FULL REVIEW
Full Review — Acceleration and Constant-Acceleration Motion — Algebra-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: Algebra-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
Recall the equation before doing arithmetic.
Which constant-acceleration equation gives final velocity from v₀, a, and t?
Reveal Answers
v = v₀ + at.
Why it works: This is the direct velocity–time equation for constant acceleration.
ACTIVITY 2
Recall Activity 2
Identify the incorrect assumption.
Why is it wrong to make acceleration positive just because an object is speeding up?
Reveal Answers
The sign of acceleration gives its direction, not whether speed grows. Speed increases when v and a have the same sign.
Why it works: Speed depends on the magnitude of velocity; direction information is carried by signs.
ACTIVITY 3
Recall Activity 3
Choose the equation that avoids an unnecessary variable.
You know v₀, v, a, and need displacement. Which kinematic equation is most direct?
Reveal Answers
v² = v₀² + 2aΔx.
Why it works: That equation includes v₀, v, a, and Δx but not t.
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
Velocity–Time Relationship
For constant acceleration, velocity changes linearly with time. The slope of a velocity–time graph equals acceleration, so the sign of the slope gives the sign of a.
v = v₀ + at
Example: From v₀ = +3 m/s with a = +4 m/s² for 2 s, v = +11 m/s.
Sensei note: A negative acceleration can increase speed when velocity is also negative.
KEY CONCEPT 2
Displacement under Constant Acceleration
Displacement accumulates while velocity changes. The constant-acceleration displacement equation combines the initial velocity contribution with the additional displacement caused by acceleration.
Δx = v₀t + ½at²
Example: From rest with a = +2 m/s² for 3 s, Δx = 9 m.
Sensei note: The ½ factor belongs to the acceleration contribution to displacement.
KEY CONCEPT 3
Time-Free Kinematic Relation
When time is not known or not needed, eliminate it to relate velocities, acceleration, and displacement directly. This is especially useful for stopping-distance and launch problems.
v² = v₀² + 2aΔx
Example: A car slows from 18 m/s to rest with a = −3 m/s²; the equation gives a stopping displacement of +54 m.
Sensei note: Squaring velocity removes its sign, so determine the physically appropriate velocity direction separately when needed.
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
Use the velocity–time equation first, then displacement.
A bike has v₀ = +6 m/s and a = +1.5 m/s² for 4 s. Find v and Δx.
Reveal Answers
v = 6 + (1.5)(4) = +12 m/s; Δx = (6)(4) + ½(1.5)(4²) = 36 m.
Why it works: Both equations use the same signed acceleration and elapsed time.
PRACTICE 2
Guided Problem
Use the time-free equation and keep the acceleration signed.
A car slows from +20 m/s to +8 m/s over 42 m. Find the constant acceleration.
Reveal Answers
Use v² = v₀² + 2aΔx: 64 = 400 + 84a, so a = −4.0 m/s².
Why it works: The time-free kinematic equation directly connects velocity change, acceleration, and displacement.
PRACTICE 3
Independent Problem
Treat vertical motion with upward positive.
A ball is thrown upward from a balcony at +12 m/s. Using a = −9.8 m/s², find its displacement when its velocity first reaches 0.
Reveal Answers
Use 0 = 12² + 2(−9.8)Δy, giving Δy ≈ +7.35 m.
Why it works: At the top, v = 0 while acceleration remains downward; the time-free relation directly gives the rise.
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
Position after Acceleration
Use the displacement equation.
An object starts at +2 m/s and accelerates at +3 m/s² for 2 s. What displacement occurs?
Reveal Answers
Δx = (2)(2) + ½(3)(2²) = 10 m.
Why it works: Initial motion contributes 4 m and acceleration contributes 6 m.
QUICK CHECK 2
Free-Fall Sign Convention
Take upward as positive.
A ball is thrown upward at +14 m/s. Using a = −9.8 m/s², what is its velocity after 1.0 s?
Reveal Answers
v = 14 − 9.8(1.0) = +4.2 m/s.
Why it works: The ball is still moving upward because its velocity remains positive after 1.0 s.
QUICK CHECK 3
Stopping Distance
Use v² = v₀² + 2aΔx.
A car at +15 m/s brakes with a = −5 m/s². What stopping displacement is required?
Reveal Answers
0 = 225 − 10Δx, so Δx = 22.5 m.
Why it works: The time-free equation is direct because initial/final velocities and acceleration are known.
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
Choose an Equation Strategically
For constant acceleration, choose the kinematic equation containing your knowns and desired unknown while minimizing extra variables.
KEY TAKEAWAY 2
Signs Carry Direction
Define the positive direction once. Keep the signs of displacement, velocity, and acceleration consistent with that axis.
KEY TAKEAWAY 3
Match the Equation to the Known Quantities
The constant-acceleration equations are equivalent descriptions. Choose the one that uses your knowns and desired unknown without introducing an unnecessary variable.
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