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Module 2: Motion in Two and Three Dimensions

 

PHYS-2325 M2L2a Kinematics in Two and Three Dimensions

 DRAFT — Pending CME Approval

This lesson page is a proposed draft created for structural review. The original M2L2 content has been split into two lessons (M2L2a and M2L2b) to allow appropriate depth for each topic area. This split is pending approval from the Course Matter Expert. Learning objectives, video titles, and final content may be revised before publication.

"Nature uses only the longest threads to weave her patterns, so each small piece of her fabric reveals the organization of the entire tapestry."
— Richard Feynman


Motion in the real world is rarely confined to a straight line. A thrown ball, an orbiting satellite, a car rounding a curve — all move in two or three dimensions simultaneously. In this lesson, you will extend the kinematic concepts you mastered in Module 1 (position, displacement, velocity, and acceleration) into 2D and 3D space using vector notation. You will express position as r(t) = x(t)î + y(t)ĵ + z(t)k̂, find velocity as the time derivative of position, v = dr/dt, and find acceleration as the derivative of velocity, a = dv/dt. These vector relationships form the mathematical backbone for everything that follows — projectile motion, circular motion, and all of Newton's Laws applied in multiple dimensions.

A 3D coordinate system showing a curved path with position vector r, velocity vector v tangent to the path, and acceleration vector a pointing toward the center of curvature.
In 2D and 3D motion, position, velocity, and acceleration are all vectors that change with time.

Course Competencies and Learning Objectives

A ★ indicates that this page contains content related to that LO.

  DRAFT: Learning objectives below are placeholders pending CME confirmation.

CC2.1 Analyze the components of motion and any applied forces

★ LO2.2.1 Express the position, velocity, and acceleration of a particle as vector functions of time using unit-vector notation

★ LO2.2.2 Calculate average and instantaneous velocity vectors from a given position function in 2D or 3D

★ LO2.2.3 Calculate average and instantaneous acceleration vectors from a given velocity function in 2D or 3D

LO2.2.4 Apply kinematic equations independently to each coordinate direction in multi-dimensional motion

Required Reading

Click the blue buttons to go to the OpenStax reading assignments. Complete all three sections before watching the videos.

4.1 Displacement and Velocity Vectors 4.2 Acceleration Vector 4.3 Projectile Motion (preview only)

 

Optional Reading

Explore More

Want additional perspectives on 2D and 3D kinematics? These free resources complement the OpenStax text and may help solidify the vector calculus approach.

Media

Watch each video in order. The pre-lecture videos connect the reading to key concepts; the mini-lectures work through the unit-vector notation in detail.

Video 1: Pre-Lecture — Displacement & Position Vectors

Pre-Lecture: Displacement and Velocity Vectors (OpenStax 4.1)

This pre-lecture video introduces the concept of position as a vector function of time and shows how displacement in 2D and 3D is calculated as the change in the position vector. The average velocity vector is derived directly from this displacement.

  • Position vector r(t) = x(t)î + y(t)ĵ + z(t)k̂
  • Displacement: Δr = r2 − r1 = Δxî + Δyĵ + Δzk̂
  • Average velocity: vavg = Δr / Δt
  • Path length vs. magnitude of displacement in 2D motion

Time: Video pending — to be recorded by instructor

Video 2: Pre-Lecture — Instantaneous Velocity Vector

Pre-Lecture: Instantaneous Velocity in 2D and 3D

Building on Section 4.1, this video defines instantaneous velocity as the derivative of the position vector with respect to time and shows how each component of velocity is independently derived from the corresponding component of position.

  • Instantaneous velocity: v(t) = dr/dt = (dx/dt)î + (dy/dt)ĵ + (dz/dt)k̂
  • The velocity vector is always tangent to the path of motion
  • Speed: the magnitude of the velocity vector, |v| = v
  • How direction of v changes even when speed is constant

Time: Video pending — to be recorded by instructor

Video 3: Pre-Lecture — Acceleration Vector

Pre-Lecture: Acceleration Vector in 2D and 3D (OpenStax 4.2)

This video extends the concept of acceleration to 2D and 3D using vector calculus. You will see how acceleration arises from both changes in speed (tangential) and changes in direction (centripetal), even when the speed is constant.

  • Average acceleration: aavg = Δv / Δt
  • Instantaneous acceleration: a(t) = dv/dt = d2r/dt2
  • Component form: a = (dvx/dt)î + (dvy/dt)ĵ + (dvz/dt)k̂
  • Acceleration can be perpendicular to velocity even when speed is constant

Time: Video pending — to be recorded by instructor

Video 4: Mini-Lecture — Position & Velocity in Unit-Vector Notation

Mini-Lecture: Writing and Differentiating Position Vectors

This worked-example video shows how to write a position function in unit-vector notation and then differentiate each component to obtain the velocity vector, step by step. Both polynomial and trigonometric position functions are demonstrated.

  • Given r(t) = (3t2 − 2t)î + (4t − 1)ĵ m, find v(t)
  • Differentiate each component independently: d/dt[x(t)], d/dt[y(t)]
  • Evaluate velocity at a specific time: v(t = 2 s) = ?
  • Find the speed and direction angle of v at that instant

Time: Video pending — to be recorded by instructor

Video 5: Mini-Lecture — Acceleration from Velocity

Mini-Lecture: Finding Acceleration by Differentiating Velocity

This video works through finding the acceleration vector from a given velocity function, evaluating it at specific times, and interpreting the physical meaning of both the magnitude and direction of the resulting acceleration vector.

  • Given v(t), differentiate to find a(t) = dv/dt
  • Distinguishing the x- and y-components of acceleration
  • Average acceleration over a time interval: Δv / Δt
  • Relating the direction of a to changes in the velocity vector

Time: Video pending — to be recorded by instructor

Video 6: Mini-Lecture — Displacement & Average Velocity

Mini-Lecture: Displacement Vector and Average Velocity in Unit-Vector Form

This video demonstrates computing displacement and average velocity directly from position vectors expressed in unit-vector notation, including cases where both x and y components are changing non-uniformly with time.

  • Displacement: Δr = r(t2) − r(t1) — subtract component by component
  • Average velocity: vavg = Δr / (t2 − t1)
  • Why average velocity direction ≠ direction of motion at every instant
  • Connecting to 1D results: each axis is an independent 1D problem

Time: Video pending — to be recorded by instructor

Practice and Apply - Conceptual

Finding Instantaneous Velocity from a Position Function

Arrange the following steps in the correct order for finding the instantaneous velocity vector from a position function given in unit-vector notation.

Order Items

Arrange the following items in the correct order.

  1. Write the position function r(t) in unit-vector notation: r(t) = x(t)î + y(t)ĵ.
  2. Differentiate each scalar component independently with respect to time.
  3. Assemble the velocity vector: v(t) = (dx/dt)î + (dy/dt)ĵ.
  4. Substitute the desired time value t to find v at that instant.
  5. Compute the speed: v = |v| = √(vx2 + vy2).
  6. Find the direction angle: θ = tan−1(vy / vx), adjusting for quadrant.

Which Statements About Velocity in 2D Are True?

From the list below, select all statements that are correct about velocity and speed in 2D motion.

Select All

Select all correct statements about velocity in two-dimensional motion.

  • The instantaneous velocity vector is always tangent to the path of motion.
  • Speed is the magnitude of the velocity vector: v = |v|.
  • Each component of velocity is found by differentiating the corresponding position component.
  • A particle can have a non-zero acceleration while moving at constant speed.
  • The direction of the average velocity vector is always the same as the instantaneous velocity.
  • If the speed is constant, the acceleration must be zero.
  • The x- and y-components of velocity are always equal to each other.

Key Concept Check

Click each card to flip it and test your understanding of kinematics in 2D and 3D.

What is the relationship between position and velocity in vector form?
Answer

Instantaneous velocity is the time derivative of the position vector: v(t) = dr/dt. In component form: vx = dx/dt and vy = dy/dt. Each axis is treated as an independent 1D problem.

 

Can a particle have a non-zero acceleration while moving at constant speed? Explain.
Answer

Yes. Acceleration is the rate of change of the velocity vector, not just its magnitude. If the direction of v is changing while its magnitude (speed) stays constant, the acceleration is non-zero but perpendicular to the velocity. Uniform circular motion is the classic example.

 

How is the average velocity vector different from instantaneous velocity?
Answer

Average velocity is vavg = Δr/Δt — it points in the direction of the displacement vector over the whole interval. Instantaneous velocity is the limit as Δt → 0: v = dr/dt. It points tangent to the path at a single instant.

 

What is the relationship between velocity and acceleration in vector form?
Answer

Instantaneous acceleration is the time derivative of the velocity vector: a(t) = dv/dt = d2r/dt2. In component form: ax = dvx/dt and ay = dvy/dt.

 

Why is the velocity vector always tangent to the trajectory?
Answer

The velocity vector is defined as the limit of Δr/Δt as Δt → 0. As the time interval shrinks, the displacement vector Δr becomes parallel to the curve at that point. Therefore v, which is in the direction of the infinitesimal displacement, points along the tangent to the path.

 

If a particle's acceleration is only in the y-direction, how does its x-motion behave?
Answer

The x-motion is uniform (constant velocity). Since ax = 0, vx = constant and x(t) = x0 + vxt. The y-motion is accelerated independently: ay ≠ 0. This is the key insight behind projectile motion — the horizontal motion is independent of the vertical.

 

Practice and Apply - Computational

Important: Working in Component Form

  • Treat each axis independently. The x- and y-equations are separate 1D problems.
  • Velocity from position: differentiate each component — vx = dx/dt, vy = dy/dt.
  • Acceleration from velocity: differentiate again — ax = dvx/dt, ay = dvy/dt.
  • Magnitude: always √(componentx2 + componenty2 + componentz2). Include SI units.
  • Direction angle: θ = tan−1(y-component / x-component). Verify the quadrant.

Practice Problem 1

A particle's position is given by r(t) = (2.0t2 − 1.0)î + (3.0t − 2.0t2)ĵ meters, where t is in seconds. (a) Find the velocity vector v(t). (b) Find the velocity and speed at t = 2.0 s. (c) Find the acceleration vector a(t). Is the acceleration constant?

Show Solution

Given: r(t) = (2.0t2 − 1.0)î + (3.0t − 2.0t2)ĵ m

Part (a) — Velocity vector:

v(t) = dr/dt = (d/dt)[2.0t2 − 1.0]î + (d/dt)[3.0t − 2.0t2]ĵ

v(t) = 4.0t î + (3.0 − 4.0t) ĵ m/s

Part (b) — Velocity at t = 2.0 s:

v(2.0) = 4.0(2.0)î + (3.0 − 4.0 × 2.0)ĵ = 8.0î + (3.0 − 8.0)ĵ = 8.0î − 5.0ĵ m/s

Speed: v = √(8.02 + (−5.0)2) = √(64 + 25) = √89 = 9.43 m/s

Part (c) — Acceleration vector:

a(t) = dv/dt = (d/dt)[4.0t]î + (d/dt)[3.0 − 4.0t]ĵ

a(t) = 4.0î − 4.0ĵ m/s2

Yes, the acceleration is constant (no t dependence). This is analogous to constant acceleration in 1D, but in 2D.

Practice Problem 2

A particle's position changes from r1 = (1.0î − 2.0ĵ + 3.0k̂) m at t1 = 1.0 s to r2 = (5.0î + 1.0ĵ + 0.0k̂) m at t2 = 3.0 s. (a) Find the displacement vector Δr. (b) Find the average velocity vector. (c) Find the magnitude of the average velocity.

Show Solution

Given: r1 = (1.0î − 2.0ĵ + 3.0k̂) m; r2 = (5.0î + 1.0ĵ + 0.0k̂) m; Δt = 2.0 s

Part (a) — Displacement:

Δr = r2 − r1 = (5.0 − 1.0)î + (1.0 − (−2.0))ĵ + (0.0 − 3.0)k̂

Δr = 4.0î + 3.0ĵ − 3.0k̂ m

Part (b) — Average velocity:

vavg = Δr / Δt = (4.0î + 3.0ĵ − 3.0k̂) / 2.0 = 2.0î + 1.5ĵ − 1.5k̂ m/s

Part (c) — Magnitude of average velocity:

|vavg| = √(2.02 + 1.52 + (−1.5)2) = √(4.0 + 2.25 + 2.25) = √8.5 = 2.92 m/s

Practice Problem 3

A particle's velocity at t = 0 s is v0 = (3.0î + 4.0ĵ) m/s, and its constant acceleration is a = (1.0î − 2.0ĵ) m/s2. (a) Write the velocity vector as a function of time. (b) Write the position vector as a function of time, given r0 = 0. (c) At what time does the y-component of the velocity equal zero?

Show Solution

Given: v0 = (3.0î + 4.0ĵ) m/s; a = (1.0î − 2.0ĵ) m/s2; r0 = 0

Part (a) — Velocity as a function of time (constant acceleration):

v(t) = v0 + at = (3.0 + 1.0t)î + (4.0 − 2.0t)ĵ m/s

Part (b) — Position as a function of time:

r(t) = r0 + v0t + ½at2

x(t) = 0 + 3.0t + ½(1.0)t2 = 3.0t + 0.5t2

y(t) = 0 + 4.0t + ½(−2.0)t2 = 4.0t − 1.0t2

r(t) = (3.0t + 0.5t2)î + (4.0t − t2)ĵ m

Part (c) — When does vy = 0?

vy(t) = 4.0 − 2.0t = 0 → t = 4.0/2.0 = 2.0 s

At t = 2.0 s the particle is moving entirely in the x-direction. The y-component of motion has momentarily stopped (this is the apex if thinking of projectile-like motion).

Ready to Move On?

Lesson Checklist

Before moving on to M2L2b, confirm you can do each of the following:

  • ☐ Write a position vector in unit-vector notation as a function of time: r(t) = x(t)î + y(t)ĵ + z(t)k̂.
  • ☐ Find the instantaneous velocity vector by differentiating each component of r(t).
  • ☐ Find the instantaneous acceleration vector by differentiating each component of v(t).
  • ☐ Compute the displacement vector Δr and average velocity vavg between two times.
  • ☐ Explain why a particle moving at constant speed can still have a non-zero acceleration.
  • ☐ Apply r(t) = r0 + v0t + ½at2 in 2D by treating x and y as independent equations.

These skills are the mathematical foundation for projectile motion, circular motion, and all of Newton's Laws in 2D and 3D. Take the time to make them automatic!