1.2 Relate position, path length, and displacement | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Particle Kinematics Foundations

EN PH 131 Engineering Mechanics: Dynamics — Study topic 1.2

In dynamics, a motion description is meaningful only after we specify what object is being followed and the reference frame used to describe it. This lesson follows a single particle, treated as an object whose size is not needed to describe its location. We will compare its position at two times, the route it follows between them, and how far its final position is from its initial position. Path length and displacement answer different questions: how much distance was travelled, and where did the particle end up relative to where it started?

What you will learn

Describe a particle’s position relative to a stated observer, reference frame, and coordinate system.
Distinguish the distance travelled along a path from the change in position.
Calculate displacement as a vector and path length as a nonnegative scalar.
Check answers using direction, units, and limiting cases.

1. Define the particle, observer, and coordinates

The system is the particle whose motion we describe. The observer is the person or measuring setup that records its location. A reference frame is the chosen origin and coordinate axes used to report those locations. In the examples, the frame is fixed to the ground.

In one dimension, choose an origin and a positive direction. A position coordinate such as

x

is signed: it is positive on one side of the origin and negative on the other. In a plane, use perpendicular axes and coordinates

x

and

y

. The position vector from the origin to the particle is

\mathbf{r}=x\mathbf{i}+y\mathbf{j}

, where

\mathbf{i}

and

\mathbf{j}

point along the positive axes.

Initial and final positions must use the same frame and origin. Denote them by

\mathbf{r}_1

and

\mathbf{r}_2

. Position tells where the particle is; position alone does not tell which route it took.

\mathbf{r}=x\mathbf{i}+y\mathbf{j}

State the system, observer, frame, origin, and positive directions before using coordinates.
Position is a vector from the origin; its components depend on the chosen axes.

2. Path length and displacement are different

The path is the actual route followed between the initial and final states. Path length, often written

s

, is the total length of that route. It is a scalar and cannot be negative. For a route made of several segments, add their lengths regardless of direction.

Displacement is the change in position from the initial state to the final state. It is a vector, so it has both magnitude and direction. In coordinates, subtract the initial position from the final position component by component. A negative component means the change is opposite to that coordinate’s positive direction.

The magnitude of displacement is the straight-line distance between the endpoints. The length of any route between those endpoints is at least this straight-line distance. Path length equals displacement magnitude for a direct route with no reversal; a detour or reversal makes the path longer.

\Delta\mathbf{r}=\mathbf{r}_2-\mathbf{r}_1,\qquad s\geq\lVert\Delta\mathbf{r}\rVert

Path length depends on the route; displacement depends only on the endpoint positions.
Path length is a nonnegative scalar; displacement is a vector.
Both quantities have units of length, such as metres.

3. Choose the calculation that matches the quantity

For motion along a line, choose the positive direction and record signed positions. Displacement is final coordinate minus initial coordinate. If the particle reverses direction, path length includes the distance before and after the reversal, while signed changes can cancel.

For motion in a plane, find the displacement components by subtracting endpoint coordinates. Its magnitude follows from the Pythagorean theorem. To find path length, divide the stated route into segments and add their lengths. Endpoints alone do not determine the route: different paths can have the same displacement but different lengths.

A useful motion sketch marks the stated route and the initial and final positions. It should not imply an unstated route. Use a path diagram when the problem gives a straight or circular route; otherwise, describe the route geometry in words and calculate from the information provided.

\lVert\Delta\mathbf{r}\rVert=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Use signed coordinates for displacement and positive segment lengths for path length.
In two dimensions, calculate displacement components before finding its magnitude.
Check that path length is at least the magnitude of displacement.

Worked example

A particle reverses along a straight line

A particle moves along a straight east–west line. Let east be positive. It starts at

x_1=2.0\,\mathrm{m}

, travels east to

x=9.0\,\mathrm{m}

, then travels west to finish at

x_2=5.0\,\mathrm{m}

. Find its path length and displacement.

Straight-line reversal

The route goes east to the turning point, then west to the final position.

Set the frame and states
The system is the particle, and the observer measures position in a ground-fixed frame. East is the positive $x$ direction. The initial and final positions are $2.0\,\mathrm{m}$ and $5.0\,\mathrm{m}$ ; the turning point is at $9.0\,\mathrm{m}$ .
Add the route segments
Path length counts both parts of the journey as positive distances. The eastward segment is $7.0\,\mathrm{m}$ , and the westward segment is $4.0\,\mathrm{m}$ .
$s=(9.0-2.0)+(9.0-5.0)=11.0\,\mathrm{m}$
Subtract endpoint positions
Displacement uses final minus initial position. The positive result means the net change is eastward.
$\Delta x=x_2-x_1=5.0-2.0=+3.0\,\mathrm{m}$

Answer: The path length is

11.0\,\mathrm{m}

. The displacement is

+3.0\,\mathrm{m}

, or

3.0\,\mathrm{m}

east.

Check: The path length exceeds the displacement magnitude, as required. Both have units of metres, and the positive displacement agrees with the final position being east of the initial position.

Worked example

Two perpendicular segments

A ground-fixed observer tracks a particle that starts at

(0,0)

, travels

6.0\,\mathrm{m}

east, then

8.0\,\mathrm{m}

north. Find the path length and displacement.

East then north

The stated route consists of an eastward segment followed by a northward segment.

Set axes and endpoints
The system is the particle. The observer uses a ground-fixed frame with the origin at the start, east as positive $x$ , and north as positive $y$ . The final position is $(6.0,8.0)\,\mathrm{m}$ .
$\mathbf{r}_1=(0,0)\,\mathrm{m},\qquad \mathbf{r}_2=(6.0,8.0)\,\mathrm{m}$
Find the route length
The particle follows two stated segments, so add their lengths. This uses the route rather than the straight-line separation of the endpoints.
$s=6.0+8.0=14.0\,\mathrm{m}$
Find the displacement
Subtract the initial position vector from the final one. The magnitude is the straight-line separation; both components are positive, so the direction is northeast.
$\Delta\mathbf{r}=(6.0\mathbf{i}+8.0\mathbf{j})\,\mathrm{m},\qquad \lVert\Delta\mathbf{r}\rVert=10.0\,\mathrm{m}$

Answer: The path length is

14.0\,\mathrm{m}

. The displacement is

6.0\,\mathrm{m}

east and

8.0\,\mathrm{m}

north, with magnitude

10.0\,\mathrm{m}

.

Check: The route length exceeds the displacement magnitude. The displacement components reproduce the final coordinates, and its magnitude has units of metres.

Worked example

One complete circular lap

A particle moves once counterclockwise around a circular track of radius

5.0\,\mathrm{m}

, starting and finishing at the same point. Find its path length and displacement.

One lap around a circle

The particle returns to its starting point after one complete lap.

Define the frame and states
The system is the particle, and the observer is fixed to the ground. The initial and final states are the same point on the circular track. The stated route is one complete circumference.
Calculate the route length
One complete lap has the circumference of a circle with the given radius.
$s=2\pi r=2\pi(5.0\,\mathrm{m})=31.4\,\mathrm{m}$
Compare endpoint positions
Because the final position equals the initial position, their vector difference is zero even though the particle travelled a nonzero distance.
$\Delta\mathbf{r}=\mathbf{r}_2-\mathbf{r}_1=\mathbf{0}$

Answer: The path length is approximately

31.4\,\mathrm{m}

. The displacement is zero.

Check: Zero displacement is correct because the particle returns to its starting point. The nonzero circumference is consistent with completing a full lap.

Common mistakes and how to avoid them

Treating path length and displacement as interchangeable.

Correction: Path length adds the distances along the route; displacement subtracts the initial position from the final position.

Making displacement negative whenever part of the motion is in the negative direction.

Correction: Displacement depends on the net change in position, not on each segment separately. Path length remains nonnegative.

Finding path length from the endpoints alone.

Correction: Endpoints determine displacement, but path length requires knowing the route.

Giving only the magnitude of displacement when direction matters.

Correction: Report the vector components or state a clear direction as well as the magnitude.

Lesson summary

Position is reported relative to a defined observer, reference frame, origin, and axes.
Displacement is final position minus initial position and is a vector.
Path length is the total distance along the route and is a nonnegative scalar.
Path length is at least the magnitude of displacement; equality holds for a direct, unreversed route.

Check your understanding

Question 1

A particle starts at

x=4.0\,\mathrm{m}

, moves to

x=10.0\,\mathrm{m}

, then returns to

x=7.0\,\mathrm{m}

. What are its path length and displacement if positive

x

points right?

Path length $9.0\,\mathrm{m}$ ; displacement $+3.0\,\mathrm{m}$
Path length $3.0\,\mathrm{m}$ ; displacement $+9.0\,\mathrm{m}$
Path length $9.0\,\mathrm{m}$ ; displacement $-3.0\,\mathrm{m}$
Path length $3.0\,\mathrm{m}$ ; displacement $+3.0\,\mathrm{m}$

Show answer and explanation

Path length

9.0\,\mathrm{m}

; displacement

+3.0\,\mathrm{m}

The route segments are

6.0\,\mathrm{m}

and

3.0\,\mathrm{m}

, giving

9.0\,\mathrm{m}

of path length. Final minus initial position is

7.0-4.0=+3.0\,\mathrm{m}

.

Question 2

A particle completes one full lap of a circular path and returns to its starting point. Which statement is correct?

Its path length is zero and its displacement is nonzero.
Its path length is nonzero and its displacement is zero.
Both its path length and displacement are zero.
Its path length equals the magnitude of its displacement.

Show answer and explanation

Its path length is nonzero and its displacement is zero.

The particle travels the circumference, so the path length is nonzero. Its final and initial positions coincide, so displacement is zero.

Key terms

Position: A vector locating a particle relative to a chosen origin and coordinate axes.
Path: The route followed by a particle between two states.
Path length: The total length of the route travelled; a nonnegative scalar.
Displacement: The vector change in position from an initial state to a final state.
Reference frame: The observer’s chosen origin and coordinate directions used to describe position.

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1.2 · Relate position, path length, and displacement

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

EN PH 131 Engineering Mechanics: Dynamics — Study topic 1.2

What you will learn

1. Define the particle, observer, and coordinates

2. Path length and displacement are different

3. Choose the calculation that matches the quantity

A particle reverses along a straight line

Two perpendicular segments

One complete circular lap

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through EN PH 131

About this lesson and its review