Vectors and motion in two dimensions

Perpendicular Components of a Vector

Any vector can be replaced by two perpendicular components that, added tip-to-tail, reproduce the original vector. This idea is the gateway to solving every two-dimensional problem on the AP exam.

Key Equations

Component resolution (x-direction):

$${A}_{x}=Acos\theta $$

Variables: $A$ = magnitude of the vector, $\theta $ = angle measured from the positive x-axis.

SI unit: same as the original vector’s unit.

Component resolution (y-direction):

$${A}_{y}=Asin\theta $$

Variables: $A$ = magnitude of the vector, $\theta $ = angle measured from the positive x-axis.

SI unit: same as the original vector’s unit.

Magnitude from components:

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

Rearrangements: not typically rearranged.

Direction from components:

$$tan\theta =\frac{{A}_{y}}{{A}_{x}}$$

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

Reference sheet status: These are standard trigonometric identities; they do not appear as named equations on the AP reference sheet, but the mathematical relationships are assumed knowledge.

Reading vector-component notation: When a symbol carries a subscript $x$ or $y$, it refers to the component [the portion of the vector that lies along that axis]. A component is a signed scalar: its sign tells you the direction along that axis, while its absolute value gives the magnitude along that axis.

A resultant vector is the single vector that has the same effect as two or more vectors acting together. Because the x- and y-components are perpendicular, they are completely independent — changing one does not affect the other. This independence is what makes component resolution so powerful.

To resolve a vector, choose a coordinate system first. The most common choice places the x-axis horizontal and the y-axis vertical, but any orientation works as long as the two axes are perpendicular. Once the axes are set, the cosine function projects the vector onto the axis closest to the angle, and the sine function projects it onto the other axis. If the angle $\theta $ is measured from the x-axis, then ${A}_{x}=Acos\theta $ and ${A}_{y}=Asin\theta $. If $\theta $ is measured from the y-axis instead, the cosine and sine assignments swap.

To reconstruct the original vector from its components, use the Pythagorean theorem for magnitude and the inverse tangent for direction. Always check the quadrant to ensure the angle makes physical sense.

MisconceptionStudents often apply $cos\theta $ and $sin\theta $ without checking which axis the angle is measured from, swapping the components. The rule: cosine goes with the axis the angle is measured from; sine goes with the other axis.

Exam TipSketch the vector and label the angle before writing any trig expression.

Worked Example: Finding Components of a Displacement Vector

Scenario

A hiker walks 240 m on a bearing that makes an angle of 35° above the positive x-axis. Determine the x- and y-components of the displacement.

Finding the x-component:

Equation used

$${A}_{x}=Acos\theta $$

Given

$$A=240\text{ m}, \theta =35^{\circ}$$

Working — substitution

$${A}_{x}=240\times cos35^{\circ}$$

$${A}_{x}=240\times 0.8192$$

$${A}_{x}=197\text{ m}$$

Finding the y-component:

Equation used

$${A}_{y}=Asin\theta $$

Working — substitution

$${A}_{y}=240\times sin35^{\circ}$$

$${A}_{y}=240\times 0.5736$$

$${A}_{y}=138\text{ m}$$

Interpretation

The hiker’s displacement has a horizontal component of 197 m and a vertical component of 138 m. Added together as perpendicular vectors, they reproduce the original 240 m displacement.

Adding Vectors Using Components

When two or more vectors act in different directions, the cleanest method is to resolve each one into components, add the components along each axis separately, and then reconstruct the resultant.

The process has three stages:

1. Resolve every vector into x- and y-components using the trig rules from the previous subtopic.
2. Add all x-components to get ${R}_{x}$ and all y-components to get ${R}_{y}$. Because addition along a single axis is just signed arithmetic, this step is straightforward.
3. Find the magnitude and direction of the resultant from ${R}_{x}$ and ${R}_{y}$.

$${R}_{x}={A}_{x}+{B}_{x}+…$$

$${R}_{y}={A}_{y}+{B}_{y}+…$$

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

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

This component-addition method works for any number of vectors at any angles. It replaces graphical tip-to-tail addition with precise algebra.

Examiner InsightAP FRQs often present three or more forces at angles and ask for the net force. The fastest, most reliable method is component addition — never attempt to add non-perpendicular vectors by simple arithmetic.

Exam TipSet up a small table with columns for ${A}_{x}$ and ${A}_{y}$ for each vector before summing.

Analyzing Two-Dimensional Motion with Components

Motion in two dimensions becomes manageable once velocity and acceleration are each split into independent perpendicular components — because each axis obeys the same one-dimensional kinematic equations already familiar from earlier units.

Key Equations

Kinematic equations applied per axis (constant acceleration):

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

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

$${v}_{y}={v}_{0y}+{a}_{y}t$$

$$\Delta y={v}_{0y}t+\frac{1}{2}{a}_{y}{t}^{2}$$

Variables: ${v}_{0x}$, ${v}_{0y}$ = initial velocity components; ${a}_{x}$, ${a}_{y}$ = acceleration components; $t$ = time; $\Delta x$, $\Delta y$ = displacement components.

SI units: m/s for velocity, m/s² for acceleration, m for displacement, s for time.

Reference sheet status: The base kinematic equations are on the reference sheet. Applying them per axis is a technique, not a separate formula.

The key insight is that the x- and y-directions are independent. A change in ${a}_{y}$ does not affect ${v}_{x}$, and a change in ${a}_{x}$ does not affect ${v}_{y}$. Time is the only variable shared by both directions — it links the two sets of equations.

To solve a two-dimensional motion problem: choose a coordinate system, resolve the initial velocity into ${v}_{0x}$ and ${v}_{0y}$, identify ${a}_{x}$ and ${a}_{y}$, then apply the kinematic equations to each axis independently. Use time to connect the two axes when needed.

Projectile Motion

Projectile motion is the most common two-dimensional problem on the AP exam. It occurs whenever the only acceleration is gravity — zero acceleration horizontally and constant acceleration $g$ vertically.

Key Equations

Horizontal motion (constant velocity, ${a}_{x}=0$):

$$\Delta x={v}_{0x}t$$

where ${v}_{0x}={v}_{0}cos\theta $.

Vertical motion (constant acceleration, ${a}_{y}=-g$):

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

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

$${v}_{y}^{2}={v}_{0y}^{2}-2g\Delta y$$

where ${v}_{0y}={v}_{0}sin\theta $ and $g\approx 10$ m/s².

ProportionalityHorizontal range is directly proportional to ${v}_{0x}$ and to the time of flight. Vertical displacement depends on ${v}_{0y}$ and on ${t}^{2}$, so doubling the flight time quadruples the vertical distance fallen.

Reference sheet status: The base kinematic equations are on the reference sheet. The substitution of $g$ for acceleration and the component forms are applied by the student.

Taking upward as positive, the acceleration components for a projectile are ${a}_{x}=0$ and ${a}_{y}=-g$. The horizontal velocity ${v}_{x}$ remains constant throughout the flight because no horizontal force acts. The vertical velocity ${v}_{y}$ changes at a steady rate of $-g$ each second.

At the highest point of a projectile’s path, ${v}_{y}=0$ but ${v}_{x}$ is unchanged. The projectile is still moving — only its vertical component is momentarily zero. The velocity at any instant is the vector sum of ${v}_{x}$ and ${v}_{y}$, so the speed at the peak equals ${v}_{0x}$.

For a projectile launched and landing at the same height, the trajectory is symmetric. The time to rise equals the time to fall, and the launch angle and landing angle are equal. Maximum horizontal range occurs at a 45° launch angle (when air resistance is negligible).

MisconceptionStudents frequently believe the acceleration is zero at the top of the trajectory because the vertical velocity is zero there. The acceleration due to gravity is $g$ downward at every point in the flight, including the peak.

Exam TipIf asked for the acceleration at the highest point, the answer is $g$ downward — not zero.

Worked Example: Projectile Launched at an Angle

Scenario

A ball is launched from ground level at 20.0 m/s at 60.0° above the horizontal. Determine the horizontal range. Use $g=10$ m/s².

Resolve the initial velocity:

$${v}_{0x}={v}_{0}cos\theta =20.0\times cos60.0^{\circ}=20.0\times 0.500=10.0\text{ m/s}$$

$${v}_{0y}={v}_{0}sin\theta =20.0\times sin60.0^{\circ}=20.0\times 0.8660=17.3\text{ m/s}$$

Finding the time of flight (the ball lands at $\Delta y=0$):

Equation used

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

Setting $\Delta y=0$:

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

Factor out $t$:

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

Discarding $t=0$ (the launch instant):

$$t=\frac{2{v}_{0y}}{g}=\frac{2\times 17.3}{10}=3.46\text{ s}$$

Finding the horizontal range:

$$\Delta x={v}_{0x}\times t=10.0\times 3.46=34.6\text{ m}$$

Interpretation

The ball travels 34.6 m horizontally before returning to ground level. The horizontal velocity stayed at 10.0 m/s throughout the flight because no horizontal acceleration acted.