3.1 Position, Displacement, and Average Velocity Notes

Position and Reference Frames

Purpose: To describe motion, you must know where something is (position), how far it has moved (displacement), how far it has traveled (distance), and how fast it is moving on average (average velocity).
Key questions when in motion:
- Where are you? (position)
- Where are you going? (displacement)
- How fast are you getting there? (average velocity)
Frame of reference (reference frame): an arbitrary set of axes from which position and motion are described.
- Earth is a common frame of reference for objects relative to the surface.
- Objects can be described relative to different frames (e.g., an airplane’s frame vs Earth).
Common notation for one-dimensional motion:
- Position along a line: commonly denoted by $x$ .
- In some contexts (e.g., vertical motion or free fall), the variable may be $y$ .
Takeaway: Displacement and position depend on the chosen frame of reference and the chosen positive direction.

Position

Definition: The position $x(t)$ describes where an object is at a given time relative to a frame of reference.
One-dimensional motion uses a coordinate axis (often the x-axis).
Examples:
- A rocket’s position relative to Earth.
- A cyclist’s position relative to nearby buildings.
- In an airplane’s frame, the position of a person would be described relative to the plane, not Earth.
Important concept: Position is a scalar value in a given frame (its numeric coordinate along the axis).

Displacement

Definition: Displacement is the change in position of an object relative to a frame of reference.
It is a vector, so it has both magnitude and direction.
Mathematical definition (one-dimensional):
- $\Delta x = xf - xi$
- where $xi$ is the initial position and $xf$ is the final position.
Sign conventions:
- If right (or east) is taken as positive, displacements to the right are positive, displacements to the left are negative.
- Example: moving 2 m to the right and then 4 m to the left yields displacements $\Delta x1 = +2\,\text{m}$ , $\Delta x2 = -4\,\text{m}$ .
Total displacement over a sequence of motions:
- $\Delta x{\text{Total}} = \sumi \Delta x_i$
Displacement vs. position:
- Displacement is the net change in position between two points.
- It is a vector and can be negative depending on direction.
Magnitude of displacement:
- Magnitude is the absolute value: $|\Delta x|$ , always nonnegative.
Example (two-step motion):
- Initial position: $x_0$
- After first step: $x1$ with displacement $\Delta x1 = x1 - x0$
- After second step: $x2$ with displacement $\Delta x2 = x2 - x1$
- Total displacement: $\Delta x{\text{Total}} = \Delta x1 + \Delta x_2$
- If $\Delta x1 = +2\text{ m}$ and $\Delta x2 = -4\text{ m}$ , then $\Delta x_{\text{Total}} = -2\text{ m}$ .

Distance Traveled (not the same as displacement)

Definition: Distance traveled is the total length of the path traveled between two positions.
It is a scalar (no direction) and is always nonnegative.
Calculation (sum of segment lengths):
- If a motion consists of successive displacements $\Delta x_i$ , the distance traveled is
- $\text{Distance} = \sumi |\Delta xi|$
Note: Distance traveled can be greater than the magnitude of displacement, especially if the path doubles back.
Example context: If a professor moves 2 m to the right and 2 m to the left, the total displacement is $0\text{ m}$ , but the distance traveled is $4\text{ m}$ .

Average Velocity

Time variable introduction: To describe motion, we must also consider time. The rate of position change defines velocity.
Average velocity (vector): the total displacement over the elapsed time.
Two-point form (for motion from point 1 to point 2):
- $v{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{x2 - x1}{t2 - t_1}$
- It is a vector and can be negative depending on the chosen positive direction.
For a multi-segment path (the entire trip):
- Total displacement: $\Delta x{\text{Total}} = \sumi \Delta x_i$
- Elapsed time: $\Delta t{\text{Total}} = t{\text{final}} - t_{\text{initial}}$
- Average velocity for the entire trip:
- $v{\text{avg}} = \frac{\Delta x{\text{Total}}}{\Delta t_{\text{Total}}}$
Relationship between displacement and distance with velocity:
- If a motion has a large distance traveled with a small net displacement, the average velocity may be small in magnitude even if the path is long.

Example: Delivering Flyers (Jill’s trip)

Scenario: Jill travels due east along a street, then retraces steps for flyers, then continues, and finally ends up 1.0 km from home, then rests after additional travel.
Plan: Build a table of time, position, and each segment’s displacement to determine total displacement and total distance traveled.
Given data (east is positive):
- Starting point: $t0 = 0\text{ min}, x0 = 0\text{ km}$
- Segment data (example from notes):
- $t1 = 9\text{ min}, x1 = 0.5\text{ km}$ , displacement $\Delta x1 = x1 - x_0 = 0.5\text{ km}$
- $t2 = 18\text{ min}, x2 = 0\text{ km}$ , displacement $\Delta x2 = x2 - x_1 = -0.5\text{ km}$
- $t3 = 33\text{ min}, x3 = 1.0\text{ km}$ , displacement $\Delta x3 = x3 - x_2 = 1.0\text{ km}$
- $t4 = 58\text{ min}, x4 = -0.75\text{ km}$ , displacement $\Delta x4 = x4 - x_3 = -1.75\text{ km}$
Calculations:
- Total displacement:
- $\Delta x{\text{Total}} = \Delta x1 + \Delta x2 + \Delta x3 + \Delta x_4 = 0.5 - 0.5 + 1.0 - 1.75 = -0.75\text{ km}$
- Magnitude of total displacement: $|\Delta x_{\text{Total}}| = 0.75\text{ km}$
- Total elapsed time: $\Delta t{\text{Total}} = t4 - t_0 = 58\text{ min}$
- Average velocity for entire trip:
- $v{\text{avg}} = \frac{\Delta x{\text{Total}}}{\Delta t_{\text{Total}}} = \frac{-0.75\text{ km}}{58\text{ min}} \approx -0.01293\ \text{km/min} \approx -0.013\ \text{km/min}$
- Total distance traveled (sum of segment magnitudes):
- $\text{Distance} = \sumi |\Delta xi| = 0.5 + 0.5 + 1.0 + 1.75 = 3.75\text{ km}$
Graphical aid: Position vs. time graph can illustrate how the slope between the endpoints relates to the average velocity.
Interpretation of Jill’s results:
- Final position relative to home: displacement of $-0.75\text{ km}$ (i.e., 0.75 km west of home).
- Average velocity of the whole trip: $-0.013\ \text{km/min}$ , indicating a net westward motion; if someone walked due west at this speed starting at the same time, they would reach the final point at the same time as Jill.
- If Jill had ended at home, displacement would be zero and the average velocity would be zero as well.
Real-world relevance: These concepts apply to navigation, sports strategies, and any scenario involving motion along a path with finite duration.

Check Your Understanding 3.1 (Cyclist)

Problem statement: A cyclist rides 3 km west and then turns around and rides 2 km east.
Solve with east as positive (the same convention used above):
- Displacement: $\Delta x_{\text{Total}} = (-3\text{ km}) + (+2\text{ km}) = -1\text{ km}$
- Distance traveled: $|\Delta x1| + |\Delta x2| = 3\text{ km} + 2\text{ km} = 5\text{ km}$
- Magnitude of displacement: $|\Delta x_{\text{Total}}| = 1\text{ km}$
Summary answers:
- (a) Displacement: $-1\text{ km}$ (1 km to the west)
- (b) Distance traveled: $5\text{ km}$
- (c) Magnitude of displacement: $1\text{ km}$

Connections and Practical Notes

Displacement is a vector quantity; velocity and force are also vectors, while distance is a scalar.
The choice of positive direction affects the sign of displacement and velocity; only magnitude is invariant.
Distinguish between displacement (net change in position) and distance traveled (total length of the path).
For complex motions, break the path into segments, compute segment displacements, then aggregate to obtain total displacement and total distance.
Units and conversions: Displacement is measured in meters (m) in SI, but kilometers (km) or other units may be used; convert to meters for calculations when required: $1\ \text{km} = 1000\ \text{m}.$