B. Dynamics · Dynamics: Forces and Motion in a Plane

Vectors in Two Dimensions: Adding, Subtracting and Changing Velocity

B2.1 B2.2 A1.12 A1.13About 70 minutes

Learning goals

I can resolve a vector into x and y components using compass directions or standard-position angles.
I can add any number of vectors by components and state the resultant with a magnitude and direction.
I can subtract vectors and find the change in velocity, Δv = v₂ − v₁.
I can find average acceleration in two dimensions and explain why a change in direction is an acceleration.
I can keep extra digits during a calculation and round the final answer to the right number of significant digits.

Words to know

vector
A quantity with both magnitude (size) and direction, such as displacement, velocity, acceleration or force.

scalar
A quantity with magnitude only, such as distance, speed, mass or time.

components
Two perpendicular vectors (usually along x and y) that add tip to tail to give the original vector.

resultant
The single vector that has the same effect as two or more vectors added together.

standard-position angle
An angle measured counterclockwise from the positive x-axis (east). [N 30° E] is 60° in standard position.

change in velocity (Δv)
The vector v₂ − v₁. It is found by adding the opposite of the first velocity to the second.

average acceleration
The change in velocity divided by the time interval: a = Δv ÷ Δt. It points the same way as Δv.

Off the boards

A hockey puck slides toward the boards at 8.0 m/s and bounces off at 6.0 m/s. It slowed down a little, but something much bigger happened: its direction changed. To the boards, that change took a huge push. To a physicist, the puck had a large acceleration, even though its speed barely changed.

This unit, Dynamics, is about forces and motion in a plane: pucks, projectiles, cars on curves, blocks on ramps and rides that spin. Every one of those problems is solved with vectors. So we start by sharpening the vector tools you already have, and adding one new one: subtracting vectors to find a change in velocity.

Remember from Grade 11

In the SPH3U Kinematics unit you wrote directions like [N 30° E] (start facing north, turn 30° toward east), drew vectors tip to tail, and added them by components: Rx = sum of the x parts, Ry = sum of the y parts, R = √(Rx² + Ry²), and θ = tan⁻¹(|Ry| ÷ |Rx|). You also met relative velocity for canoes on rivers. Everything in this lesson builds on those skills.

Two ways to write a direction

Physics texts and university courses use two systems, and you should be fluent in both.

Compass form: [N 30° E], [S 20° W], [W 15° N]. Good for maps and navigation.
Standard position: one angle θ measured counterclockwise from +x (east). Then the components are always Ax = A cos θ and Ay = A sin θ, and the signs come out automatically. [N 30° E] is θ = 60°, [W 15° N] is θ = 165°, and [S 20° W] is θ = 250°.

To convert compass form to standard position, sketch the vector first. Then measure from east, going counterclockwise. With a sketch, you will catch most sign mistakes before they happen.

The same vector described two ways: 30° east of north, or 60° counterclockwise from +x. Its components are Ax = 250 cos 60° = 125 km and Ay = 250 sin 60° = 217 km.

Worked example

Example 1: Adding by components

A bush plane flies 250 km [N 30° E] from Timmins and then 180 km [W]. Find its resultant displacement.

Let east be +x and north be +y. In standard position, the first leg is at 60° and the second is at 180°.
Leg 1: x = 250 cos 60° = 125.0 km; y = 250 sin 60° = 216.5 km.
Leg 2: x = 180 cos 180° = −180.0 km; y = 0.
Rx = 125.0 − 180.0 = −55.0 km. Ry = 216.5 km.
R = √((−55.0)² + 216.5²) = √(3025 + 46 872) = √49 897 ≈ 223 km.
Rx is negative (west) and Ry is positive (north), so the resultant is in the northwest quadrant. Measured from north: θ = tan⁻¹(55.0 ÷ 216.5) = 14.3°.

223 km [N 14° W] (or 104° in standard position).

Subtracting vectors

To subtract a vector, add its opposite. The opposite of B has the same magnitude and points the other way:

A − B = A + (−B)

In components it is even simpler: subtract the x parts and subtract the y parts.

The most important subtraction in dynamics is the change in velocity:

Δv = v₂ − v₁ and the average acceleration is a = Δv ÷ Δt

Acceleration points in the direction of Δv, not in the direction of motion. If an object changes direction, Δv is not zero even when the speed stays the same. A car driving around a corner at a steady speed is accelerating.

To find Δv, place the opposite of v₁ at the tip of v₂. Δv runs from the tail of v₂ to the tip of −v₁. For a car turning left from east to north, Δv points into the turn (northwest).

Worked example

Example 2: A corner at constant speed

A car drives at 15 m/s [E], turns left onto a road going north, and is soon moving at 15 m/s [N]. The turn takes 4.0 s. Find the average acceleration.

Δv = v₂ − v₁ = 15 m/s [N] + 15 m/s [W] (the opposite of 15 m/s [E]).
Components: Δvx = 0 − 15 = −15 m/s; Δvy = 15 − 0 = +15 m/s.
|Δv| = √(15² + 15²) = 21.2 m/s. The direction is 45° between west and north: [N 45° W].
a = Δv ÷ Δt = 21.2 ÷ 4.0 = 5.3 m/s² [N 45° W].

5.3 m/s² [N 45° W]. The speed never changed, yet the car accelerated toward the inside of the turn. You will meet this again in Uniform Circular Motion, later in this unit.

Worked example

Example 3: The puck off the boards

A puck moving at 8.0 m/s [E 30° N] hits the boards (which run north–south) and leaves at 6.0 m/s [W 30° N]. It touches the boards for 0.025 s. Find the average acceleration.

Standard angles: v₁ is at 30°, v₂ is at 150°.
v₁: x = 8.0 cos 30° = 6.93 m/s; y = 8.0 sin 30° = 4.00 m/s.
v₂: x = 6.0 cos 150° = −5.20 m/s; y = 6.0 sin 150° = 3.00 m/s.
Δvx = −5.20 − 6.93 = −12.12 m/s; Δvy = 3.00 − 4.00 = −1.00 m/s.
|Δv| = √(12.12² + 1.00²) = 12.2 m/s. Both components are negative, so Δv points west and slightly south: θ = tan⁻¹(1.00 ÷ 12.12) = 4.7° south of west.
a = 12.2 ÷ 0.025 = 487 m/s², which rounds to 4.9 × 10² m/s² (two significant digits).

About 4.9 × 10² m/s² [W 4.7° S]. The boards push the puck almost straight west. The small southward part comes from the puck losing some of its northward speed.

Precision and significant digits

Keep at least one or two extra digits in every intermediate step (write 12.12, not 12), and round only the final answer. The answer gets as many significant digits as the least precise measurement you were given: in Example 3, 8.0 and 6.0 have two, so the answer has two. Angles are usually given to the nearest degree or tenth of a degree.

Just for fun

Why did the vector break up with the scalar?

It wanted a partner with a sense of direction.

Hands-on activity

Map a walk with vectors

You need: a measuring tape or metre stick, a phone compass app (or a magnetic compass), paper, a pencil and a protractor.

Measure your normal pace: walk 10 steps, measure the distance, and divide by 10.
Outdoors with a grown-up (a park or quiet sidewalk), or indoors through several rooms, walk three straight legs in different directions. For each leg, count paces and read the compass direction. Record each leg as a vector, such as 24 m [N 20° E].
At home, find the resultant by components. Then draw a scale diagram (for example 1 cm = 2 m) and measure the resultant with a ruler and protractor.
Test it: from the end point, face the opposite of your resultant's direction and pace back. How close to your starting point do you land? List two sources of error (pace length, compass reading, walking off line).

Stay on sidewalks and paths, and watch for traffic.

Your turn

Resolve a force of 40 N [S 25° W] into x (east) and y (north) components.

Add 5.0 km [E] and 12 km [N].

Write the direction [W 20° S] as a standard-position angle.

A ball moving at 20 m/s [N] is caught and thrown back at 20 m/s [S]. Find Δv.

An object's velocity changes from 6.0 m/s [E] to 8.0 m/s [N] in 2.0 s. Find the average acceleration.

Explain how a car moving at a constant speed of 60 km/h can be accelerating.

More practice: warm-up

Is speed a scalar or a vector?

Give the opposite of 5.0 m/s [N 30° E].

Write [N 30° E] in the form [E ? N].

What are the x and y components of 10 m [E]?

A vector has components x = 3.0 and y = 4.0. What is its magnitude?

More practice: on your own

A cyclist rides 3.0 km [N] and then 4.0 km [E 30° S]. Find her resultant displacement.

Find A − B, where A = 7.0 m/s [E] and B = 7.0 m/s [N 60° E].

A tennis ball hits a wall at 12 m/s [E] and rebounds at 10 m/s [W]. Contact lasts 0.040 s. Find the average acceleration.

A plane flying at 300 km/h [N] turns until it flies at 300 km/h [E]. The turn takes 30 s. Find the average acceleration in m/s².

Three ropes pull on a crate: 20 N [E], 30 N [N] and 10 N [W 45° S]. Find the resultant force.

Challenge

An object moves in a circle at a constant 10 m/s. In 0.50 s its velocity turns through 60°. Find the magnitude of the average acceleration. (Hint: what kind of triangle do v₁, v₂ and Δv make?)

Vector A = 15 units [N 20° E]. A + B = 25 units [E]. Find B.

Show with an example that |A − B| is usually not equal to |A| − |B|. When are they equal?

Check your understanding

Quiz

The direction [N 30° E] written as a standard-position angle (counterclockwise from +x) is:

A car drives around a curve at a constant 20 m/s. Which statement is true?

To find v₂ − v₁ with a diagram, you:

A vector has Rx = −6.0 m and Ry = −8.0 m. Its direction is:

A ball hits a wall at 5.0 m/s [E] and bounces back at 4.0 m/s [W]. The magnitude of its change in velocity is:

In a calculation, the given values are 8.0 m/s, 6.0 m/s and 0.025 s. The final answer should have:

For parents and teachers

Big idea: Vectors are the language of two-dimensional dynamics (B2.2). This lesson reviews SPH3U vector addition and adds vector subtraction and change in velocity, which the rest of the unit (and the circular-motion lessons in particular) depends on. It also practises vector diagrams (A1.12) and significant digits (A1.13).

How to help: Ask your teen to sketch every vector before calculating and to say aloud which quadrant the answer is in. A sketch catches most sign mistakes.

Common difficulties:

Using sine where cosine belongs because the compass angle is measured from north, not from east. Converting to a standard-position angle avoids this.
Subtracting speeds instead of velocities (a ball that bounces back has a large Δv, not a small one).
Thinking acceleration must point in the direction of motion. It points along Δv.
Rounding too early and getting a final answer that is a little off.