B. Kinematics · Kinematics: Motion in One and Two DimensionsVector Components and Two-Dimensional Motion Problems
B2.5 B2.7 B3.2 A1.12About 75 minutes
Learning goals- I can resolve a vector into x and y components using sine and cosine.
- I can add several vectors algebraically by adding their components.
- I can find average velocity, change in velocity and average acceleration in two dimensions.
- I can solve relative-velocity problems such as a boat crossing a river or a plane in a crosswind.
Words to know
component
The part of a vector that points along one axis (x: east–west; y: north–south).
resolving a vector
Splitting a vector into its x and y components.
resultant velocity (ground velocity)
The velocity relative to the ground, found by adding the object's velocity relative to the air or water and the velocity of the air or water.
heading
The direction a boat or plane is pointed, which can differ from the direction it actually travels.
change in velocity (Δv)
Δv = v₂ − v₁ = v₂ + (−v₁). In two dimensions, a change in direction alone produces a Δv.
From drawings to calculations
Scale diagrams are great for seeing a problem, but they are slow and only good to about two significant digits. Navigators, engineers and video-game programmers need something faster and more precise. The trick is to break every vector into an east–west part and a north–south part. Then two-dimensional problems turn into two one-dimensional problems, which you already know how to solve.
Resolving a vector into components
Let east be +x and north be +y. Any vector A can be replaced by two perpendicular components, Ax and Ay, that add tip to tail to give A. They form a right triangle with A as the hypotenuse.
If θ is the angle between A and the x-axis: Ax = A cos θ and Ay = A sin θ.
Compass directions measure angles from different axes, so draw a sketch every time and ask which side is adjacent to the angle. For [N 30° W], the 30° is measured from north (the y-axis), so the y component uses cosine and the x component uses sine. Then give each component its sign: west and south are negative.
Adding vectors by components
- Resolve every vector into x and y components (with signs).
- Add all the x components to get Rx. Add all the y components to get Ry.
- Magnitude: R = √(Rx² + Ry²).
- Direction: find the angle with θ = tan⁻¹(|Ry| ÷ |Rx|), then use the signs of Rx and Ry to decide the quadrant and write the direction in compass form.
This works for any number of vectors and any angles, and it is as precise as your calculator.
Worked exampleExample 1: A three-leg snowmobile trip
A snowmobile travels 12.0 km [E], then 8.0 km [N 30° W], then 5.0 km [S]. The trip takes 2.0 h. Find (a) the resultant displacement and (b) the average velocity.
- Leg 1: x = +12.0, y = 0.
- Leg 2 (from the diagram above): x = −8.0 sin 30° = −4.0; y = +8.0 cos 30° = +6.93.
- Leg 3: x = 0, y = −5.0.
- Rx = 12.0 − 4.0 + 0 = 8.0 km. Ry = 0 + 6.93 − 5.0 = 1.93 km.
- R = √(8.0² + 1.93²) = √(64 + 3.72) ≈ 8.2 km. θ = tan⁻¹(1.93 ÷ 8.0) ≈ 14°. Both components are positive (east and north), so Δd = 8.2 km [E 14° N].
- (b) vₐᵥ = Δd ÷ Δt = 8.2 km [E 14° N] ÷ 2.0 h ≈ 4.1 km/h [E 14° N]. (The average speed is much larger: 25 km ÷ 2.0 h = 12.5 km/h.)
(a) 8.2 km [E 14° N]; (b) 4.1 km/h [E 14° N]
Relative velocity: rivers and winds
A canoe paddled across a river is carried downstream by the current. A plane flying through moving air drifts with the wind. In both cases:
velocity relative to the ground = velocity relative to the water (or air) + velocity of the water (or air)
The velocities add as vectors, tip to tail. The direction a boat is pointed (its heading) is not always the direction it actually moves. A useful fact: the time to cross a river depends only on the velocity component straight across the river, because the current pushes only along it.
Worked exampleExample 2: Crossing a river
A river 120 m wide flows east at 1.5 m/s. A canoeist points her canoe straight across (north) and paddles at 2.0 m/s relative to the water. Find (a) her velocity relative to the ground, (b) the time to cross, and (c) how far downstream she lands.
- (a) v(ground) = 2.0 m/s [N] + 1.5 m/s [E]. These are perpendicular: |v| = √(2.0² + 1.5²) = √6.25 = 2.5 m/s; θ = tan⁻¹(1.5 ÷ 2.0) ≈ 37° from north toward east. v(ground) = 2.5 m/s [N 37° E].
- (b) Only the northward part (2.0 m/s) carries her across: Δt = 120 m ÷ 2.0 m/s = 60 s.
- (c) During those 60 s the current carries her east: Δd = 1.5 m/s × 60 s = 90 m downstream.
(a) 2.5 m/s [N 37° E]; (b) 60 s; (c) 90 m downstream
Example 2. The canoe is aimed straight across, but the current carries it east, so it actually travels along the diagonal and lands 90 m downstream. (The paths are drawn to scale; the current arrow is not.)Change in velocity and acceleration in two dimensions
In two dimensions, an object can accelerate without changing speed, just by changing direction. As always, Δv = v₂ − v₁ = v₂ + (−v₁), and aₐᵥ = Δv ÷ Δt points in the direction of Δv.
Worked exampleExample 3: Turning a corner
A car travelling at 15 m/s [N] turns a corner and is then travelling at 15 m/s [E]. The turn takes 3.0 s. Find the change in velocity and the average acceleration.
- Δv = v₂ + (−v₁) = 15 m/s [E] + 15 m/s [S].
- These are perpendicular and equal: |Δv| = √(15² + 15²) ≈ 21.2 m/s, at 45°. Δv ≈ 21 m/s [S 45° E].
- aₐᵥ = Δv ÷ Δt = 21.2 ÷ 3.0 ≈ 7.1 m/s² [S 45° E].
- The speed never changed, yet the car accelerated, because its direction changed. The acceleration points toward the inside of the turn.
Δv ≈ 21 m/s [S 45° E]; aₐᵥ ≈ 7.1 m/s² [S 45° E]
To find Δv, add v₂ and the opposite of v₁ tip to tail. Even though both velocities have the same magnitude, Δv is far from zero.Component checklist
- Calculator in degree mode.
- Draw each vector and mark which axis its angle is measured from.
- Signs: east and north +, west and south −.
- Keep unrounded components until the end.
- Check with a rough sketch: if your answer points north-east but the sketch ends up south-west, a sign is wrong.
- Check the Lesson 10 canoe: 2.0 km [N] + 3.0 km [N 60° E] gives Rx = 3.0 sin 60° = 2.60 km, Ry = 2.0 + 3.0 cos 60° = 3.50 km, R = 4.36 km at tan⁻¹(2.60 ÷ 3.50) = 37° from north: 4.4 km [N 37° E], matching the scale diagram.
Just for funPlanes flying east across Canada often get a boost from the jet stream, a band of fast wind high in the atmosphere. That's why a cross-country flight going east can be noticeably shorter than the same trip going west. So if your trip home is westbound and it feels longer, it probably is.
Hands-on activityA river on the kitchen table
You need: a large sheet of paper (or taped-together sheets), a toy car or a small object, a ruler, a stopwatch, and a partner.
- The paper is the "river". Your partner slowly and steadily slides the paper sideways across the table (this is the current). Practise until they can move it about 20 cm in 4 s.
- While the paper moves, push the toy car straight "across" the paper (perpendicular to its motion), at a steady pace, also taking about 4 s to cross 20 cm. Hold a pencil tip against the paper beside the car's path if you can, so it traces the car's path relative to the paper.
- Look down from above (or film from above). Measure how far the car moved across and how far it drifted sideways relative to the table.
- Calculate the car's velocity relative to the table by components and compare with what you measured.
What to look for: relative to the paper, the path is straight across. Relative to the table, it is diagonal. With 20 cm across and 20 cm of drift in 4 s, each component is 5 cm/s, so the table-velocity is about √(5² + 5²) ≈ 7 cm/s at 45° to the edge.
Your turn
1Find the components of 20 m/s [N 40° E] (east and north positive).
x = 20 sin 40° ≈ +12.9 m/s; y = 20 cos 40° ≈ +15.3 m/s.
2Add 6.0 km [W] and 10.0 km [N 60° W].
x = −6.0 − 10.0 sin 60° = −14.66; y = 10.0 cos 60° = 5.00. R = √(14.66² + 5.00²) ≈ 15.5 km; θ = tan⁻¹(5.00 ÷ 14.66) ≈ 19°: 15.5 km [W 19° N].
3A plane flies with an airspeed of 250 km/h heading [N]. A wind blows at 50 km/h [E]. Find its ground velocity.
√(250² + 50²) ≈ 255 km/h; θ = tan⁻¹(50 ÷ 250) ≈ 11°: 255 km/h [N 11° E].
4A boat heads straight across an 80 m river at 1.6 m/s. The current is 1.2 m/s. How long does it take to cross, and how far downstream does it land?
Δt = 80 ÷ 1.6 = 50 s; drift = 1.2 × 50 = 60 m.
5A ball moving at 6.0 m/s [E] is deflected so it moves at 6.0 m/s [N]. The deflection takes 0.50 s. Find Δv and aₐᵥ.
Δv = 6.0 [N] + 6.0 [W]: √72 ≈ 8.5 m/s [N 45° W]. aₐᵥ = 8.5 ÷ 0.50 ≈ 17 m/s² [N 45° W].
6A student walks 300 m [S] then 400 m [E] in 500 s. Find her average speed and average velocity.
Average speed = 700 ÷ 500 = 1.4 m/s. Δd = √(300² + 400²) = 500 m at tan⁻¹(400 ÷ 300) ≈ 53° from south toward east; vₐᵥ = 500 ÷ 500 = 1.0 m/s [S 53° E].
More practice: warm-up
1With east as +x, what is the sign of the x component of a vector pointing [W 10° S]?
Negative (and its y component is also negative).
2A vector of 10 m points [E]. What are its components?
3Components are x = 3.0 m and y = −4.0 m. What is the magnitude?
4Can an object accelerate while its speed stays the same? Explain in one sentence.
Yes, if its direction changes, its velocity changes, so it accelerates.
5A boat heading across a river has a crossing velocity of 3.0 m/s. The river is 90 m wide. How long does it take to cross?
More practice: on your own
1A hiker walks 5.0 km [N] and then 7.0 km [E 30° S]. Find her displacement.
x = 7.0 cos 30° = 6.06; y = 5.0 − 7.0 sin 30° = 1.50. R = √(6.06² + 1.50²) ≈ 6.2 km; θ = tan⁻¹(1.50 ÷ 6.06) ≈ 14°: 6.2 km [E 14° N].
2Find the components of 45 m [S 20° W].
South (y) = −45 cos 20° ≈ −42.3 m; west (x) = −45 sin 20° ≈ −15.4 m.
3A swimmer swims at 1.0 m/s [N] across a 50 m river flowing at 0.60 m/s [W]. Find her ground velocity, crossing time and drift.
|v| = √(1.0² + 0.60²) ≈ 1.2 m/s, tan⁻¹(0.60 ÷ 1.0) ≈ 31°: [N 31° W]. Δt = 50 ÷ 1.0 = 50 s. Drift = 0.60 × 50 = 30 m [W].
4A pilot wants to fly due north. The plane's airspeed is 300 km/h and the wind blows at 60 km/h [W]. Which heading should she use, and what is her ground speed?
She must point partly east to cancel the wind: sin θ = 60 ÷ 300, θ ≈ 11.5°. Heading [N 12° E]. Ground speed = √(300² − 60²) ≈ 294 km/h [N].
5Explain why pointing a canoe straight across a river is the fastest way across, even though you land downstream.
The crossing time depends only on the velocity component across the river. Pointing straight across puts all of your paddling speed into that component, so Δt is as small as possible. The current only adds a sideways component.
Challenge
1The canoeist in Example 2 (2.0 m/s in still water; current 1.5 m/s [E]; river 120 m) wants to land directly across from her start. Which heading must she use, and how long does the crossing take?
Her eastward drift must be cancelled: 2.0 sin θ = 1.5, θ = sin⁻¹(0.75) ≈ 49°. Heading [N 49° W]. Her northward component is √(2.0² − 1.5²) ≈ 1.32 m/s, so Δt = 120 ÷ 1.32 ≈ 91 s (longer than 60 s).
2A car moving at 20 m/s [E] turns and is moving at 20 m/s [S] after 4.0 s. Find its average acceleration.
Δv = 20 [S] + 20 [W] = √800 ≈ 28.3 m/s [S 45° W]. aₐᵥ = 28.3 ÷ 4.0 ≈ 7.1 m/s² [S 45° W].
3A ship's position is 5.0 km [N 30° E] of a port. Later it is 8.0 km [E] of the port. Find its displacement.
d₁: x = 5.0 sin 30° = 2.50, y = 5.0 cos 30° = 4.33. d₂: x = 8.0, y = 0. Δd = d₂ − d₁: x = 5.50, y = −4.33. R = √(5.50² + 4.33²) ≈ 7.0 km; θ = tan⁻¹(4.33 ÷ 5.50) ≈ 38°: 7.0 km [E 38° S].
Quiz
1. The x component (east +) of 10 m [E 30° N] is:
2. A resultant has Rx = −6.0 km and Ry = +8.0 km. Its direction is:
3. A boat aims straight across a river at 4.0 m/s; the current is 3.0 m/s. Its speed relative to the ground is:
4. A car moves at constant speed around a curve. Its acceleration is:
5. A river is 60 m wide. A swimmer aims straight across at 1.5 m/s while the current flows at 2.0 m/s. How long does she take to cross?
6. To find Δv in two dimensions, you calculate:
Big idea: Breaking vectors into components (B2.5) turns two-dimensional motion into two one-dimensional problems, which students solve algebraically (B2.7). This also prepares directly for projectile motion in the next two lessons.
How to help: For every component problem, ask to see the sketch with the angle marked. Most errors are sine/cosine mix-ups or missing negative signs, and the sketch catches both.
Common difficulties:
- Always using cos for x and sin for y, even when the angle is measured from north.
- Forgetting negative signs for west and south.
- Adding magnitudes instead of components.
- In river problems, using the diagonal speed to find the crossing time instead of the across component.