Combining Vectors (College Board AP® Physics 1: Algebra-Based)

Study Guide

Leander Oates

Written by: Leander Oates

Reviewed by: Caroline Carroll

Combining vectors

Vector sums in one dimension

  • In one-dimensional coordinate systems, vectors of opposing directions are represented as positive and negative

Two pairs of vectors labeled A and -A, B and -B, showing equal magnitudes with opposing directions. Text below reads "Vectors of equal magnitude with opposing directions."
Vectors A and -A are equal in magnitude but opposite in direction. Vectors B and -B are equal in magnitude and opposite in direction. Vectors B and -B are smaller in magnitude than vectors A and -A, and have a different direction.

Scalar multiplication

  • Vectors can be multiplied by a scalar

  • The effect is either a change in magnitude or a reversal of the direction

  • For example, vector B with rightwards arrow on top is multiplied by 2

    • B with rightwards arrow on top is the vector

    • 2 is the scalar

  • The resultant vector is 2 B with rightwards arrow on top

    • The magnitude of the vector has doubled

    • The direction has not changed

Vector B and its scalar multiple 2B, with B represented by a shorter arrow and 2B by a longer arrow to indicate increased magnitude.
When the vector B is multiplied by the scalar 2, the magnitude of the vector doubles but the direction stays the same
  • When vector B with rightwards arrow on top is multiplied by negative 1

    • B with rightwards arrow on top is the vector

    • negative 1 is the scalar

  • The resultant vector is negative B with rightwards arrow on top

    • The magnitude of the vector has not changed

    • The direction has reversed

Diagram showing vector B pointing right and negative vector B pointing left, with text below stating "SCALAR MULTIPLICATION RESULTING IN A REVERSE OF DIRECTION".
When the vector B is multiplied by the scalar -1, the magnitude of the vector stays the same, but the direction reverses

Vector addition

  • Vectors can be added together using a vector sum

  • The effect is a resultant vector that describes the combined effect of both vectors

  • For example, vector A with rightwards arrow on top is added to vector B with rightwards arrow on top

  • The resultant vector is A with rightwards arrow on top space plus space B with rightwards arrow on top

A with rightwards arrow on top space plus space B with rightwards arrow on top space equals space A with rightwards arrow on top space plus space B with rightwards arrow on top

  • There are two methods to adding vectors graphically:

    • The tip-to-tail method

    • The parallelogram method

Tip-to-tail method of adding vectors

Diagram illustrating vector addition: Vector A and Vector B are shown individually and then added together, forming Vector A + B. Text below reads "VECTOR ADDITION: A + B = A + B"
Vector A and vector B are added together, the resultant vector is A + B
  • The tip-to-tail method places the vector arrows tip-to-tail to find the resultant vector

Parallelogram method of adding vectors

Vector addition using the parallelogram method. Two vectors A and B form a parallelogram, with the resultant vector A+B shown in red.
The parallelogram method gives the same resultant vector A + B. Notice that the dotted line for vector B is positioned in the exact same way as the tip to tail method
  • The parallelogram method places the vectors tail-to-tail to find the resultant vector

Vector subtraction

  • Vectors can also be subtracted using a vector sum and scalar multiplication

  • For example, vector B with rightwards arrow on top is subtracted from vector A with rightwards arrow on top

  • Vector B with rightwards arrow on top is first multiplied by the scalar negative 1

    • This reverses the direction of vector B with rightwards arrow on top to give negative B with rightwards arrow on top

  • The vectors are then added by placing them tip to tail

  • The resultant vector is A with rightwards arrow on top space plus space open parentheses negative B with rightwards arrow on top close parentheses

A with rightwards arrow on top space minus space B with rightwards arrow on top space equals space A with rightwards arrow on top space plus space open parentheses negative B with rightwards arrow on top close parentheses

Diagram illustrating vector subtraction. Vector A and B are shown separately. On the right, vector A and negative B combine to form resultant vector A + (-B).
Vector B is subtracted from vector A by using scalar multiplication to reverse the direction of B and using vector addition by placing vectors A and -B tip to tail

Vectors in one and two dimensions

  • A one-dimensional vector has magnitude and direction along a single straight line

    • This can be along the x-axis or y-axis in a coordinate system

  • A two-dimensional vector has magnitude and direction in a plane

    • This means it has two component vectors: one along the x-axis and one along the y-axis

Resolving vectors in two dimensions

  • Two-dimensional vectors can be resolved into two perpendicular one-dimensional vectors

    • The vector along the horizontal x-axis

    • The vector along the vertical y-axis

  • These are called component vectors

Graph depicting a 2D vector A with components Ax and Ay. Ax and Ay are 1D vectors along x and y axes respectively. Direction of A indicated by arrows.
Two-dimensional vectors can be resolved into their component vectors along the x and y axes

Finding the magnitude of the vector

  • The magnitude of a two-dimensional vector can be found using Pythagoras' theorem

a squared space plus space b squared space equals space c squared

A right-angled triangle labeled with sides 'a', 'b', and hypotenuse 'c' and a note saying 'Side c is always the hypotenuse'. Below is the equation a^2 + b^2 = c^2.
Pythagoras' theorem is used to find the length of an unknown side of a right angled triangle
  • The two-dimensional vector is the hypotenuse of the triangle, therefore:

open parentheses A with rightwards arrow on top close parentheses squared space equals space open parentheses A with rightwards arrow on top subscript y close parentheses squared space plus space open parentheses A with rightwards arrow on top subscript x close parentheses squared

A with rightwards arrow on top space equals space square root of open parentheses A subscript y close parentheses squared space plus space open parentheses A subscript x close parentheses squared end root

Finding the direction of the vector

  • The direction of the vector can be described with respect to an axis

    • This is usually the horizontal x-axis, unless otherwise directed

  • The symbol theta theta is used for an unknown angle

A vector diagram showing vector A with components Ax and Ay, making an angle theta with the x-axis. Text indicates the angle is measured from the x-axis.
The angle is generally measured from the x-axis unless otherwise stated
  • Trigonometry can then be used to find:

    • the angle, if the magnitude of the vectors are known

    • the magnitude of a vector, if the angle and the magnitude of one vector are known

Diagram of a right triangle with sides a, b, and hypotenuse c. Includes trigonometric ratios: sin θ = a/c, cos θ = b/c, tan θ = a/b, and mnemonic SOHCAHTOA.
Trigonometry is used for calculating an angle in a right angled triangle

Worked Example

A kayaker rows from the north bank to the south bank of a river flowing due east at speed r. The kayaker knows they can maintain an average speed of k when rowing in still water.

(a) State an equation relating the velocities k and r.

(b) Give an expression for the direction of the kayaker as they cross the river in terms of k and r.

(c) Before the kayaker sets off, the river flow slows to half its initial speed. Derive an equation for the new velocity that the kayaker needs to cross the river.

Answer:

Analyze the scenario

  • Sketch a diagram to help visualize the situation

Diagram showing a kayaker on the North Bank aiming for the South Bank with velocity vector v0 at an angle θ0. Additional vectors k, r0, and r direction are labeled.
  • The kayaker travels from the north bank to the south bank, therefore:

    • The kayaker is positioned on the north bank

    • The kayaker travels south across the river

  • The kayaker can maintain a speed of k in still water, therefore:

    • The kayaker's velocity in still water can be written as k with rightwards arrow on top to the south

  • The river flows due east at an initial speed of r, therefore:

    • The initial velocity of the river can be written as r with rightwards arrow on top subscript 0 to the east

  • The kayaker will be pulled to the east by the current, therefore:

    • The initial resultant velocity of the kayaker can be written as v with rightwards arrow on top subscript 0 at angle theta subscript 0 east from south

Part (a)

State an equation relating the velocities k and r

Step 1: Identify the relevant fundamental principle

  • The initial velocity vectors k with rightwards arrow on top and space r with rightwards arrow on top subscript 0 are the components of vector v with rightwards arrow on top subscript 0

  • Therefore, the relevant fundamental principle is vector addition

Step 2: Apply the fundamental principle to the scenario

  • The sum of vectors k with rightwards arrow on top and r with rightwards arrow on top subscript 0 is equal to the resultant vector v with rightwards arrow on top subscript 0

v with rightwards arrow on top subscript 0 space equals space k space plus space r with rightwards arrow on top subscript 0

Part (b)

Give an expression for the direction of the kayaker as they cross the river in terms of k and r

Step 1: Identify the direction of the resultant vector v with rightwards arrow on top subscript 0

  • The direction of a resultant vector is described by an angle

  • The value of the initial angle is unknown, so it can be written as theta subscript 0

Step 2: Determine the trigonometric ratio for the resultant vector v with rightwards arrow on top subscript 0

Diagram of a kayaker paddling from the north bank to the south bank forming a right triangle. Labels include SOH CAH TOA, adjacent, opposite, hypotenuse, theta0, and vectors.
  • Use the mnemonic SOH CAH TOA to recall the trigonometric ratios

  • Choose the ratio that includes the terms of k with rightwards arrow on top and r with rightwards arrow on top subscript 0

    • k with rightwards arrow on top is the side adjacent to the angle theta subscript 0

    • r with rightwards arrow on top subscript 0 is the side opposite to the angle theta subscript 0

    • Therefore, the correct trigonometric ratio is tangent

theta subscript 0 space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator r with rightwards arrow on top subscript 0 over denominator k with rightwards arrow on top end fraction close parentheses

Part (c)

Before the kayaker sets off, the river flow slows to half its initial speed. Derive an equation for the new velocity that the kayaker needs to cross the river

Step 1: Add the information to the sketch

Diagram of a kayaker paddling from the north bank to the south bank. Various labeled vectors illustrate the kayaker's velocity, direction, and distance traveled.

Step 2: Determine an expression for the new river speed vector

  • The initial velocity vector for river speed is r with rightwards arrow on top subscript 0 east

  • The new river speed is half of the initial speed

  • Therefore, the new velocity of the river can be written as fraction numerator stack r subscript 0 with rightwards arrow on top over denominator 2 end fraction

Step 3: Determine the magnitude of the new resultant vector

  • The new resultant velocity vector can be written as v with rightwards arrow on top

  • The magnitude of this vector is the sum of the vectors k with rightwards arrow on top and r with rightwards arrow on top subscript 0 over 2

v with rightwards arrow on top space equals space k space plus space r with rightwards arrow on top subscript 0 over 2

Step 4: Determine the direction of the new resultant vector

  • The direction of the new resultant vector is the new angle, which can be written as theta

  • The expression for the initial direction from part (B) can be applied to the new situation

theta space equals space tan to the power of negative 1 end exponent open parentheses r with rightwards arrow on top subscript 0 over 2 times fraction numerator 1 over denominator k with rightwards arrow on top end fraction close parentheses space

theta space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator r with rightwards arrow on top subscript 0 over denominator 2 k end fraction close parentheses

Step 5: State the magnitude and direction of the new resultant vector

v with rightwards arrow on top space equals space k space plus space r with rightwards arrow on top subscript 0 over 2 at an angle theta space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator r with rightwards arrow on top subscript 0 over denominator 2 k with rightwards arrow on top end fraction close parentheses east from south

Examiner Tips and Tricks

The problem above is an example of the Qualitative/Quantitative Translation style of free-response question. This is where you are expected to make and justify a claim about a scenario, derive an equation related to the scenario, and then use consistent reasoning to formulate your answer.

You are expected to use algebra as a mode of communication, but it is always a good idea to also use worded statements to show the examiner your thought process. The examiner can only award marks where your logic is clear enough for them to follow. For this reason, it is always a good idea to read through your answer and check for clarity, adding in any notes that you think it needs.

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Leander Oates

Author: Leander Oates

Expertise: Physics

Leander graduated with First-class honours in Science and Education from Sheffield Hallam University. She won the prestigious Lord Robert Winston Solomon Lipson Prize in recognition of her dedication to science and teaching excellence. After teaching and tutoring both science and maths students, Leander now brings this passion for helping young people reach their potential to her work at SME.

Caroline Carroll

Author: Caroline Carroll

Expertise: Physics Subject Lead

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.