2.1 Position vs Displacement vs Average Velocity – What's the Difference?

If you cannot tell the difference between how far you traveled and where you ended up, you will lose marks on every kinematics question from A-Level to the IPhO. This is the foundation of all mechanics. In this guide you will master how physicists locate an object on an axis, how to calculate the exact change in its position regardless of the path it took, and how to extract both average velocity and average speed from words, numbers, and x(t) graphs. We will build everything from the particle model to the slope meaning of v_avg, with exam-style examples that make the difference between distance and displacement automatic.

Key Insights

• Position x locates a particle relative to an origin on an x-axis with positive to the right and negative to the left.
• Displacement Δx = x2 - x1 is a vector that depends only on initial and final positions, not on total path length.
• Average velocity v_avg = Δx/Δt is the slope of the straight line connecting two points on an x(t) graph and its sign gives direction.
• Average speed s_avg = total distance / Δt is a scalar with no sign and can be much larger than the magnitude of average velocity.
• One-dimensional motion in Fundamentals of Physics is restricted to a straight line, treated as a particle, and studied without forces as kinematics.

Position, Displacement, and Average Velocity

This module is the entry to kinematics, the classification and comparison of motions. You will learn to treat extended objects as point-like particles when all parts move together, to read position from a scaled axis, to compute displacement, and to relate displacement and time interval to average velocity and average speed.

Learning Objectives you will master:

1. Identify when an object can be treated as a particle
2. Read position x from an x-axis
3. Apply Δx = x2 - x1 and interpret its sign
4. Apply v_avg = Δx/Δt = (x2 - x1)/(t2 - t1)
5. Apply s_avg = total distance / Δt
6. Determine v_avg from an x(t) graph as a slope

Callout Box: The Particle ModelIf every portion of an object moves in the same direction at the same rate, we can treat the whole object as if it were a single point-like particle. A stiff board sliding down a straight slide qualifies. A tumbling tumbleweed that rotates as it translates does not, because different parts move in different directions.

What Is Physics?

One purpose of physics is to study the motion of objects — how fast they move and how far they move in a given time. This is not abstract. NASCAR engineers use it to determine the performance of their cars before and during a race. Geologists use it to measure tectonic-plate motion to attempt to predict earthquakes. Medical researchers need it to map blood flow through a partially closed artery when diagnosing a patient. Motorists use it to decide how much they must slow down when their radar detector sounds a warning.

In this chapter we study the basic physics of motion where the object — a race car, a tectonic plate, a blood cell, or any other object — moves along a single axis. Such motion is called one-dimensional motion.

Motion

The world and everything in it moves. Even seemingly stationary things, such as a roadway, move with Earth's rotation, Earth's orbit around the Sun, the Sun's orbit around the center of the Milky Way galaxy, and that galaxy's migration relative to other galaxies. The classification and comparison of motions called kinematics is often challenging: what exactly do you measure and what do you compare?

Before we attempt an answer, we shall examine some general properties of motion that is restricted in three ways.

1. The motion is along a straight line only. The line may be vertical, horizontal, or slanted, but it must be straight.
2. Forces cause motion but will not be discussed until Chapter 5. In this chapter we discuss only the motion itself and changes in the motion. Does the moving object speed up, slow down, stop, or reverse direction? If the motion does change, how is time involved in the change?
3. The moving object is either a particle or an object that moves like a particle. By particle we mean a point-like object where every portion moves in the same direction at the same rate.

Pro Tip for Exams:When a problem says "a car" or "a runner" and gives only positions along a road, immediately apply the particle model. You do not need the car's length unless the question explicitly asks about two ends of the object passing a point.

Position and Displacement

To locate an object means to find its position relative to some reference point, often the origin or zero point of an axis such as the x-axis. The positive direction of the axis is in the direction of increasing numbers, which is to the right. The opposite is the negative direction.

For example, a particle might be located at x = 5 m, which means it is 5 m in the positive direction from the origin. If it were at x = -5 m, it would be just as far from the origin but in the opposite direction. On the axis, a coordinate of -5 m is less than a coordinate of -1 m, and both are less than a coordinate of +5 m. A plus sign for a coordinate need not be shown, but a minus sign must always be shown.

3D photorealistic x-axis ruler from -3 m to +3 m showing origin, positive direction arrow in red to the right, negative direction arrow in blue to the left, and metallic particle spheres at different positions

A change from position x1 to position x2 is called a displacement Δx, where

Δx = x2 - x1

The symbol Δ, the Greek uppercase delta, represents a change in a quantity and it means the final value minus the initial value. When numbers are inserted for x1 and x2, a displacement in the positive direction always comes out positive, and a displacement in the opposite direction always comes out negative.

For example, if the particle moves from x1 = 5 m to x2 = 12 m, then the displacement is Δx = (12 m) - (5 m) = +7 m. The positive result indicates motion in the positive direction. If instead the particle moves from x1 = 5 m to x2 = 1 m, then Δx = (1 m) - (5 m) = -4 m. The negative result indicates motion in the negative direction.

The actual number of meters covered for a trip is irrelevant; displacement involves only the original and final positions. For example, if the particle moves from x = 5 m out to x = 200 m and then back to x = 5 m, the displacement from start to finish is Δx = (5 m) - (5 m) = 0.

Photorealistic 3D displacement diagram showing metallic spheres moving from x1=5 m to x2=12 m with green arrow Δx=+7 m, and from 5 m to 1 m with red arrow Δx=-4 m on ruler

Signs and Magnitude: A plus sign for a displacement need not be shown, but a minus sign must always be shown. If we ignore the sign, we are left with the magnitude or absolute value of the displacement. For example, a displacement of Δx = -4 m has a magnitude of 4 m.

Displacement is an example of a vector quantity, which is a quantity that has both a direction and a magnitude. We explore vectors more fully in Chapter 3, but here all we need is the idea that displacement has two features: (1) Its magnitude is the distance between the original and final positions. (2) Its direction, from an original position to a final position, can be represented by a plus sign or a minus sign if the motion is along a single axis.

Concept	Depends On	Symbol	Vector or Scalar?	Can Be Zero Even If You Moved?
Position	Origin	x	Scalar with sign indicating side of origin	Yes, if at origin
Displacement	x1 and x2 only	Δx = x2 - x1	Vector	Yes, if you return to start
Distance	Entire path	d	Scalar	No, always >=0 and increases with path

Callout Box: Checkpoint 2.1.1 SolvedWhich pairs give a negative displacement: (a) -3 m, +5 m; (b) -3 m, -7 m; (c) 7 m, -3 m? Apply Δx = x2 - x1. (a) 5 - (-3) = +8 m positive. (b) -7 - (-3) = -4 m negative. (c) -3 - 7 = -10 m negative. So (b) and (c) are negative. This is a classic AP Physics 1 trap where students pick (a) because they see a negative initial coordinate.

Exam Trap:Students confuse position and displacement. Position is where you are. Displacement is how far your position changed. If the question says "how far from the origin" it wants position magnitude. If it says "how far did you move from your starting point" it wants displacement. Always write x1, x2 explicitly before subtracting.

Average Velocity and Average Speed

A compact way to describe position is with a graph of position x plotted as a function of time t — a graph of x(t). The notation x(t) represents a function x of t, not the product x times t.

As a simple example, the graph of x(t) for a stationary object like an armadillo treated as a particle over a 7 s interval is a horizontal line. The animal's position stays at x = -2 m. The value of x is -2 m for all times t. Same position for any time means zero velocity.

The graph becomes interesting when it involves motion. The armadillo is apparently first noticed at t = 0 when it is at x = -5 m. It moves toward x = 0, passes through that point at t = 3 s, and then moves on to increasingly larger positive values of x. The graph also depicts the straight-line motion of the animal at three times and is something like what you would see in a stroboscopic photograph.

Actually, several quantities are associated with the phrase "how fast." One of them is the average velocity v_avg, which is the ratio of the displacement Δx that occurs during a particular time interval Δt to that interval:

v_avg = Δx / Δt = (x2 - x1) / (t2 - t1)

The notation means that the position is x1 at time t1 and then x2 at time t2. A common unit for v_avg is meter per second (m/s). You may see other units like km/h, but they are always in the form of length/time.

Photorealistic 3D position-time graph showing curved blue x(t) from -5 m at 0 s to 2 m at 4 s, with magenta straight secant line slope v_avg = Δx/Δt, rise 6 m run 3 s, and realistic armadillo models

Graphs. On a graph of x versus t, v_avg is the slope of the straight line that connects two particular points on the x(t) curve: one is the point that corresponds to x2 and t2 and the other is the point that corresponds to x1 and t1. Like displacement, v_avg has both magnitude and direction — it is another vector quantity. Its magnitude is the magnitude of the line's slope. A positive v_avg and positive slope tells us that the line slants upward to the right; a negative v_avg and negative slope tells us that the line slants downward to the right. The average velocity v_avg always has the same sign as the displacement Δx because Δt is always positive.

To find v_avg for the time interval t = 1 s to t = 4 s, we draw the straight line that connects the point on the position curve at the beginning of the interval and the point on the curve at the end of the interval. Then we find the slope Δx/Δt of the straight line. For the given time interval, the average velocity is

v_avg = 6 m / 3 s = 2 m/s

Photorealistic 3D comparison of average velocity displacement 13 km vs total distance 16 km with car and jogger models

Average speed s_avg is a different way of describing "how fast" a particle moves. Whereas average velocity involves the particle's displacement Δx, average speed involves the total distance covered independent of direction:

s_avg = total distance / Δt

Because average speed does not include direction, it lacks any algebraic sign. Sometimes s_avg is the same except for the absence of a sign as v_avg. However, the two can be quite different.

Sample Problem 2.1.1 – The definitive worked example you must know:

You get a lift from a car service to take you to a state park along a straight road due east for 10.0 km at an average velocity of 40.0 km/h. From the drop-off point, you jog along a straight path due east for 3.00 km, which takes 0.500 h.

(a) What is your overall displacement from your starting point to the point where your jog ends?

Assume you move in the positive direction of an x-axis from a first position x1 = 0 to a second position x2 = 10.0 km + 3.00 km = 13.0 km at the end of the jog. Displacement is second minus first: Δx = x2 - x1 = 13.0 - 0 = 13.0 km in the positive x direction.

(b) What is the time interval Δt from the beginning of your movement to the end of the jog?

We know jogging interval Δt_jog = 0.500 h, but we lack Δt_car. We know displacement for the ride is Δx_car = 10.0 km and v_avg,car = 40.0 km/h. Average velocity is displacement over time interval for the ride, so v_avg,car = Δx_car / Δt_car. Rearranging: Δt_car = Δx_car / v_avg,car = 10.0 km / 40.0 km/h = 0.250 h. So Δt = Δt_car + Δt_jog = 0.250 + 0.500 = 0.750 h.

(c) What is your average velocity from start to end of the jog? Find it numerically and graphically.

From Eq. v_avg for entire trip is ratio of displacement 13.0 km for entire trip to time interval 0.750 h. Calculation: v_avg = 13.0 km / 0.750 h = 17.3 km/h. Graphically, plot x(t) with "Riding" segment steeply rising to 10 km at 0.25 h, then "Jogging" more slowly to 13 km at 0.75 h. The slope of the straight line joining origin and point labeled "Stop" is the average velocity from start to stop, rise 13.0 km over run 0.750 h = 17.3 km/h.

(d) Suppose you then jog back to the drop-off point for another 0.500 h. What is your average speed from beginning to that return?

Average speed is total distance over total time. Total distance = 10.0 km + 3.00 km + 3.00 km = 16.0 km. Total time = 0.250 + 0.500 + 0.500 = 1.25 h. So s_avg = 16.0 km / 1.25 h = 12.8 km/h. Note that average velocity for this extended trip would be Δx = 10 km / 1.25 h = 8.0 km/h, very different from average speed.

Average Velocity vavg	Average Speed savg
vavg = displacement / time = Δx/Δt	savg = total distance / time
Vector, has + or - sign indicating direction	Scalar, always >=0, no direction
Can be zero if you return to start	Zero only if you never moved
Slope of secant line on x(t) graph	Not a slope on x(t) graph
AP, A-Level, IB test sign heavily	IB and College Physics love round-trip traps

Pro Tip for Exams:On any x-t graph, draw the secant line first, then read rise and run. If the secant slopes up to the right, v_avg is positive. If it slopes down, negative. For average speed, break the trip into segments, sum absolute distances, never subtract.

Callout Box: Stationary vs Moving x(t) GraphsA stationary object gives a horizontal line on x(t). Same position for any time means Δx = 0, so v_avg = 0. A moving object gives a line that is not horizontal. If the object moves at constant velocity, x(t) is a straight line. If it speeds up, x(t) curves upward with increasing slope.

Exam Trap:Students use v_avg = (v1 + v2)/2. That only works if time intervals are equal and it is still risky. The definition is always displacement over total time. For non-uniform motion, never average the velocities unless the problem explicitly says to.

Practice Questions

MCQs (Multiple Choice Questions)

01. A particle starts at x = -3 m and ends at x = +5 m. What is its displacement?

1. +2 m
2. +8 m
3. -8 m
4. -2 m

Correct: B
Δx = x2 - x1 = 5 - (-3) = 8 m. Positive means motion in +x direction. Option A confuses addition of magnitudes.

02. A particle moves from x = 5 m to x = 200 m and back to x = 5 m. Its displacement is:

1. 0 m
2. 195 m
3. 390 m
4. 200 m

Correct: A
Displacement depends only on start and end. 5 - 5 = 0. Total distance would be 390 m, which is what C traps.

03. Which x(t) graph represents an object at rest at x = -2 m?

1. Horizontal line at x = -2
2. Line through origin with positive slope
3. Curved upward line
4. Vertical line

Correct: A
Same position for any time = horizontal. Vertical line would mean infinite velocity, unphysical.

04. For t = 1 s to t = 4 s, x changes from -4 m to 2 m. v_avg is:

1. 2 m/s
2. -2 m/s
3. 6 m/s
4. 1.5 m/s

Correct: A
Δx = 2 - (-4) = 6 m, Δt = 3 s, v_avg = 2 m/s. Option D divides by 4 instead of 3.

05. Average speed for a round trip that starts and ends at same point in 2 h with 10 km out and 10 km back is:

1. 0 km/h
2. 5 km/h
3. 10 km/h
4. 20 km/h

Correct: CTotal distance = 20 km / 2 h = 10 km/h. Average velocity would be 0. Classic distinction.

06. On x(t), average velocity between two times is:

1. Slope of tangent at midpoint
2. Slope of secant connecting the two points
3. Area under curve
4. Intercept

Correct: B
Tangent gives instantaneous velocity. Secant gives average.

07. Which pair gives negative displacement? -3 m to +5 m, -3 m to -7 m, 7 m to -3 m?

1. First only
2. Second and third
3. All three
4. None

Correct: B
Calculated above: +8, -4, -10. Second and third negative.

08. A car goes 10 km at 40 km/h. Time taken is:

1. 0.25 h
2. 4 h
3. 400 h
4. 0.4 h

Correct: A
Δt = Δx / v_avg = 10/40 = 0.25 h = 15 min.

09. Position -5 m is:

1. 5 m in positive direction
2. 5 m in negative direction
3. Closer to origin than -1 m
4. Same as +5 m

Correct: B
Negative sign means opposite side. -5 is farther from origin than -1.

10. If v_avg is negative, the particle:

1. Is at negative position
2. Moved in negative x-direction on average
3. Is speeding up
4. Has negative speed

Correct: B
Sign of v_avg matches sign of Δx, indicates direction of net displacement, not location.

FRQs (Free Response Questions)

FRQ 1: AP 1 style: Interpreting x(t)
An armadillo's x(t) is given by a curve starting at -5 m at t=0, passing 0 at t=3 s, reaching 2 m at t=4 s.
(a) Find v_avg from 0 to 3 s.
(b) From 0 to 4 s.
(c) Explain why v_avg is smaller in magnitude than the maximum instantaneous slope.

Rubric:

• (a) Δx = 0 - (-5)=5 m, Δt=3 s => 1.67 m/s (2 pts: formula + calculation)
• (b) Δx=2-(-5)=7 m, Δt=4 s =>1.75 m/s (2 pts)
• (c) Curve is steep early then flattens, average secant slope less than early tangent slope because speed decreases (2 pts for mentioning decreasing slope / slowing)

FRQ 2: A-Level style: Displacement vs Distance
A student walks 3 m east, then 4 m west, then 5 m east in 10 s. Calculate
(i) displacement
(ii) total distance
(iii) average velocity
(iv) average speed

Rubric:

• (i) Choose east +, displacement = 3-4+5=4 m east (2 pts)
• (ii) Distance = 3+4+5=12 m (1 pt)
• (iii) v_avg=4/10=0.4 m/s east (1 pt)
• (iv) s_avg=12/10=1.2 m/s (1 pt)

Must show signs

FRQ 3: IB HL style: Two-stage trip
Car stage: Δx1=10 km, v1=40 km/h. Jog stage: Δx2=3 km, Δt2=0.5 h. Find total Δt, v_avg, and if jogger returns 3 km in 0.5 h, s_avg for whole 16 km trip.

Rubric:

• Δt1=0.25 h (2 pts)
• total Δt=0.75 h (1 pt)
• v_avg=13/0.75=17.3 km/h (2 pts)
• s_avg=16/1.25=12.8 km/h (2 pts)

Units required

FRQ 4: Graphical analysis
Sketch x(t) for: object at rest at x=2 m from t=0-2 s, then moves to x=6 m at constant velocity from t=2-5 s, then remains at 6 m. Find v_avg 0-5 s and describe how to get it from graph.

Rubric:

• Correct horizontal then rising then horizontal (2 pts)
• Δx=4 m, Δt=5 s =>0.8 m/s (2 pts)
• explanation: secant from (0,2) to (5,6) slope (2 pts)

FRQ 5: USAPhO style conceptual
Prove that average speed can be greater than magnitude of average velocity but never less. Give an example where s_avg = |v_avg|.

Rubric:

• Proof: total distance >= |Δx| by triangle inequality, divide by same Δt (3 pts)
• Example: motion in one direction without reversal, e.g., x monotonically increasing, distance = displacement (2 pts)

Frequently Asked Questions

Q1: What is the difference between position and displacement?

Position x is your location relative to an origin. Displacement Δx = x2 - x1 is the change in that location. Position tells where you are, displacement tells how far and in which direction you moved from start to finish.

Q2: Is displacement a vector or scalar?

Displacement is a vector. It has magnitude, the distance between start and end, and direction indicated by + or - along an axis. Distance is the scalar counterpart with no direction.

Q3: How do you calculate average velocity?

Average velocity is displacement divided by time interval: v_avg = (x2 - x1)/(t2 - t1). On an x(t) graph it is the slope of the straight line connecting the points at t1 and t2. Its sign shows direction.

Q4: How is average speed different from average velocity?

Average speed is total distance divided by Δt, a scalar always positive. Average velocity is displacement divided by Δt, a vector that can be negative or zero. For a round trip, average velocity is zero but average speed is not.

Q5: Can displacement be zero if you moved?

Yes. If you start at 5 m, go to 200 m, and return to 5 m, total distance is 390 m but displacement is 0 because x2 = x1. Displacement cares only about initial and final positions.

Q6: What does a horizontal line mean on a position-time graph?

It means the object is stationary. Position x does not change as time t increases, so Δx = 0 and average and instantaneous velocity are zero. Same position for any time.

Q7: Why is Δt always positive?

Time moves forward. t2 is later than t1, so Δt = t2 - t1 > 0. Therefore v_avg always has the same sign as Δx. A negative Δt would imply going backward in time.

Q8: When does average speed equal magnitude of average velocity?

Only when the object moves in one direction without reversing and without detours. Then total distance equals |Δx|, so s_avg = |v_avg|. Any reversal makes s_avg larger.

Q9: What is a particle in physics?

A particle is an idealized point-like object where all parts move in the same direction at the same rate. We use it when size and rotation do not affect the motion we study, like a car moving along a straight road.

Q10: How do you find average velocity from a position-time graph?

Draw a straight secant line between the points at t1 and t2 on the x(t) curve. Its slope is rise over run, Δx/Δt, which equals v_avg. A steeper upward secant means larger positive average velocity.

Exam Mapping

Concept	A-Level Physics (Edexcel/AQA)	AP Physics 1 & 2	IB Physics HL	College Physics 1 & 2	USAPhO / IPhO
Particle model & 1D restriction	3.4.1.1 Kinematics: modelling with particles	Unit 1 Kinematics: particle approximation	2.1 Motion: particle model	Ch 2: One-dimensional kinematics	Essential for defining system
Position, origin, +/- direction	3.4.1.2 Distance and displacement	Learning Obj 3.A.1.1: position vs distance	2.1: Position vs displacement	Position coordinate system	Sign convention mastery
Displacement Δx = x2 - x1	Required calculation, vector nature	Essential Science Practice: calculating displacement	Assessment: calculate Δx from graph	Δx problems, round-trip zero	Vector displacement in 1D limit
Average velocity as slope of x(t) secant	3.4.1.3 Velocity from x-t gradient	LO 3.A.1.2: vavg = Δx/Δt, slope interpretation	2.1: Determine vavg from x-t graph	Graphical analysis lab	Graphical kinematics
Average speed vs average velocity	3.4.1.4 Speed vs velocity, savg = distance/Δt	Distinguish scalar vs vector, traps	2.1: Average speed vs average velocity	Conceptual comparison questions	Inequality savg ≥ vavg proofs
Sample multi-stage problem	M1 kinematics with car/jogging context	Free response: multi-stage trip calculation	Data-based question with two phases	Homework: total time, vavg, savg	Complex path integration

Conclusion

You now have the full toolkit from Fundamentals of Physics for this opening module: the particle approximation that lets you ignore rotation, the x-axis with its origin and signed positions, the definition Δx = x2 - x1 that makes displacement depend only on endpoints, the vector nature of displacement, the definition v_avg = Δx/Δt as a signed slope on x(t), and the scalar s_avg = total distance/Δt that never cares about direction. With the armadillo graphs, the car-plus-jog example, and the distinction between zero displacement and non-zero distance, you can attack any A-Level, AP, IB, or college question that starts with "a particle moves along a straight line..."

Practice drawing the secant, writing x1, x2, t1, t2 before any numbers, and always checking the sign. That habit turns kinematics from a formula hunt into a story about where you started, where you ended, and how long it took.

References

• Fundamentals of Physics by Halliday, Resnick, and Walker, 12th Edition, Chapter 2 Motion Along a Straight Line, Module 2.1 Position, Displacement, and Average Velocity
• Supporting concepts: kinematics, one-dimensional motion, particle model