships before worrying about solving problems.
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.
KEY CONCEPT 3
Graphical and Integral Consistency
Derivative and integral views are complementary. The slope of v(t) is a(t), while the signed area under a(t) gives Δv. Likewise, the signed area under v(t) gives displacement.
Δv = ∫ₜ₁ᵗ² a(t) dt
Example: For constant a, the integral reduces to a(t₂ − t₁), reproducing the linear velocity relation.
Sensei note: Signed area matters: portions below the time axis contribute negatively.
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 the position function.
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
Guided Problem
Use initial conditions after integrating.
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.
PRACTICE 3
Independent Problem
Connect the graph area to velocity change.
Acceleration is +4 m/s² from t = 0 to 2 s, then −2 m/s² from t = 2 to 5 s. If v(0) = +1 m/s, find v(5 s).
Reveal Answers
Δv₁ = (+4)(2) = +8 m/s; Δv₂ = (−2)(3) = −6 m/s. Therefore v(5) = 1 + 8 − 6 = +3 m/s.
Why it works: Velocity change is the signed area under a(t), so piecewise-constant intervals add algebraically.
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.
QUICK CHECK 3
Piecewise Acceleration
Add signed acceleration–time areas.
Starting from v = +2 m/s, acceleration is +3 m/s² for 1 s and then −1 m/s² for 4 s. What is the final velocity?
Reveal Answers
v = 2 + (3)(1) + (−1)(4) = +1 m/s.
Why it works: Each interval contributes a signed Δv equal to its acceleration–time area.
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
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).
KEY TAKEAWAY 3
Slope and Area Must Agree
Use derivatives for local rates and integrals for accumulated changes. Constant acceleration is the simplest case where both views reduce to the familiar kinematic equations.
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