FULL REVIEW
Full Review — Position, Displacement, and Velocity — Calculus-Based
Review the essential ideas, relationships, and problem-solving tools for Position, Displacement, and Velocity.
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-T01 | TOPIC: Position, Displacement, and Velocity | PARENT UNIT: MEC-U03 — Motion in One Dimension | COURSE LEVEL: Calculus-Based introductory physics
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 displacement to a function.
Write displacement from t₁ to t₂ when position is x(t).
Reveal Answers
Δx = x(t₂) − x(t₁).
Why it works: Displacement is still final position minus initial position, even when position is given as a function.
ACTIVITY 2
Recall Activity 2
Differentiate a polynomial.
If x(t) = 2t³ − 5t, find v(t).
Reveal Answers
v(t) = 6t² − 5.
Why it works: Instantaneous velocity is the derivative of position with respect to time.
ACTIVITY 3
Recall Activity 3
Interpret derivative sign.
What does v(t) < 0 imply about x(t) at that instant?
Reveal Answers
x(t) is decreasing with time; the tangent slope is negative.
Why it works: The derivative sign gives the instantaneous direction along the coordinate axis.
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 functions and displacement
A differentiable position function x(t) specifies the coordinate of the particle at each time. Displacement over an interval depends only on the two endpoint values of x(t), not on the detailed path between them.
Δx = x(t₂) − x(t₁)
Example: For x(t) = t² − 4t, from 0 s to 5 s the displacement is x(5) − x(0) = 5 m.
Sensei note: Do not integrate |v| when the question asks only for displacement; endpoint positions are sufficient.
KEY CONCEPT 2
Average velocity is the secant slope
Average velocity over [t₁,t₂] is the ratio of the finite position change to the finite time change. Geometrically it is the slope of the secant line connecting the two points on the x–t curve.
v̄ = [x(t₂) − x(t₁)]/(t₂ − t₁)
Example: For x(t) = t², from 2 s to 5 s: v̄ = (25 − 4)/3 = 7 m/s.
Sensei note: For nonlinear x(t), average velocity generally differs from the instantaneous velocity at either endpoint.
KEY CONCEPT 3
Instantaneous velocity is the derivative
Instantaneous velocity is the limiting slope of the position curve as the time interval shrinks to zero. The derivative v(t) = dx/dt gives both local rate and signed direction. Zeros of v correspond to horizontal tangents of x(t).
v(t) = dx/dt
Example: If x(t) = t³ − 6t² + 9t, then v(t) = 3t² − 12t + 9.
Sensei note: A point where v = 0 is a candidate turning point; confirm a direction change by checking the sign of v on each side.
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
Compute a finite-interval velocity.
For x(t) = 2t² − 3t + 1, find displacement and average velocity from t = 1 s to t = 4 s.
Reveal Answers
x(1) = 0 m, x(4) = 21 m; Δx = 21 m; v̄ = 7 m/s.
Why it works: Evaluate position at both endpoints, subtract, then divide by the elapsed time.
PRACTICE 2
Guided Problem
Compare average and instantaneous velocity.
For x(t) = t³ − 3t, find v̄ from 0 s to 2 s and v at t = 2 s.
Reveal Answers
x(0) = 0 m, x(2) = 2 m, so v̄ = 1 m/s. v(t) = 3t² − 3, so v(2) = 9 m/s.
Why it works: The secant slope over the interval and tangent slope at the endpoint describe different rates.
PRACTICE 3
Independent Problem
Find and classify direction changes.
For x(t) = t³ − 3t², find the times when v = 0 and determine the direction of motion just before and after t = 2 s.
Reveal Answers
v(t) = 3t(t − 2). Around t = 2 s, v is negative just before and positive just after, so the particle reverses from negative to positive direction.
Why it works: A zero of v signals a horizontal tangent; the sign change confirms the reversal.
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
Endpoint displacement
Use x(t₂) − x(t₁).
For x(t) = 5 − 2t, find displacement from t = 1 s to t = 4 s.
Reveal Answers
x(1) = 3 m, x(4) = −3 m, so Δx = −6 m.
Why it works: Displacement depends on endpoint positions, not on the form of the function between them.
QUICK CHECK 2
Differentiate for velocity
Find the derivative.
For x(t) = 2t³ + t², find v at t = 1 s.
Reveal Answers
v(t) = 6t² + 2t, so v(1) = 8 m/s.
Why it works: Instantaneous velocity is the derivative evaluated at the requested time.
QUICK CHECK 3
Turning-point reasoning
Check the sign around the zero.
Suppose v(t) changes from positive to negative as t passes through t₀ where v(t₀) = 0. What happens to x(t)?
Reveal Answers
x(t) has a local maximum at t₀ and the particle reverses from positive to negative direction.
Why it works: The derivative changes from positive slope to negative slope across the horizontal tangent.
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
Displacement remains an endpoint difference
Even for x(t), use Δx = x(t₂) − x(t₁).
KEY TAKEAWAY 2
Average velocity uses finite change
The secant slope over an interval is v̄ = Δx/Δt.
KEY TAKEAWAY 3
Instantaneous velocity is local change
Use v(t) = dx/dt; its sign gives direction and its zeros locate horizontal tangents.
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?
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