B2.5 Solve distance, position, and displacement problems with vectors

Ontario Grade 11 Physics

Kinematics

Ontario Grade 11 Physics — expectation B2.5

A person can walk a long route and still finish close to where they started. Distance describes the route travelled. Displacement describes the change from the starting position to the ending position. Position tells where an object is relative to a chosen reference point. This lesson uses vectors to find and compare these quantities. First, choose the object, reference point, and coordinate directions. Then use units and directions consistently.

What you will learn

Distinguish scalar distance from vector position and displacement.
Choose and use a clear reference point and positive direction.
Find displacement from initial and final positions.
Add vector components to solve one- and two-dimensional motion problems.

1. Prerequisite bridge: scalars, vectors, and reference points

A scalar has size but no direction. Distance is a scalar. For example, a route length of

12\ \mathrm{m}

does not say which way the motion went.

A vector has both size and direction. Position and displacement are vectors. A position of

+12\ \mathrm{m}

on a straight line means

12\ \mathrm{m}

in the chosen positive direction from the reference point. The plus sign matters because it states direction.

The physical system is the object whose motion is being described. A reference point is the chosen place used to describe position. A coordinate direction is the direction called positive. For motion along a straight line, choose one direction as positive; the opposite direction is negative. For motion on a flat surface, use perpendicular horizontal and vertical directions.

Position depends on the reference point and chosen directions. Displacement does not depend on the route taken, but it does depend on the initial and final positions. Use metres, symbol

\mathrm{m}

, for position, distance, and displacement in SI units.

\text{vector}=(\text{magnitude},\text{direction})

Distance: scalar route length; never negative.
Position: vector location relative to a reference point.
Displacement: vector change from initial to final position.
State the reference point and positive direction before using signed coordinates.

2. The physical situation and the vector rules

Imagine a straight sidewalk with a signpost at the reference point. Call the direction east positive. A position east of the signpost is positive. A position west of it is negative. A labelled number line can show this: west is to the left, the signpost is at

0\ \mathrm{m}

, and east is to the right.

The position of an object is written as

x

. The initial position is

x_i

, and the final position is

x_f

. Displacement is written as

\Delta x

. Subtract the initial position from the final position. This subtraction keeps the direction: a negative result means motion toward the negative direction.

Distance is found by adding the lengths of all parts of the route. When the route changes direction, do not simply treat the signed displacement as the distance. For example, travelling

5\ \mathrm{m}

east and then

2\ \mathrm{m}

west gives a distance of

7\ \mathrm{m}

but a displacement of

3\ \mathrm{m}

east.

In two dimensions, split a displacement into horizontal and vertical components. Components are the parts of a vector along the chosen coordinate directions. Add east-west parts together and north-south parts together, keeping their signs. The resulting horizontal and vertical components describe the net displacement.

When components form a right triangle, use the Pythagorean relationship to find the displacement magnitude. Use basic trigonometry to find its direction. Give the angle a reference direction, such as north of east, so the direction is unambiguous.

\Delta x=x_f-x_i

Signed position coordinates encode direction along a line.
Displacement depends only on the start and finish.
Distance adds every part of the route, regardless of direction.
For perpendicular components, combine the components to find the vector magnitude and direction.

3. Solving and checking vector problems

Begin by identifying the object and reference point. Write the known positions or route segments, with units. State which direction is positive. Identify whether the question asks for distance, position, or displacement.

For a straight-line problem, use the signed positions in the displacement relationship. Keep metres in the substitution. A positive result points in the chosen positive direction, and a negative result points in the opposite direction.

For a route with turns, add segment lengths to find distance. Separately add signed components to find displacement. In two dimensions, calculate the horizontal and vertical net components first. Then find the magnitude and direction of the resulting vector.

Finish with a reasonableness check. Distance should not be negative and cannot be shorter than the magnitude of the displacement for the same route. Displacement may be zero if the object returns to its starting position. Check that the final answer has units and a clear direction whenever it is a vector.

d_{\text{route}}\geq |\Delta x|

Read the requested quantity carefully before choosing a relationship.
Keep distance and displacement separate when a route reverses or turns.
Report vector magnitude and direction, not magnitude alone.
Check units, sign, direction, and whether the result fits the described route.

Worked example

1. Displacement along a straight line

A cart moves along a straight track. The reference marker is at

0\ \mathrm{m}

, and east is positive. The cart starts at

x_i=-3.2\ \mathrm{m}

and finishes at

x_f=+4.5\ \mathrm{m}

. Find its displacement.

Set the system and direction
The system is the cart. The reference marker defines position zero, and east is positive. The known values are the initial and final positions; the unknown is the displacement.
Choose the relationship
Displacement is final position minus initial position. This subtraction gives the change in coordinate and its sign gives direction.
$\Delta x=x_f-x_i$
Substitute with units
Subtract the signed initial coordinate. Subtracting a negative value increases the result.
$\Delta x=(+4.5\ \mathrm{m})-(-3.2\ \mathrm{m})=+7.7\ \mathrm{m}$

Answer: The cart's displacement is

7.7\ \mathrm{m}

east.

Check: The answer has units of metres and is positive, so it points east. The cart finishes east of where it started, and the displacement magnitude is reasonable for the two positions.

Worked example

2. Distance and displacement on a route that reverses

A student walks

8.0\ \mathrm{m}

east from a doorway, then walks

3.0\ \mathrm{m}

west. Let the doorway be the reference point and east be positive. Find the total distance, final position, and displacement.

Describe the route
The system is the student. East is positive and west is negative. The route has two segments, so add their lengths for distance. Keep their signs when finding final position and displacement from the doorway.
Find total distance
Distance counts the full path, including the part walked back west. It is a scalar, so the direction does not change the addition.
$d=8.0\ \mathrm{m}+3.0\ \mathrm{m}=11.0\ \mathrm{m}$
Find final position and displacement
The student ends at the net signed coordinate relative to the doorway. Since the starting position is zero, the final position also equals the displacement.
$x_f=\Delta x=(+8.0\ \mathrm{m})+(-3.0\ \mathrm{m})=+5.0\ \mathrm{m}$

Answer: The distance is

11.0\ \mathrm{m}

. The final position is

+5.0\ \mathrm{m}

, or

5.0\ \mathrm{m}

east of the doorway. The displacement is

5.0\ \mathrm{m}

east.

Check: All results have units of metres. The distance is greater than the displacement magnitude, as expected for a route that includes a return segment. The positive position and displacement point east.

Worked example

3. Displacement with perpendicular components

A hiker walks

6.0\times10^2\ \mathrm{m}

east, then

8.0\times10^2\ \mathrm{m}

north. Take east as positive horizontal and north as positive vertical. Find the displacement magnitude and direction.

Set the components
The system is the hiker. The reference point is the starting point. The net horizontal component is east, and the net vertical component is north. These perpendicular components form a right triangle.
$\Delta x=+6.0\times10^2\ \mathrm{m},\quad \Delta y=+8.0\times10^2\ \mathrm{m}$
Find the magnitude
Use the right-triangle relationship because the east and north components are perpendicular. The result is the straight-line displacement from start to finish.
$|\Delta \vec r|=\sqrt{(6.0\times10^2\ \mathrm{m})^2+(8.0\times10^2\ \mathrm{m})^2}=1.0\times10^3\ \mathrm{m}$
Find the direction
Measure the angle from east toward north. The tangent ratio compares the north component with the east component.
$\theta=\tan^{-1}\left(\frac{8.0\times10^2}{6.0\times10^2}\right)=53^\circ$

Answer: The hiker's displacement is

1.0\times10^3\ \mathrm{m}

at

53^\circ

north of east.

Check: The magnitude has units of metres and is greater than either component but less than their sum of

1.4\times10^3\ \mathrm{m}

. Both components are positive, so the direction must be northeast. The reported magnitude uses two significant figures, consistent with the given distances.

Common mistakes and how to avoid them

Treating distance and displacement as the same quantity.

Correction: Distance adds the lengths of the route. Displacement compares only the final and initial positions.

Ignoring the sign of a position or component.

Correction: Choose a positive direction first. Use negative coordinates for positions or components in the opposite direction.

Giving a vector magnitude without its direction.

Correction: State the direction in words or with a clear signed coordinate. For an angle, name the reference direction.

Adding perpendicular component magnitudes as if they were along one line.

Correction: Keep horizontal and vertical components separate, then combine them as perpendicular sides of a right triangle.

Lesson summary

Distance is the total route length and is a scalar.
Position is a vector location measured from a reference point.
Displacement is the change from initial to final position.
For straight-line motion, subtract the initial signed position from the final position.
For two-dimensional motion, combine perpendicular components and report both magnitude and direction.
Check units, sign, direction, significant figures, and physical reasonableness.

Check your understanding

Question 1

A marker is the reference point, and east is positive. An object moves from

-2.0\ \mathrm{m}

to

+5.0\ \mathrm{m}

. What is its displacement?

$3.0\ \mathrm{m}$ east
$7.0\ \mathrm{m}$ east
$7.0\ \mathrm{m}$ west
$3.0\ \mathrm{m}$ west

Show answer and explanation

7.0\ \mathrm{m}

east

Subtract the initial position from the final position:

+5.0\ \mathrm{m}-(-2.0\ \mathrm{m})=+7.0\ \mathrm{m}

. The positive sign means east.

Question 2

A person walks

4.0\ \mathrm{m}

north and then

4.0\ \mathrm{m}

south. What are the distance and displacement?

Distance $0\ \mathrm{m}$ ; displacement $8.0\ \mathrm{m}$ north
Distance $8.0\ \mathrm{m}$ ; displacement $0\ \mathrm{m}$
Distance $8.0\ \mathrm{m}$ ; displacement $8.0\ \mathrm{m}$ south
Distance $4.0\ \mathrm{m}$ ; displacement $0\ \mathrm{m}$

Show answer and explanation

Distance

8.0\ \mathrm{m}

; displacement

0\ \mathrm{m}

The route length is

4.0\ \mathrm{m}+4.0\ \mathrm{m}=8.0\ \mathrm{m}

. The person returns to the starting point, so the displacement is zero.

Question 3

A displacement has a

3.0\ \mathrm{m}

east component and a

4.0\ \mathrm{m}

north component. What is its magnitude?

$1.0\ \mathrm{m}$
$5.0\ \mathrm{m}$
$7.0\ \mathrm{m}$
$12\ \mathrm{m}$

Show answer and explanation

5.0\ \mathrm{m}

The components are perpendicular, so the magnitude is

\sqrt{(3.0\ \mathrm{m})^2+(4.0\ \mathrm{m})^2}=5.0\ \mathrm{m}

. The result is larger than either component and smaller than their sum.

Key terms

Scalar: A quantity with magnitude but no direction.
Vector: A quantity with both magnitude and direction.
Reference point: The chosen location from which position is described.
Position: An object's location relative to a reference point, including direction.
Distance: The total length of the route travelled.
Displacement: The change from an object's initial position to its final position.
Component: The part of a vector along one chosen coordinate direction.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation B2.5. 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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B2.5 · Solve distance, position, and displacement problems with vectors

Ontario Grade 11 Physics

Ontario Grade 11 Physics — expectation B2.5

What you will learn

1. Prerequisite bridge: scalars, vectors, and reference points

2. The physical situation and the vector rules

3. Solving and checking vector problems

1. Displacement along a straight line

2. Distance and displacement on a route that reverses

3. Displacement with perpendicular components

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