FULL REVIEW
Full Review: Motion in Two Dimensions
Review the essential ideas, relationships, graphs, and algebra-based problem-solving tools for motion in two dimensions.
TIME
60 minutes
BEST FOR
A complete unit review
FINISH WITH
A readiness check
After this full review, you'll be able to...
interpret two-dimensional motion, apply algebra-based kinematics to representative problems, and determine what to study next.
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-U03 — Motion in Two Dimensions
SCOPE: Unit Review
PHYSICS LEVEL: Algebra-Based
BEST USED
✓ Before homework on two-dimensional kinematics
✓ Before a quiz or exam
✓ When components, launch angles, or relative velocity feel uncertain
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 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
Use vector displacement to calculate average velocity.
A drone moves 12 m east and 5.0 m north in 4.0 s. Find the magnitude and direction of its average velocity.
Reveal Answers
3.25 m/s at 22.6° north of east.
Why it works: v⃗avg=⟨12,5⟩/4=⟨3.0,1.25⟩ m/s. Its magnitude is 3.25 m/s and its direction is 22.6° north of east.
ACTIVITY 2
Recall Activity
Subtract velocity vectors component by component.
Velocity changes from ⟨8.0,2.0⟩ m/s to ⟨2.0,8.0⟩ m/s in 3.0 s. Find average acceleration.
Reveal Answers
⟨−2.0,2.0⟩ m/s².
Why it works: a⃗avg=(⟨2,8⟩−⟨8,2⟩)/3=⟨−2,2⟩ m/s².
ACTIVITY 3
Recall Activity
Use independent horizontal and vertical equations.
A ball launches horizontally at 10.0 m/s from a 19.6 m ledge. Find flight time and range.
Reveal Answers
2.00 s; 20.0 m.
Why it works: Vertical motion gives 19.6=½gt², so t=2.00 s. Horizontal motion gives x=(10.0)(2.00)=20.0 m.
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
Vectors, Components, Velocity, and Acceleration
Choose x and y axes first. Resolve vectors into components, analyze components independently, then recombine them for magnitude and direction.
Δr⃗=⟨Δx,Δy⟩ and v⃗avg=Δr⃗/Δt. A vector ⟨6,8⟩ has magnitude 10 and direction 53.1° above +x.
Sensei Note: Component signs describe axis directions; vector magnitudes are nonnegative.
KEY CONCEPT 2
Constant Acceleration Applies Component by Component
Apply constant-acceleration equations separately to x and y. One shared time connects the component motions.
x=x₀+v₀ₓt+½aₓt² and y=y₀+v₀ᵧt+½aᵧt². For projectiles, aₓ=0 and aᵧ=−g.
Sensei Note: One shared time connects both component equations.
KEY CONCEPT 3
Projectile Motion and Relative Velocity Use Vector Rules
In ideal projectile motion, horizontal velocity is constant while vertical velocity changes under gravity. Relative velocities add as vectors.
v₀ₓ=v₀cosθ, v₀ᵧ=v₀sinθ, and vᵧ=v₀ᵧ−gt. For relative motion, v⃗shore=v⃗boat/water+v⃗water/shore.
Sensei Note: At the top, only vᵧ is zero; vₓ and downward acceleration remain.
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
Resolve the launch velocity, then use the shared flight time.
A projectile launches at 20.0 m/s, 30.0° above horizontal from level ground. Find flight time, maximum height, and range.
Reveal Answers
Time 2.04 s; height 5.10 m; range 35.3 m.
Why it works: v₀ₓ=17.3 m/s and v₀ᵧ=10.0 m/s. Then T=2v₀ᵧ/g=2.04 s, H=v₀ᵧ²/(2g)=5.10 m, and R=v₀ₓT=35.3 m.
PRACTICE 2
Guided Problem
Add perpendicular relative-velocity vectors.
A boat moves north at 4.0 m/s relative to water while the river flows east at 3.0 m/s. Find velocity relative to shore.
Reveal Answers
5.0 m/s at 53.1° north of east.
Why it works: v⃗shore=⟨3,4⟩ m/s. Its magnitude is 5.0 m/s and its direction is tan⁻¹(4/3)=53.1° north of east.
PRACTICE 3
Independent Problem
Use vertical motion for time, then horizontal motion for range.
A ball launches horizontally at 15.0 m/s from a 44.1 m cliff. Find impact time, range, and impact speed.
Reveal Answers
3.00 s; 45.0 m; 33.0 m/s.
Why it works: 44.1=½gt² gives t=3.00 s and x=45.0 m. Impact velocity is ⟨15.0,−29.4⟩ m/s, with speed 33.0 m/s.
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
Identify the Constant Component
Use the acceleration components.
Which velocity component stays constant in ideal projectile motion, and why?
Reveal Answers
vₓ; because aₓ=0.
Why it works: Zero horizontal acceleration means the horizontal velocity component is constant.
QUICK CHECK 2
Recombine Components
Use magnitude and direction.
A vector has components ⟨6.0,8.0⟩ m. Find magnitude and direction above +x.
Reveal Answers
10.0 m at 53.1°.
Why it works: Magnitude is √(6²+8²)=10.0 m; direction is tan⁻¹(8/6)=53.1°.
QUICK CHECK 3
Resolve a Launch Vector
Use sine and cosine with the launch angle.
A projectile launches at 14.0 m/s at 45°. Find its initial velocity components.
Reveal Answers
v₀ₓ=v₀ᵧ=9.90 m/s.
Why it works: v₀cos45°=v₀sin45°=9.90 m/s.
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 Both Axes Before the Algebra
Resolve all vectors in one coordinate system before substituting numbers.
KEY TAKEAWAY 2
Match Each Component Equation to the Model
Use one time variable and the correct acceleration component in each axis.
KEY TAKEAWAY 3
Use Trajectories and Vector Checks
Check component signs, vector magnitude, direction, units, and trajectory shape.
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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