Unit 1 · Lesson 2a

2aDistance and Displacement

Understand the critical difference between distance (total path) and displacement (change in position), and master scalar vs vector quantities.

Distinguishing distance from displacement is the first step in kinematics. Navigation systems, sports analytics, and spacecraft trajectory planning all depend on correctly computing displacement — not just total distance traveled.

Lesson Overview

Motion begins with describing where an object is and how far it has traveled. In this lesson you will distinguish between distance — the total path length — and displacement — the straight-line change in position with direction. Understanding the difference between these scalar and vector quantities is the foundation of all kinematics.

Key Concepts

Scalar

A quantity with magnitude only (e.g., distance, speed, mass)

Vector

A quantity with both magnitude and direction (e.g., displacement, velocity, force)

Distance (d)

Total length of the path traveled — always positive

Displacement (Δx)

Change in position: Δx = x_f − x_i — can be positive, negative, or zero

Reference Point

The origin from which position is measured

Position (x)

Location of an object relative to a reference point, with sign indicating direction

Example 1

A student walks 4 m east, then 3 m west. What is (a) the total distance traveled and (b) the displacement?

Answer:(a) Distance = 4 + 3 = 7 m (total path length, always positive). (b) Displacement = 4 m east − 3 m east = 1 m east (or +1 m if east is positive).
Example 2

A runner completes one full lap of a 400 m circular track. What is (a) the distance and (b) the displacement?

Answer:(a) Distance = 400 m — the runner traveled the full perimeter. (b) Displacement = 0 m — the runner returned to the starting position, so the change in position is zero.
Example 3

An object starts at position x = −5 m and moves to x = +8 m. Find the displacement.

Answer:Δx = x_f − x_i = 8 − (−5) = 13 m in the positive direction.
Example 4

A car travels 60 km north, then 80 km east. Find (a) the total distance and (b) the magnitude of the displacement.

Answer:(a) Distance = 60 + 80 = 140 km. (b) The displacement is the straight-line distance from start to finish: |Δr| = √(60² + 80²) = √(3600 + 6400) = √10 000 = 100 km, directed northeast.
Example 5

Is it possible for displacement to be greater than distance? Explain.

Answer:No. Displacement is the straight-line distance between start and end points; distance is the total path length. The shortest path between two points is a straight line, so displacement ≤ distance. They are equal only when motion is in a straight line without reversing direction.
Guided Problem 1

A dog runs 12 m north, then 5 m south. Find the distance and displacement.

Hint: Distance adds all path segments. Displacement = final position − initial position (use a sign convention).

Guided Problem 2

An object moves from x = +10 m to x = +3 m. Is the displacement positive or negative? What is its magnitude?

Hint: Δx = x_f − x_i. Check the sign of the result.

Guided Problem 3

Can an object have zero displacement but non-zero distance? Give an example.

Hint: Think about a round trip.

Guided Problem 4

A hiker walks 3 km east, 4 km north, then 3 km west. What is the total distance? What is the magnitude of the displacement?

Hint: For displacement, find the straight-line distance from start to finish using the Pythagorean theorem if needed.

Guided Problem 5

Why is displacement a vector quantity but distance is a scalar?

Hint: Think about what additional information displacement carries that distance does not.

Key Vocabulary

Distance

The total length of the path traveled by an object, regardless of direction. A scalar quantity.

Example: A car driving around a 2 km loop has traveled a distance of 2 km.

Displacement

The change in position of an object: Δx = x_f − x_i. A vector quantity with magnitude and direction.

Example: If you walk 5 m east then 5 m west, your displacement is 0 m.

Scalar

A physical quantity described by magnitude (size) only, with no direction.

Example: Temperature, mass, speed, and distance are all scalars.

Vector

A physical quantity described by both magnitude and direction.

Example: Displacement, velocity, acceleration, and force are all vectors.

Interactive Practice — 5 Questions

1

A person walks 6 m east and then 6 m west. What is the displacement?

2

Which of the following is a vector quantity?

3

An object moves from x = −3 m to x = +7 m. What is the displacement?

4

A runner completes a 200 m race in a straight line. The distance and displacement are:

5

Which statement about displacement is always true?

Independent Practice

1

A cyclist rides 15 km north, then 20 km south. Calculate (a) the total distance and (b) the displacement, including direction.

2

Explain in your own words why a marathon runner who finishes at the starting line has zero displacement but a distance of 42.2 km.

3

An ant starts at position x = +2 cm, walks to x = −6 cm, then returns to x = +1 cm. Find (a) total distance and (b) displacement.

4

A ship sails 30 km east and then 40 km north. Find the magnitude and direction of the displacement using the Pythagorean theorem and trigonometry.

5

★ In three dimensions, a drone flies 10 m east, 10 m north, and 10 m straight up. Calculate the magnitude of the total displacement. How does this compare to the total distance traveled?

Challenge
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Common Mistakes

Treating displacement and distance as the same thing.

Distance is the total path length (scalar); displacement is the straight-line change in position (vector). They are equal only for straight-line motion without reversal.

Forgetting the sign (direction) when calculating displacement.

Always use Δx = x_f − x_i and assign positive/negative based on your chosen direction convention.

Saying displacement is always positive.

Displacement is negative when the final position is in the negative direction relative to the initial position.

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Math Tips

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Always define a positive direction first (e.g., east = positive, north = positive). Then assign signs consistently. Displacement = x_final − x_initial.

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For 2D displacement, use the Pythagorean theorem: |Δr| = √(Δx² + Δy²). For direction, use θ = arctan(Δy/Δx).