AP Physics C: Mechanics • Kinematics

Motion in Two or Three Dimensions

Extending kinematics beyond straight lines to describe projectiles, circular paths, and arbitrary trajectories using vector calculus.

A motion gauge with symbols for mass and movement.
Inside this lesson 10 sections 5 questions 10 figures

SECTION 1

Historical Context & Motivation

For most of human history, the motion of objects through the air—arrows, catapulted stones, celestial bodies—remained deeply mysterious. Ancient Greek natural philosophers like Aristotle proposed that projectiles required a continuous "motive force" to stay aloft, an idea that persisted for nearly two millennia. It was not until the Renaissance that systematic observation and mathematical reasoning began to replace philosophical speculation, culminating in the realization that motion in multiple dimensions could be decomposed into independent, analyzable components. This principle of superposition—the idea that horizontal and vertical motions proceed independently—became one of the most powerful tools in all of classical mechanics.

1638 • Galileo's Two New Sciences

Galileo Galilei demonstrated that projectile trajectories are parabolic by decomposing motion into uniform horizontal displacement and uniformly accelerated vertical fall, overturning Aristotelian impetus theory.

1687 • Newton's Principia Mathematica

Isaac Newton unified terrestrial and celestial mechanics with his three laws of motion and the law of universal gravitation, providing the force-based framework that governs multidimensional kinematics.

1788 • Lagrange's Mécanique Analytique

Joseph-Louis Lagrange reformulated mechanics using generalized coordinates, enabling elegant treatment of constrained motion in arbitrary dimensions without requiring force diagrams.

1880s • Vector Notation Matures

Josiah Willard Gibbs and Oliver Heaviside independently developed modern vector algebra, giving physicists a compact and powerful language for expressing position, velocity, and acceleration in any number of dimensions.

The central question motivating this topic is deceptively simple: how do we precisely describe the path, speed, and acceleration of an object that is free to move in a plane or in three-dimensional space? The answer lies in treating the position of a particle as a vector-valued function of time, then differentiating to extract velocity and acceleration vectors whose components can be analyzed independently. Mastering this framework is essential not only for projectile motion and circular dynamics, but for every subsequent topic in mechanics.

SECTION 2

Core Principles & Definitions

Multidimensional kinematics rests on a handful of foundational ideas that extend naturally from the one-dimensional case. The key conceptual leap is that every kinematic quantity—position, displacement, velocity, acceleration—becomes a vector, and the relationships among them are governed by vector calculus rather than simple algebra. Each component of these vectors obeys the familiar one-dimensional equations of motion independently, a fact that allows us to break complex trajectories into manageable pieces.

1

Position Vector r⃗(t)

The position vector locates a particle relative to a chosen origin. In Cartesian coordinates: r⃗ = x(t) î + y(t) ĵ + z(t) k̂. Its time dependence encodes the complete trajectory.
2

Velocity Vector v⃗(t)

Velocity is the time derivative of position: v⃗ = dr⃗/dt. Its magnitude |v⃗| is the speed, and its direction is always tangent to the trajectory at every instant.
3

Acceleration Vector a⃗(t)

Acceleration is the time derivative of velocity: a⃗ = dv⃗/dt = d²r⃗/dt². It can change a particle's speed, direction, or both simultaneously.
4

Independence of Components

In a Cartesian frame the x-, y-, and z-equations of motion are decoupled. Horizontal and vertical motions of a projectile, for instance, evolve independently.
5

Relative Motion

Velocities measured in different inertial frames are related by vector addition: v⃗₍A/C₎ = v⃗₍A/B₎ + v⃗₍B/C₎. This principle underlies river-crossing and moving-platform problems.
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KEY TAKEAWAY

Think of a GPS tracker logging your car's latitude, longitude, and altitude every millisecond. Each coordinate evolves according to its own one-dimensional equation, yet together they trace a smooth three-dimensional curve. Multidimensional kinematics works the same way: decompose into independent components, solve each one, then recombine. This "divide and conquer" strategy transforms a seemingly intractable vector problem into several familiar scalar ones.

SECTION 3

Visual Explanation — Projectile Trajectory

xyProjectile Motion — Decomposed Componentst=0t₁vxvyv⃗t₂ (apex)vxvy = 0t₃v⃗t₄HR/2R (Range) = Horizontal (constant) = Vertical (accelerating)
A projectile launched at angle θ follows a parabolic arc (dashed cyan curve). At each sample time, the velocity vector v⃗ (violet) is decomposed into a constant horizontal component vₓ (amber) and a time-varying vertical component vᵧ (pink). At the apex, vᵧ = 0 and the speed equals vₓ alone.

The diagram above captures the essential insight of two-dimensional projectile kinematics. Because the only acceleration is gravitational (directed downward), the horizontal component of velocity remains constant throughout the flight, while the vertical component decreases on the way up, passes through zero at the apex, and increases in magnitude on the way down. The resultant velocity vector, shown in violet, is always tangent to the parabolic path and rotates continuously as vᵧ changes. Notice that the range R and maximum height H are emergent properties of the initial conditions—they fall out directly from the independent horizontal and vertical equations of motion.

SECTION 4

Mathematical Framework

The vector description of motion begins with the position vector r⃗(t) and proceeds through successive differentiation with respect to time. Each derivative reveals a new layer of kinematic information: first the velocity, then the acceleration. For constant-acceleration problems—including all projectile motion near Earth's surface—these derivatives can be integrated to yield closed-form expressions that serve as the workhorse equations of AP Physics C.

POSITION VECTOR

r⃗(t)=x(t)ı^+y(t)ȷ^+z(t)k^r⃗(t) = x(t) î + y(t) ĵ + z(t) k̂

where î, ĵ, k̂ are unit vectors along the x, y, and z axes respectively, and x(t), y(t), z(t) are the scalar component functions of time.

VELOCITY VECTOR

v⃗(t)=dr⃗/dt=(dx/dt)ı^+(dy/dt)ȷ^+(dz/dt)k^v⃗(t) = dr⃗/dt = (dx/dt) î + (dy/dt) ĵ + (dz/dt) k̂

The velocity vector is tangent to the trajectory at every point. Its magnitude |v⃗| = √(vₓ² + vᵧ² + vz²) gives the instantaneous speed.

ACCELERATION VECTOR

a⃗(t)=dv⃗/dt=d2r⃗/dt2=(d2x/dt2)ı^+(d2y/dt2)ȷ^+(d2z/dt2)k^a⃗(t) = dv⃗/dt = d²r⃗/dt² = (d²x/dt²) î + (d²y/dt²) ĵ + (d²z/dt²) k̂

Acceleration can change speed (tangential component), direction (centripetal/normal component), or both. In free fall near Earth's surface, a⃗ = −g ĵ with g ≈ 9.8 m/s².

CONSTANT-ACCELERATION KINEMATICS (2D)

r⃗(t)=r⃗0+v⃗0t+½a⃗t2r⃗(t) = r⃗₀ + v⃗₀t + ½ a⃗ t²

This vector equation encapsulates two (or three) scalar equations simultaneously. For projectile motion with r⃗₀ = 0: x(t) = v₀ cos θ · t and y(t) = v₀ sin θ · t − ½gt².

💡

Deriving the Trajectory Equation

Eliminating t between x(t) and y(t) yields y = x tan θ − (g / 2v₀²cos²θ) x², which is the equation of a parabola opening downward. This derivation, obtained by substituting t = x / (v₀ cos θ) into the y-equation, is a classic AP Physics C free-response technique. The range R = v₀² sin 2θ / g and maximum height H = v₀² sin²θ / 2g follow directly.

SECTION 5

Circular & Curvilinear Motion

Projectile motion under constant gravitational acceleration is only one special case of two-dimensional kinematics. A far richer structure emerges when we consider objects moving along curved paths with varying speed. For any smooth trajectory, the acceleration vector can be decomposed into a tangential component at (along the direction of motion) that changes the particle's speed, and a centripetal (normal) component ac (perpendicular to the direction of motion, toward the center of curvature) that changes the particle's direction. Uniform circular motion is the special case where at = 0 and only centripetal acceleration is present.

Uniform Circular MotionNon-Uniform Curvilinear MotionOrv⃗a⃗cv⃗a⃗cac = v²/r (constant speed)at = 0v⃗a⃗ca⃗ta⃗a⃗ = a⃗t + a⃗c|a⃗| = √(at² + ac²)tangentialcentripetalnet a⃗
Left: In uniform circular motion, the speed is constant, so only centripetal acceleration (pink, directed inward) is present. Right: In general curvilinear motion, both a tangential component (amber, along the path) and a centripetal component (pink, perpendicular to the path) contribute to the net acceleration (violet).

CENTRIPETAL ACCELERATION

ac=v2/r=ω2ra_c = v² / r = ω²r

where v is the instantaneous speed, r is the radius of curvature, and ω is the angular speed. This acceleration is always directed toward the center of curvature.

TANGENTIAL ACCELERATION

at=d∣v⃗∣/dt=dv/dta_t = d|v⃗|/dt = dv/dt

The tangential acceleration is the rate of change of speed (a scalar derivative). It is nonzero whenever the particle speeds up or slows down along its path.

For uniform circular motion, expressing the position in terms of the angular coordinate θ(t) = ωt gives r⃗(t) = r cos(ωt) î + r sin(ωt) ĵ. Differentiating once yields v⃗ = −rω sin(ωt) î + rω cos(ωt) ĵ, and differentiating again yields a⃗ = −rω² cos(ωt) î − rω² sin(ωt) ĵ = −ω²r⃗, confirming that the acceleration points radially inward with magnitude ω²r. These derivations, requiring comfort with differentiation of trigonometric functions, are standard fare on the AP Physics C exam.

A red wheeled cart rests on the low curved part of a smooth metal track, supported by blue stands over a laboratory bench.Velocity follows the local tangent to the path. Normal acceleration points toward the center of curvature, while tangential acceleration changes the speed.

Normal and tangential acceleration. Velocity follows the local tangent to the path. Normal acceleration points toward the center of curvature, while tangential acceleration changes the speed.Velocity follows the local tangent to the path. Normal acceleration points toward the center of curvature, while tangential acceleration changes the speed.

SECTION 6

Worked Example — Projectile Launched from a Cliff

A ball is launched from the edge of a 45.0-m-high cliff with an initial speed of 30.0 m/s at an angle of 37.0° above the horizontal. We wish to find (a) the time of flight until the ball hits the ground below the cliff, (b) the horizontal range, and (c) the speed at impact. Take g = 9.80 m/s².

Projectile from an Elevated Launch Point

  1. 1

    Step 1 — Resolve Initial Velocity

    Using the launch angle θ = 37.0°: v₀ₓ = v₀ cos θ = 30.0 × cos 37.0° = 30.0 × 0.7986 ≈ 23.96 m/s, and v₀ᵧ = v₀ sin θ = 30.0 × sin 37.0° = 30.0 × 0.6018 ≈ 18.05 m/s. The horizontal component remains constant throughout the flight.
    v₀ₓ ≈ 23.96 m/s, v₀ᵧ ≈ 18.05 m/s
  2. 2

    Step 2 — Establish Coordinate System

    Place the origin at the launch point with +x horizontal (in the direction of launch) and +y upward. The initial position is (0, 0), and the ground is at y = −45.0 m. The vertical equation of motion becomes: y(t) = v₀ᵧ t − ½gt² = 18.05t − 4.90t².
    y = −45.0 m at impact
  3. 3

    Step 3 — Solve for Time of Flight

    Set y(t) = −45.0: −45.0 = 18.05t − 4.90t². Rearranging: 4.90t² − 18.05t − 45.0 = 0. Apply the quadratic formula: t = [18.05 ± √(18.05² + 4 × 4.90 × 45.0)] / (2 × 4.90). The discriminant is 325.8 + 882.0 = 1207.8, so √1207.8 ≈ 34.75. Taking the positive root: t = (18.05 + 34.75) / 9.80 ≈ 52.80 / 9.80 ≈ 5.39 s.
    t ≈ 5.39 s
  4. 4

    Step 4 — Compute Horizontal Range

    Since horizontal motion is uniform: x = v₀ₓ × t = 23.96 × 5.39 ≈ 129.1 m.
    R ≈ 129 m
  5. 5

    Step 5 — Find Speed at Impact

    The horizontal velocity at impact is still vₓ = 23.96 m/s. The vertical velocity at impact: vᵧ = v₀ᵧ − gt = 18.05 − 9.80 × 5.39 = 18.05 − 52.82 ≈ −34.77 m/s (negative = downward). The speed is |v⃗| = √(vₓ² + vᵧ²) = √(23.96² + 34.77²) = √(574.1 + 1209.0) = √1783.1 ≈ 42.2 m/s. Alternatively, energy conservation gives the same result: ½mv² = ½mv₀² + mgh → v = √(v₀² + 2gh) = √(900 + 882) ≈ 42.2 m/s.
    v_impact ≈ 42.2 m/s

SECTION 7

Coordinate Systems & Representations Compared

A powerful aspect of multidimensional kinematics is the freedom to choose the coordinate system that best matches the geometry of the problem. Different representations can simplify the mathematics dramatically. The table below compares three common coordinate frameworks encountered in AP Physics C: Mechanics.

Coordinate SystemBest ForPosition VectorKey Advantage
Cartesian (x, y, z)Projectile motion, linear acceleration problemsr⃗ = x î + y ĵ + z k̂Components decouple when acceleration is along an axis
Polar (r, θ)Circular motion, central-force orbitsr⃗ = r r̂Natural for radially symmetric problems; centripetal term arises automatically
Tangential-Normal (s, n)General curvilinear paths, road designPath parameterized by arc length sSeparates speed changes (tangential) from direction changes (normal) cleanly

Comparison of coordinate systems for multidimensional kinematics

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KEY TAKEAWAY

Choosing the right coordinate system is like choosing the right tool in engineering design: a wrench and a screwdriver both apply torque, but one fits the fastener far better. In kinematics, Cartesian coordinates are the "universal wrench," but polar or tangential-normal coordinates often turn a multi-step algebraic slog into a one-line calculation. On the AP exam, recognizing which framework naturally matches the problem geometry can save critical minutes.

SECTION 8

Connections to Advanced Dynamics

Everything we have developed so far assumes either constant acceleration (projectile motion) or a specific constraint like circular motion. In reality, forces—and therefore accelerations—often vary with position, velocity, or time. The machinery of vector kinematics extends seamlessly to these cases, but closed-form solutions are no longer guaranteed and numerical integration or more advanced analytical methods become necessary. The table below contrasts the constant-acceleration regime with the general case.

FeatureConstant Acceleration (AP C Focus)Variable Acceleration (Advanced)
Equation of motionr⃗(t) = r⃗₀ + v⃗₀t + ½a⃗t²r⃗(t) = r⃗₀ + ∫v⃗ dt, solved via differential equations or numerical methods
Trajectory shapeParabola (projectile), circle (uniform circular)Ellipses (Kepler orbits), spirals, chaotic paths
Solving techniqueAlgebraic (quadratic formula, trigonometry)Separation of variables, Lagrangian mechanics, Runge-Kutta integration
AP C exam relevanceDirectly tested in MCQ and FRQOccasionally tested when a(t) or a(v) is given and you must integrate

Constant vs. variable acceleration in multidimensional kinematics

On the AP Physics C exam, you may encounter problems where acceleration is a given function of time, such as a⃗(t) = (αt) î + (β) ĵ. In these cases, the strategy is to integrate a⃗(t) with respect to time to obtain v⃗(t), applying the initial condition v⃗(0) = v⃗₀, and then integrate again to obtain r⃗(t). This direct application of the fundamental theorem of calculus is the bridge between the constant-acceleration toolkit and the more general framework of Newtonian dynamics that dominates the remainder of the AP C: Mechanics curriculum.

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Exam Strategy

When given a position function r⃗(t) with non-polynomial terms (e.g., involving sin, cos, or exponentials), differentiate to find v⃗(t) and a⃗(t). When given a⃗(t), integrate to find v⃗(t) and r⃗(t), using initial conditions to determine constants of integration. Always check units and limiting cases to verify your result.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A projectile is launched at 60° above the horizontal. At the highest point of its trajectory, which of the following correctly describes the velocity and acceleration vectors?
  • A. Both velocity and acceleration are zero.
  • B. Velocity is horizontal and nonzero; acceleration is zero.
  • C. Velocity is horizontal and nonzero; acceleration is g downward.
  • D. Velocity is zero; acceleration is g downward.
Show answer (C)
The correct answer is C. At the apex, the vertical component of velocity is zero, but the horizontal component v₀ cos 60° persists because there is no horizontal acceleration (ignoring air resistance). The acceleration due to gravity acts downward at all points in the trajectory, including the apex—it never becomes zero during free flight. Choice A is wrong because velocity is not zero (horizontal component remains). Choice B is wrong because gravitational acceleration does not vanish at any point. Choice D is wrong because the velocity is not zero; only the vertical component is zero.

Question 2: BASIC CALCULATION

A particle has position vector r⃗(t) = (3t²) î + (4t − t²) ĵ, where r is in meters and t is in seconds. What is the magnitude of the particle's acceleration?
  • A. 2 m/s²
  • B. 2√10 m/s²
  • C. 6 m/s²
  • D. 10 m/s²
Show answer (B)
The correct answer is B. Differentiate r⃗(t) to find v⃗(t) = (6t) î + (4 − 2t) ĵ. Differentiate again: a⃗ = (6) î + (−2) ĵ. The acceleration is constant. Its magnitude is |a⃗| = √(6² + (−2)²) = √(36 + 4) = √40 = 2√10 ≈ 6.32 m/s². Choice A (2 m/s²) only accounts for the y-component. Choice C (6 m/s²) only accounts for the x-component. Choice D (10 m/s²) is the incorrect sum 6 + (−2)² = 10 rather than the Pythagorean combination.

Question 3: INTERMEDIATE

A car moves along a circular track of radius 50.0 m. At a certain instant its speed is 20.0 m/s and its speed is increasing at a rate of 3.00 m/s². What is the magnitude of the car's total acceleration at that instant?
  • A. 3.00 m/s²
  • B. 8.00 m/s²
  • C. 8.54 m/s²
  • D. 11.0 m/s²
Show answer (C)
The correct answer is C. The centripetal acceleration is aca_c = v²/r = (20.0)²/50.0 = 400/50.0 = 8.00 m/s². The tangential acceleration is ata_t = 3.00 m/s². Since these two components are perpendicular, the total acceleration magnitude is |a⃗| = √(aca_c² + ata_t²) = √(8.00² + 3.00²) = √(64.0 + 9.00) = √73.0 ≈ 8.54 m/s². Choice A ignores centripetal acceleration entirely. Choice B gives only the centripetal component. Choice D incorrectly adds the components arithmetically (8.00 + 3.00 = 11.0) instead of using the Pythagorean theorem.

Question 4: APPLIED

10 pts
A rocket-propelled drone has a time-dependent acceleration given by a⃗(t) = (2.0t) î + (6.0 − 2.0t) ĵ (m/s²). The drone starts from rest at the origin at t = 0. (a) Derive an expression for the velocity v⃗(t). (b) Derive an expression for the position r⃗(t). (c) At what time does the drone's y-component of velocity first reach zero, and what is the drone's position at that time? (d) Determine the speed of the drone at t = 4.0 s.
Show answer
Point-by-point rubric: (a) Velocity derivation [3 points]: • [1 pt] Correctly sets up the integral: v⃗(t) = ∫a⃗ dt = ∫(2.0t î + (6.0 − 2.0t) ĵ) dt • [1 pt] Correctly integrates: v⃗(t) = (1.0t²) î + (6.0t − 1.0t²) ĵ + C⃗ • [1 pt] Applies initial condition v⃗(0) = 0⃗ to find C⃗ = 0⃗, giving v⃗(t) = (t²) î + (6.0t − t²) ĵ m/s (b) Position derivation [3 points]: • [1 pt] Sets up the integral: r⃗(t) = ∫v⃗ dt = ∫(t² î + (6.0t − t²) ĵ) dt • [1 pt] Correctly integrates: r⃗(t) = (t³/3) î + (3.0t² − t³/3) ĵ + C⃗ • [1 pt] Applies initial condition r⃗(0) = 0⃗ to confirm C⃗ = 0⃗, giving r⃗(t) = (t³/3) î + (3.0t² − t³/3) ĵ m (c) y-component of velocity zero [2 points]: • [1 pt] Sets vᵧ = 6.0t − t² = t(6.0 − t) = 0, giving t = 0 or t = 6.0 s. The first nonzero time is t = 6.0 s. • [1 pt] Substitutes t = 6.0 s into r⃗: x = 6.0³/3 = 72 m, y = 3.0(36) − 216/3 = 108 − 72 = 36 m. Position is (72 m, 36 m). (d) Speed at t = 4.0 s [2 points]: • [1 pt] Calculates components: vₓ(4) = 16 m/s, vᵧ(4) = 24 − 16 = 8.0 m/s • [1 pt] Computes speed: |v⃗| = √(16² + 8.0²) = √(256 + 64) = √320 ≈ 17.9 m/s

Question 5: CRITICAL THINKING

7 pts
A small ball is launched horizontally with speed v₀ from the top of a hemisphere of radius R, as shown. The ball leaves the surface tangentially at the very top. (a) Using the coordinate system with the origin at the center of the hemisphere and +y upward, write expressions for x(t) and y(t) after launch. (b) The ball lands on the hemisphere's surface when x² + y² = R². Derive an equation (in terms of v₀, R, g, and t) whose positive solution gives the time at which the ball returns to the hemisphere's surface. (c) For R = 2.0 m and g = 9.8 m/s², determine the launch speed v₀ such that the ball lands at an angular position 45° from the top of the hemisphere. (d) Explain qualitatively why there is a maximum angle beyond which the ball cannot land on the hemisphere regardless of v₀.
Show answer
Point-by-point rubric: (a) Parametric equations [2 points]: • [1 pt] x(t) = v₀t (horizontal, constant velocity) • [1 pt] y(t) = R − ½gt² (launched from top of hemisphere at height R above center, no initial vertical velocity) (b) Condition for landing on hemisphere [2 points]: • [1 pt] Substitutes into x² + y² = R²: (v₀t)² + (R − ½gt²)² = R² • [1 pt] Expands and simplifies: v₀²t² + R² − Rgt² + ¼g²t⁴ = R², yielding v₀²t² − Rgt² + ¼g²t⁴ = 0. Dividing by t² (since t > 0): v₀² − Rg + ¼g²t² = 0, so t² = 4(Rg − v₀²)/g². (c) Numerical calculation [2 points]: • [1 pt] At 45° from the top: x = R sin 45° = R/√2 and y = R cos 45° = R/√2. So x(t) = v₀t = R/√2 and y(t) = R − ½gt² = R/√2. • [1 pt] From y-equation: ½gt² = R − R/√2 = R(1 − 1/√2), so t² = 2R(1 − 1/√2)/g. From x-equation: v₀ = R/(t√2). Substituting R = 2.0 m: t² = 2(2.0)(0.2929)/9.8 = 1.1716/9.8 = 0.1195 s², t = 0.3457 s, v₀ = 2.0/(0.3457 × √2) = 2.0/0.4889 ≈ 4.09 m/s. (d) Qualitative explanation [1 point]: • [1 pt] The ball follows a parabolic trajectory that curves downward due to gravity. For the ball to reach a point far around the hemisphere, it would need to curve inward toward the sphere's surface. Beyond a certain angle, the parabolic path always passes below the hemisphere's surface before reaching that angular position—the ball hits the ground (or falls away from the hemisphere) first. Equivalently, increasing v₀ flattens the trajectory horizontally, moving the landing point farther around, but gravity eventually pulls the ball below the sphere's curvature no matter how fast it is launched, establishing a geometric limit.

SUMMARY

Key Concepts in Motion in Two or Three Dimensions

Summary

Motion in multiple dimensions is described by treating the position vector r⃗(t) as a vector-valued function of time. The velocity v⃗ = dr⃗/dt is tangent to the trajectory, and the acceleration a⃗ = dv⃗/dt governs how both speed and direction change. The independence of components in Cartesian coordinates allows each dimension to be solved separately and then recombined. For projectile motion (constant a⃗ = −g ĵ), this yields parabolic trajectories described by r⃗(t) = r⃗₀ + v⃗₀t + ½a⃗t².

For circular and curvilinear motion, the acceleration decomposes into a tangential component at = dv/dt (changing speed) and a centripetal component ac = v²/r (changing direction). When acceleration varies with time, integration of a⃗(t) with appropriate initial conditions yields velocity and position. Choosing the right coordinate system—Cartesian, polar, or tangential-normal—can dramatically simplify the mathematical analysis.

Varsity Tutors • AP Physics C: Mechanics • Motion in Two or Three Dimensions