B3.2 Distinguish scalar and vector quantities in motion

Ontario Grade 11 Physics

Kinematics

SPH3U B3.2: Tell apart quantities that have magnitude alone and quantities that also have direction.

A number can tell you how much motion occurred without telling you which way it occurred. For example, “

5\ \mathrm{m}

” gives a length, while “

5\ \mathrm{m}

east” gives a length and a direction. Physics uses the terms scalar and vector to make this distinction clear. In this lesson, we will use these terms to describe motion. We will define the object being studied and choose a direction before using signed values.

What you will learn

Define scalar and vector quantities in the context of motion.
Identify distance, speed, displacement, and velocity as scalars or vectors.
Use direction and a sign convention to describe motion.
Explain why two quantities with the same magnitude may still describe different motion.

1. Prerequisite bridge: amount and direction

A measurement includes a number and a unit. The unit tells us what kind of quantity is being measured. For example, metres measure length and seconds measure time. A direction tells us which way something points or moves, such as east, west, left, or right.

A scalar is a quantity described by a magnitude and a unit. Magnitude means the size or amount of the quantity. A scalar does not include direction. Distance, time, and speed are scalars.

A vector is a quantity described by a magnitude, a unit, and a direction. Displacement and velocity are vectors. A vector’s direction is part of its meaning, not an optional detail.

Before describing motion, identify the system: the object whose motion you are considering. Then choose a reference direction. A reference direction is the direction defined as positive for the description. For a straight path, we might choose east as positive. West is then negative. The signs are a way to keep track of direction; they do not make a distance or a time negative.

Scalar: magnitude and unit, with no direction.
Vector: magnitude, unit, and direction.
Define the object and positive direction before using signed vector values.

2. Motion quantities: distance and displacement

Distance is the total length of the path travelled. It is a scalar, so it has no direction. Distance is zero or positive.

Displacement describes the change in position from the starting point to the ending point. It is a vector, so it needs a direction. For straight-line motion, we can represent displacement with a signed value after choosing a positive direction. A positive displacement points in the positive direction; a negative displacement points in the opposite direction.

The route matters for distance, but only the starting and ending positions matter for displacement. An object can travel a long distance and still have zero displacement if it returns to its starting point.

Consider a person who walks east and then turns around and walks west. A motion sketch can show the path as

\text{start} \;\longrightarrow\; \text{east} \;\longrightarrow\; \text{turn} \;\longleftarrow\; \text{west}

. The arrows show directions. The total path length is the distance; the arrow from the start directly to the finish represents displacement.

\Delta d = d_f - d_i

Distance adds the lengths of the path taken.
Displacement compares the final position with the initial position.
A return to the starting position gives zero displacement, even if the distance is not zero.

3. Motion quantities: speed and velocity

Speed describes how much distance is travelled per unit of time. It is a scalar. Average speed is found by dividing total distance by the time taken. Its SI unit is metres per second, written

\mathrm{m/s}

.

Velocity describes displacement per unit of time. It is a vector. Average velocity includes a direction, or a sign that represents direction on a chosen straight-line axis. Its SI unit is also

\mathrm{m/s}

. The same unit does not make speed and velocity the same quantity.

For straight-line motion, use a signed displacement when calculating average velocity. A positive result points in the chosen positive direction. A negative result points in the opposite direction. The result for average speed cannot be negative because distance and elapsed time are non-negative.

A useful way to classify a quantity is to ask: “Would I need to know which way for this description to be complete?” If yes, it is a vector. “The object moved at

3\ \mathrm{m/s}

” gives speed. “The object moved at

3\ \mathrm{m/s}

east” gives velocity.

\text{average speed} = \frac{\text{total distance}}{\text{elapsed time}},\quad \text{average velocity} = \frac{\text{displacement}}{\text{elapsed time}}

Average speed uses total distance and elapsed time.
Average velocity uses displacement and elapsed time.
Both use $\mathrm{m/s}$ , but velocity also requires direction.

4. A reliable way to tell them apart

Start by naming the quantity, rather than looking only at its unit. Metres can describe distance or displacement. Metres per second can describe speed or velocity. In each pair, the unit alone cannot tell you whether the quantity is a scalar or a vector.

Next, check whether the description includes direction. If it does, make sure the direction matches the chosen reference direction. For a one-dimensional path, write the direction in words or use a positive or negative sign after stating what positive means.

Finally, check what information the quantity represents. Distance and speed describe amounts without direction. Displacement and velocity include direction. Do not use “distance” when you mean a change in position, or “speed” when you need to report a direction.

For this lesson, a motion quantity is not a vector just because it has a sign in an unrelated context. The sign is useful here because we defined it to represent direction along our chosen line. Always state that convention.

The unit alone does not identify a quantity as scalar or vector.
Direction is required for a complete vector description.
State what positive means before interpreting positive or negative values.

Motion quantities at a glance

Quantity	What it describes	Scalar or vector?	Direction needed?
Distance	Total path length	Scalar	No
Displacement	Change from initial to final position	Vector	Yes
Speed	Distance travelled per time	Scalar	No
Velocity	Displacement per time	Vector	Yes

Worked example

1. One trip, two different descriptions

A cart is the system. It rolls

12.0\ \mathrm{m}

east, then

5.0\ \mathrm{m}

west, in

10.0\ \mathrm{s}

. Find its total distance and displacement. Then find its average speed and average velocity. Choose east as positive.

Set the direction
East is positive, so west is negative. The cart’s two path lengths are both positive when adding distance, but the second part has a negative direction when finding displacement.
$\text{east}=+,\quad \text{west}=-$
Find distance and displacement
Distance is the total path length. Displacement is the signed change in position, so the westward part subtracts from the eastward part.
$d=12.0\ \mathrm{m}+5.0\ \mathrm{m}=17.0\ \mathrm{m},\quad \Delta d=12.0\ \mathrm{m}-5.0\ \mathrm{m}=+7.0\ \mathrm{m}$
Find average speed and velocity
Use total distance for average speed and displacement for average velocity. Keep the time unit in each division. The positive velocity means east under the chosen convention.
$v_{\mathrm{avg,speed}}=\frac{17.0\ \mathrm{m}}{10.0\ \mathrm{s}}=1.70\ \mathrm{m/s},\quad v_{\mathrm{avg}}=\frac{+7.0\ \mathrm{m}}{10.0\ \mathrm{s}}=+0.70\ \mathrm{m/s}$

Answer: The distance is

17.0\ \mathrm{m}

, a scalar. The displacement is

7.0\ \mathrm{m}

east, a vector. The average speed is

1.70\ \mathrm{m/s}

, a scalar. The average velocity is

0.70\ \mathrm{m/s}

east, a vector.

Check: Each speed or velocity has units of metres divided by seconds, or

\mathrm{m/s}

. The displacement is smaller than the distance because the cart reverses direction. The velocity is eastward because the cart ends east of where it started. The values use three significant figures for distance and time, and two for displacement and velocity, limited by subtracting the measured path lengths.

Worked example

2. Same magnitude, opposite directions

A cyclist is the system. In one part of a trip, the cyclist’s displacement is

18.0\ \mathrm{m}

north in

6.00\ \mathrm{s}

. In another part, the cyclist’s displacement is

18.0\ \mathrm{m}

south in

6.00\ \mathrm{s}

. Find the average velocity for each part. Choose north as positive.

Assign signs
North is the positive direction. Therefore, the northward displacement is positive and the southward displacement is negative.
$\Delta d_{\mathrm{north}}=+18.0\ \mathrm{m},\quad \Delta d_{\mathrm{south}}=-18.0\ \mathrm{m}$
Calculate each velocity
Average velocity is displacement divided by elapsed time. The equal time intervals and equal displacement magnitudes give equal velocity magnitudes, but the signs show opposite directions.
$v_{\mathrm{north}}=\frac{+18.0\ \mathrm{m}}{6.00\ \mathrm{s}}=+3.00\ \mathrm{m/s},\quad v_{\mathrm{south}}=\frac{-18.0\ \mathrm{m}}{6.00\ \mathrm{s}}=-3.00\ \mathrm{m/s}$

Answer: The average velocity is

3.00\ \mathrm{m/s}

north in the first part and

3.00\ \mathrm{m/s}

south in the second part. These are vectors with equal magnitudes and opposite directions.

Check: Both results have units of

\mathrm{m/s}

. Their equal magnitudes fit the equal displacement magnitudes and equal times. Their signs correctly show opposite directions. Three significant figures match the given values.

Worked example

3. Returning to the starting point

A student is the system. The student walks

40.0\ \mathrm{m}

east and then

40.0\ \mathrm{m}

west, taking

32.0\ \mathrm{s}

in total. Find the distance, displacement, average speed, and average velocity. Choose east as positive.

Describe the path and final position
The total distance includes both parts of the path. The displacement adds the signed changes in position. Since the student returns to the starting point, the two changes cancel.
$d=40.0\ \mathrm{m}+40.0\ \mathrm{m}=80.0\ \mathrm{m},\quad \Delta d=+40.0\ \mathrm{m}-40.0\ \mathrm{m}=0\ \mathrm{m}$
Calculate the averages
Divide total distance by elapsed time for average speed. Divide displacement by elapsed time for average velocity. Zero displacement means zero average velocity for the whole trip.
$v_{\mathrm{avg,speed}}=\frac{80.0\ \mathrm{m}}{32.0\ \mathrm{s}}=2.50\ \mathrm{m/s},\quad v_{\mathrm{avg}}=\frac{0\ \mathrm{m}}{32.0\ \mathrm{s}}=0\ \mathrm{m/s}$

Answer: The distance is

80.0\ \mathrm{m}

, the displacement is

0\ \mathrm{m}

, the average speed is

2.50\ \mathrm{m/s}

, and the average velocity is

0\ \mathrm{m/s}

. Distance and speed are scalars; displacement and velocity are vectors.

Check: The speed has units of

\mathrm{m/s}

and is positive because the student travelled a non-zero path. Zero average velocity is reasonable because the student ended at the starting position. The zero displacement has no direction to report.

Common mistakes and how to avoid them

Calling distance and displacement the same thing.

Correction: Distance is the full path length. Displacement compares the final position with the initial position and includes direction.

Assuming that speed and velocity are identical because both use

\mathrm{m/s}

.

Correction: Speed is scalar. Velocity is vector. Velocity also tells the direction of motion.

Giving a negative distance or a negative speed when an object moves west.

Correction: Distance and speed are non-negative scalars. Use direction words or signed values for displacement and velocity after defining the positive direction.

Reporting a velocity magnitude without direction.

Correction: Give the direction in words or state a sign convention and include the signed result.

Thinking that zero displacement means the object did not move.

Correction: Zero displacement means the final position matches the initial position. The object may still have travelled a non-zero distance.

Lesson summary

A scalar has magnitude and a unit but no direction.
A vector has magnitude, a unit, and a direction.
Distance and speed are scalars; displacement and velocity are vectors.
Average speed uses total distance divided by elapsed time. Average velocity uses displacement divided by elapsed time.
Define the system and positive direction before interpreting a signed displacement or velocity.

Check your understanding

Question 1

A runner completes one full lap and stops at the starting point. Which statement must be true for the full lap?

The distance and displacement are both zero.
The distance is greater than zero and the displacement is zero.
The distance is zero and the displacement is greater than zero.
The distance and displacement are equal and greater than zero.

Show answer and explanation

The distance is greater than zero and the displacement is zero.

The runner travelled along a path, so the distance is greater than zero. The runner ended at the starting position, so the displacement is zero.

Question 2

Which description is a vector?

$12\ \mathrm{s}$
$4.0\ \mathrm{m}$ of distance
$6.0\ \mathrm{m/s}$ east
$20\ \mathrm{m}$ of path length

Show answer and explanation

6.0\ \mathrm{m/s}

east

The direction east is included, so

6.0\ \mathrm{m/s}

east describes velocity, a vector. The other choices give amounts without direction.

Question 3

East is chosen as positive. An object has displacement

-9.0\ \mathrm{m}

. What does the sign tell you?

The object travelled a negative distance.
The object moved west by $9.0\ \mathrm{m}$ from its starting position.
The object’s speed was $-9.0\ \mathrm{m/s}$ .
The object ended $9.0\ \mathrm{m}$ east of its starting point.

Show answer and explanation

The object moved west by

9.0\ \mathrm{m}

from its starting position.

A negative displacement points opposite to the positive direction. Since east is positive, the displacement is

9.0\ \mathrm{m}

west. The sign describes direction, not negative distance.

Key terms

Magnitude: The size or amount of a quantity, without its direction.
Scalar: A quantity described by magnitude and unit, with no direction.
Vector: A quantity described by magnitude, unit, and direction.
System: The object whose motion is being studied.
Reference direction: The direction chosen as positive when describing motion along a line.
Distance: The total length of the path travelled.
Displacement: The change from an object's initial position to its final position, including direction.
Speed: Distance travelled per unit of time.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation B3.2. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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B3.2 · Distinguish scalar and vector quantities in motion

Ontario Grade 11 Physics

SPH3U B3.2: Tell apart quantities that have magnitude alone and quantities that also have direction.

What you will learn

1. Prerequisite bridge: amount and direction

2. Motion quantities: distance and displacement

3. Motion quantities: speed and velocity

4. A reliable way to tell them apart

Motion quantities at a glance

1. One trip, two different descriptions

2. Same magnitude, opposite directions

3. Returning to the starting point

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SPH3U

About this lesson and its review