Solve problems involving velocity, displacement and force by representing each quantity as a vector

Learning goal: Solve problems involving velocity and other quantities that can be represented by vectors, such as displacement and force, by writing each quantity in component form, combining the components, and converting the result back to a magnitude and a direction with units.

Scope note. N.VM.3 is marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate a standard "for students who are planning on taking advanced courses". It is taught here in full, as extension content for students on that path.

Why velocity is a vector. Speed tells you how fast; velocity tells you how fast and which way. An aircraft's airspeed indicator reads 240 miles per hour, but where the aircraft actually goes depends on the air it is flying through. If the air itself is moving, the aircraft's velocity relative to the ground is the sum of two vectors: its velocity relative to the air, and the air's velocity relative to the ground (the wind). The same reasoning applies to a boat in a current, a swimmer in a river, or a person walking along a moving walkway. Whenever two motions combine, the resulting velocity is the vector sum, never the sum of the speeds, unless both motions are in the same direction.

The method. Every velocity problem in this lesson follows the same four steps. First, choose axes: usually east is the positive x-direction and north is the positive y-direction. Second, write each velocity in component form ⟨v_x, v_y⟩. A velocity of magnitude s pointing at angle θ measured counterclockwise from east has components ⟨s cos θ, s sin θ⟩. Third, add the components: horizontal with horizontal, vertical with vertical. Fourth, convert the resulting vector back to a magnitude, √(v_x² + v_y²), and a direction, using the arctangent of the ratio of the components together with the quadrant. State the units on every quantity and say what the direction is measured from.

A fully worked model: aircraft and wind. An aircraft flies with airspeed 240 mph heading due east. A steady wind of 70 mph blows toward the north. What is the aircraft's ground speed and its true direction of travel? Axes: east is x, north is y. The aircraft's velocity relative to the air is ⟨240, 0⟩ mph. The wind's velocity is ⟨0, 70⟩ mph. Ground velocity: ⟨240 + 0, 0 + 70⟩ = ⟨240, 70⟩ mph. Ground speed: √(240² + 70²) = √(57,600 + 4,900) = √62,500 = 250 mph. Direction: the vector points east and a little north; tan θ = 70/240, so θ = arctan(70/240) ≈ 16.26° north of east. Notice that the ground speed is 250 mph, not 240 + 70 = 310 mph: the wind is at right angles to the heading, so it does not add its whole magnitude to the forward speed. The 240 and 70 are the legs of a right triangle and 250 is its hypotenuse. Keep the units: the answer is a velocity of 250 mph in the direction 16.26° north of east.

Describing a wind. Weather reports usually describe wind by the direction it comes from. A "north wind" blows from the north toward the south. In vector work you need the direction the air moves toward, so a north wind of 70 mph is the vector ⟨0, −70⟩. Read the problem wording carefully; a reversed wind direction silently changes every later number.

Resolving a velocity into components. A cyclist rides at 60 km/h in a direction 25° north of east. Components: ⟨60 cos 25°, 60 sin 25°⟩ ≈ ⟨54.38, 25.36⟩ km/h. The eastward component, 54.38 km/h, is how fast the cyclist's east coordinate grows, and 25.36 km/h is how fast the north coordinate grows. Check: √(54.38² + 25.36²) ≈ √(2,957 + 643) = √3,600 = 60 km/h. Components are the tool that makes the arithmetic possible; magnitudes at different angles cannot simply be added.

Other vector quantities. Displacement is a change of position: walk 2 km north and then 3 km at 60° north of east and the total displacement is the vector sum of the two legs, while the distance walked is the scalar 5 km. Force is a push or pull with a direction; two forces acting on the same object combine as vectors, and their vector sum is called the resultant. Treating the object as a single point (a particle), the resultant has the same effect as the separate forces; for a real object of some size, where each force is applied also decides whether it turns, which this lesson does not model. A 30 N force toward the east and a 40 N force toward the north together act like one force of √(30² + 40²) = 50 N directed at arctan(40/30) ≈ 53.13° north of east. If the two forces were aligned the resultant would be 70 N; if opposite, 10 N. Direction matters as much as size.

Choosing a heading to compensate for a current. Sometimes the resultant is known and one of the parts is wanted. A pilot wants to travel due north with airspeed 200 mph while a wind of 40 mph blows toward the west. Heading straight north would give a ground velocity ⟨−40, 200⟩, which drifts west. To cancel the drift the aircraft must aim east of north by an angle θ with 200 sin θ = 40, so sin θ = 0.2 and θ ≈ 11.54°. The eastward component of the airspeed, 200 sin θ = 40 mph, cancels the wind exactly, and the northward component is the ground speed: 200 cos θ = √(200² − 40²) = √38,400 ≈ 195.96 mph. The aircraft moves north more slowly than its airspeed, because part of its effort is spent holding position against the wind.

The misconception: add the speeds. A boat that makes 5 m/s in still water heading straight across a river with a 2 m/s current is not moving at 7 m/s and not at 3 m/s. Its velocity relative to the bank is ⟨2, 5⟩ m/s (taking downstream as x and across as y), with magnitude √29 ≈ 5.39 m/s. The speeds add only when the directions are the same; they subtract only when the directions are opposite. In every other case you must add components. A second common error is to compute the crossing time with the resultant speed. The time to cross depends only on the component perpendicular to the bank: a river 100 m wide is crossed in 100 ÷ 5 = 20 s regardless of the current, and during those 20 s the current carries the boat 2 × 20 = 40 m downstream.

Checking a result. A velocity answer should pass three checks: the magnitude lies between the difference and the sum of the magnitudes being combined; for nonzero vectors that are not opposite, the direction lies inside the smaller angle between the two directions being combined, closer to the larger one; and converting the final magnitude and direction back to components reproduces the component sum. For the aircraft, 250 lies between 170 and 310, 16.26° lies between 0° (east) and 90° (north) and is nearer to east because the 240 mph part is larger, and ⟨250 cos 16.26°, 250 sin 16.26°⟩ ≈ ⟨240, 70⟩.

Worked, river crossing: A boat heads straight across a river at 5 m/s relative to the water. The current runs at 2 m/s. The river is 100 m wide. Take downstream as x and across as y. Boat relative to water: ⟨0, 5⟩ m/s. Current: ⟨2, 0⟩ m/s. Velocity relative to the bank: ⟨2, 5⟩ m/s, magnitude √(4 + 25) = √29 ≈ 5.39 m/s. The drift angle from straight across is arctan(2/5) ≈ 21.80° downstream. Crossing time uses the across component only: 100 m ÷ 5 m/s = 20 s. Downstream drift in that time: 2 m/s × 20 s = 40 m. The boat lands 40 m downstream after 20 s, having moved √(100² + 40²) ≈ 107.7 m relative to the bank (through the water it covered only the 100 m it steered across).

Worked, resultant of two forces: Two ropes pull a crate along a floor: 30 N toward the east and 40 N toward the north. Resultant: ⟨30, 40⟩ N, magnitude √(900 + 1,600) = √2,500 = 50 N, direction arctan(40/30) ≈ 53.13° north of east. Treating the crate as a particle and considering only these two pulls, they have the same combined effect as a single 50 N force in that direction (where the ropes are attached would also matter for whether a real crate twists, and friction would change how it actually moves). If instead both ropes pulled east, the resultant would be 70 N east; the 50 N answer is smaller because the pulls are at right angles.

Worked, displacement along two legs: A hiker walks 2.0 km due north, then 3.0 km in a direction 60° north of east. First leg: ⟨0, 2⟩ km. Second leg: ⟨3 cos 60°, 3 sin 60°⟩ ≈ ⟨1.5, 2.598⟩ km. Total displacement: ⟨1.5, 4.598⟩ km, to three decimal places. Magnitude: √(1.5² + 4.598²) ≈ √(2.25 + 21.142) ≈ √23.392 ≈ 4.84 km. Direction: arctan(4.598/1.5) ≈ 71.93° north of east. The hiker is 4.84 km from the start after walking 5.0 km; distance walked and displacement magnitude differ because the legs were not in a straight line.

Worked, heading to cancel a wind: A pilot wants to fly due north. Airspeed is 200 mph and the wind blows toward the west at 40 mph. Let the heading be θ east of north. The eastward component of the aircraft's air velocity is 200 sin θ, and it must cancel the wind's ⟨−40, 0⟩, so 200 sin θ = 40, sin θ = 0.2, θ ≈ 11.54°. Ground speed is the northward component, 200 cos θ = √(200² − 40²) = √38,400 ≈ 195.96 mph. Check the components: ⟨200 sin 11.54° − 40, 200 cos 11.54°⟩ ≈ ⟨40 − 40, 195.96⟩ = ⟨0, 195.96⟩, due north as intended.

Recording task: Draw the aircraft problem to scale, 1 cm for 20 mph: a 12 cm arrow pointing east for the 240 mph airspeed, then from its tip a 3.5 cm arrow pointing north for the 70 mph wind, and the resultant from the start of the first arrow to the tip of the second. Measure the resultant with a ruler; it should be 12.5 cm, representing 250 mph. Measure its angle from east with a protractor; it should be about 16°. Record a table with three rows (air velocity, wind velocity, ground velocity), each with its components, magnitude and direction, and write one sentence explaining why the ground speed is 250 mph and not 310 mph.

Independent performance: Write and solve three original problems of your own: one aircraft or boat problem in which the wind or current is at right angles to the heading; one displacement problem with two legs at different angles, at least one of which is not along an axis; and one force problem with two forces whose resultant you must find, where the two forces are neither aligned nor at right angles. For each problem state the axes you chose, write every quantity in component form with units, add the components, convert the result to a magnitude and a direction, and run the three checks: the magnitude lies between the difference and the sum of the parts, the direction lies inside the smaller angle between the directions of the parts, and converting your answer back to components reproduces the sum. Success criteria: units on every quantity; every direction stated with its reference ("north of east", "downstream from straight across"); every magnitude computed with the Pythagorean theorem from components and not by adding speeds; and all three checks written out and passing.