Motion in two or three dimensions

Separating Motion into Components

Motion in two or three dimensions becomes manageable when each dimension is treated as its own independent one-dimensional problem. The kinematic equations from earlier topics apply separately along each chosen axis.

Vector component notation: A vector in two dimensions can be written as v⃗ = vₓî + vyĵ, where î and ĵ are unit vectors [vectors of magnitude 1 pointing along the positive x- and y-axes, respectively]. The subscript x or y tells you which axis that component belongs to. In three dimensions, add a k̂ component along the z-axis.

To analyze 2D or 3D motion, choose perpendicular axes. Then write the position, velocity, and acceleration along each axis separately. Each axis obeys the same kinematic equations used for 1D motion:

$$x(t)={x}_{0}+{v}_{0x}t+\frac{1}{2}{a}_{x}{t}^{2}$$

$${v}_{x}(t)={v}_{0x}+{a}_{x}t$$

$${v}_{x}^{2}={v}_{0x}^{2}+2{a}_{x}(x-{x}_{0})$$

Identical equations apply along y (and z, if needed), with their own initial conditions and accelerations. This is the power of component analysis: one vector problem becomes two or three scalar problems.

MisconceptionStudents sometimes add x- and y-components directly. Components along perpendicular axes never combine by simple addition; use the Pythagorean theorem for magnitudes and inverse tangent for direction.

Exam TipAlways state which axis is which before writing any equation.

Nonuniform Velocity and Acceleration

Velocity and acceleration need not be the same along every axis. In many real situations, the object accelerates in one direction while moving at constant velocity in another.

The velocity in each dimension is the time derivative of position along that axis. The acceleration in each dimension is the time derivative of velocity along that axis:

$${v}_{x}=\frac{dx}{dt}, {a}_{x}=\frac{d{v}_{x}}{dt}=\frac{{d}^{2}x}{d{t}^{2}}$$

$${v}_{y}=\frac{dy}{dt}, {a}_{y}=\frac{d{v}_{y}}{dt}=\frac{{d}^{2}y}{d{t}^{2}}$$

Because each component has its own derivative, the acceleration vector can point in a completely different direction from the velocity vector. An object can speed up along x while slowing down along y. The net velocity magnitude at any instant is found from the components:

$$v=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}$$

The direction of v⃗ relative to the x-axis is:

$$\theta ={tan}^{-1}\left(\frac{{v}_{y}}{{v}_{x}}\right)$$

Examiner InsightFRQs often give x(t) and y(t) as functions and ask for velocity or acceleration at a specific time. Differentiate each component separately, then combine.

Exam TipShow the derivative step explicitly — stating "I took the derivative" without showing it loses credit.

Independence of Perpendicular Dimensions

A core principle of multidimensional kinematics: motion in one dimension may be changed without causing a change in a perpendicular dimension. A horizontal force does not affect vertical motion, and a vertical force does not affect horizontal motion.

This independence holds because Newton's second law applies separately along each axis. A force along x produces acceleration along x only. No component of that force appears along y or z.

A practical consequence: two objects dropped from the same height at the same instant hit the ground at the same time, even if one is also moving horizontally. The horizontal velocity has zero effect on the time to fall.

MisconceptionStudents often believe a faster horizontal speed makes an object fall slower. Horizontal speed has no effect on vertical free fall because perpendicular components are independent.

Exam TipState "horizontal and vertical motions are independent" explicitly in justifications.

Projectile Motion

Projectile motion is a special case of two-dimensional motion. The only acceleration is g directed downward. No acceleration acts horizontally.

Key Equations

Horizontal position:

$$x={x}_{0}+{v}_{0x}t$$

Variables:

• $x$ = horizontal position (m)
• ${x}_{0}$ = initial horizontal position (m)
• ${v}_{0x}$ = initial horizontal velocity component (m/s)
• $t$ = time (s)

ProportionalityHorizontal displacement is directly proportional to time.

Reference sheet status: Derived from general kinematic equation with aₓ = 0.

Vertical position:

$$y={y}_{0}+{v}_{0y}t-\frac{1}{2}g{t}^{2}$$

Variables:

• $y$ = vertical position (m)
• ${y}_{0}$ = initial vertical position (m)
• ${v}_{0y}$ = initial vertical velocity component (m/s)
• $g$ = magnitude of gravitational acceleration (m/s²)

SI unit: m

ProportionalityVertical displacement depends on t linearly (from v₀ᵧt) and on t² (from the ½gt² term).

Reference sheet status: On the reference sheet (general kinematic equation with a = −g).

Vertical velocity:

$${v}_{y}={v}_{0y}-gt$$

Calculus form:

$${v}_{y}=\frac{dy}{dt}$$

Reference sheet status: On the reference sheet.

Choose positive y upward. Then aₓ = 0 and aᵧ = −g. The horizontal velocity vₓ stays constant throughout the flight because no horizontal force acts. The vertical velocity changes at rate g downward.

To solve any projectile problem, resolve the initial velocity into components:

$${v}_{0x}={v}_{0}cos\theta , {v}_{0y}={v}_{0}sin\theta $$

Then apply the kinematic equations independently to each axis. The time variable t links the two dimensions — find t from whichever axis gives enough information, then substitute into the other axis.

Examiner InsightA common FRQ task: "Derive an expression for the range." Start from the kinematic equations, find the time of flight from the vertical equation (set y = 0), then substitute into the horizontal equation.

Exam TipAlways define your positive direction and origin before writing equations.

Worked Example: Finding the Range of a Projectile

Scenario

A ball is launched from ground level at 20 m/s at 30° above the horizontal. Find the horizontal range. Use g = 10 m/s².

Step-by-step solution

Resolve initial velocity:

$${v}_{0x}=20cos30^{\circ}=20(0.866)=17.3\text{ m/s}$$

$${v}_{0y}=20sin30^{\circ}=20(0.500)=10.0\text{ m/s}$$

Find time of flight (set y = 0, with y₀ = 0):

$$0={v}_{0y}t-\frac{1}{2}g{t}^{2}$$

$$0=t\left({v}_{0y}-\frac{1}{2}gt\right)$$

Discard t = 0 (launch):

$$t=\frac{2{v}_{0y}}{g}=\frac{2(10.0)}{10}=2.00\text{ s}$$

Find horizontal range:

$$x={v}_{0x}t=17.3\times 2.00=34.6\text{ m}$$

Interpretation

The ball lands 34.6 m from the launch point. The horizontal distance depends on both the horizontal speed and the time aloft, which is controlled entirely by the vertical motion.

Derivation: General Range Equation (Symmetric Projectile)

Starting point: Kinematic equations with aₓ = 0 and aᵧ = −g, launched from and landing at the same height (y = 0).

Step 1 --- Find time of flight from vertical equation

Set y = 0:

$$0={v}_{0}sin\theta \cdot t-\frac{1}{2}g{t}^{2}$$

$$t=\frac{2{v}_{0}sin\theta }{g}$$

Step 2 --- Substitute into horizontal equation

$$R={v}_{0}cos\theta \cdot t={v}_{0}cos\theta \cdot \frac{2{v}_{0}sin\theta }{g}$$

Step 3 --- Simplify using the identity 2 sin θ cos θ = sin 2θ

$$R=\frac{{v}_{0}^{2}sin2\theta }{g}$$

Result

$$R=\frac{{v}_{0}^{2}sin2\theta }{g}$$

The range is maximized when sin 2θ = 1, which occurs at θ = 45°.

BoundaryAP Physics C: Mechanics expects quantitative analysis in two dimensions. AP Physics C: E&M also expects qualitative description in three dimensions.

Know these cases for the AP exam.