AshPhys

Combining displacements

AshPhys · Revision cheat sheet
AS Level9702 AS Ch.1Chapter 1: KinematicsLesson 1.5Sections 1.5

Goal: Add two vectors tip to tail, and split one vector into two perpendicular components.

velocity

01The big picture

To add vectors, draw them tip to tail: the resultant runs from the start of the first to the end of the last. For two perpendicular vectors use Pythagoras and tan⁡θ\tan\theta; for any other angle use a scale drawing.

Resolving does the reverse: one vector at angle θ to the x direction is replaced by two perpendicular components that together have the same effect.

Resultant [paraphrase]
The single vector with the same effect as two or more vectors together.
Components [paraphrase]
Two perpendicular vectors that add up to the original vector.
Fig. 1 · Resolving into components
Resolving into componentsA velocity v at angle theta above the horizontal, with a dashed horizontal component v cos theta and a dashed vertical component v sin theta forming a right-angled triangle.θvv cos θv sin θ
v cos θ is adjacent to the angle; v sin θ is opposite it.

02Key equations

Resultant of perpendicular vectors
R=a2+b2R = \sqrt{a^{2} + b^{2}}
Its direction
tan⁡θ=ba\tan\theta = \frac{b}{a}
Components
vx=vcos⁡θ,  vy=vsin⁡θv_x = v\cos\theta,\ \ v_y = v\sin\theta
Symbols and units
  • a, ba,\ btwo perpendicular vectors(their unit)\text{(their unit)}
  • RRresultant(same unit)\text{(same unit)}
  • θ\thetaangle to the x direction∘^\circ

03Sample problems with solutions

Sample 1 · Resolving a velocity
A ball is kicked at 20 m s⁻¹ at 30° above the horizontal. Calculate the horizontal and vertical components of its velocity.

  1. Horizontal (adjacent to θ):

    vx=20cos⁡30∘=17.3 m s−1v_x = 20\cos 30^\circ = 17.3\ \mathrm{m\,s^{-1}}

  2. Vertical (opposite θ):

    vy=20sin⁡30∘=10.0 m s−1v_y = 20\sin 30^\circ = 10.0\ \mathrm{m\,s^{-1}}
Answer: 17 m s⁻¹ horizontally, 10 m s⁻¹ vertically
Unit check: components keep the vector’s unit: m s−1 ✓\text{components keep the vector's unit: } \mathrm{m\,s^{-1}}\ \checkmark

04Practice problems

  1. P1
    Find the resultant of 12 N east and 5.0 N north.
    Final answer13 N at 23° north of east
    Answer: 13 N at 23° north of east
  2. P2
    Resolve a 50 N force at 40° above the horizontal.
    Final answer38 N horizontal, 32 N vertical
    Answer: 38 N horizontal, 32 N vertical
  3. P3
    Two forces of 3 N and 4 N act on a point. What are the largest and smallest possible resultants?
    Final answer7 N and 1 N
    Answer: 7 N and 1 N
  4. P4
    Find the vertical component of a velocity of 8.0 m s⁻¹ at 60° to the horizontal.
    Final answer6.9 m s⁻¹
    Answer: 6.9 m s⁻¹

05Other useful information

Common mistakes

  • ✗ Wrong: Add the sizes of two vectors at an angle (3 N + 4 N = 7 N).

    ✓ Right: Only when they point the same way. At 90°: √(3² + 4²) = 5 N.

  • ✗ Wrong: Use sin for the side next to the angle.

    ✓ Right: Adjacent side: cos θ. Opposite side: sin θ.

Exam tips

  • Always draw the vector triangle and mark θ before choosing sin or cos.

What the syllabus asks

  • 9702 topic 1: add and subtract coplanar vectors; represent a vector as two perpendicular components. Coursebook 1.5.

Explore: Vector Addition Sandbox (simulation) · Practice questions for this lesson