1.4 Displacement, Path Length, and Trajectory

Hook

A runner completes exactly one lap of a 400 m track. When she crosses the finish line, a coach asks her two questions:

1. How far did you run?
2. How far are you from where you started?

The first answer is 400 m. The second answer is zero.

Both answers are correct. They are answering different questions about the same motion. The fact that these two perfectly reasonable measurements give such different results tells us something deep about how we describe motion mathematically. The rest of this section is about understanding exactly what each one measures and why they are not the same thing.

Prediction

Before we formalize anything, look at this path traced by a drone flying over a field.

[Interactive: A 2D grid (10 x 10) displays a complex, curvy path from point A near the bottom-left to point B near the upper-right. The path loops around, doubles back once, and meanders before arriving at B. The start and end points are clearly marked. Two input fields appear below: "Estimate the displacement (straight-line distance from A to B)" and "Estimate the total distance traveled (length along the path)." A submit button locks in the student's answers before revealing the computed values.]

Before you continue: Estimate both numbers. Which one is larger? Will they ever be equal? Commit to your answers before scrolling down.

Take a moment with this. The path clearly covers a lot of ground, but A and B are not that far apart. Your intuition about the relationship between these two numbers is the seed of everything in this section.

Exploration

Drawing your own path

Now it is your turn to build intuition by experimenting.

[Interactive: A drawing canvas on a grid. The student clicks to place a starting point, then draws or clicks waypoints to create a path. Two quantities update in real time as they draw: - Displacement is shown as a straight arrow from the start to the current endpoint, with its magnitude displayed (e.g., "Displacement: 6.3 units"). - Path length is shown as a running total measured along the curve (e.g., "Path length: 14.7 units"). A reset button lets the student start over. Below the canvas, three guided challenges appear in sequence.]

Try these three challenges:

Challenge 1: Draw a path where the displacement and path length are equal.

What kind of path does this require? You should find that only a perfectly straight path from start to end gives equal values. Any deviation --- any curve, any turn --- makes the path length larger than the displacement.

Challenge 2: Draw a path where the displacement is zero but the path length is as large as you can make it.

You need to end up exactly where you started. A closed loop does the trick. The longer and more winding the loop, the larger the path length, but the displacement stays stubbornly at zero.

Challenge 3: Draw two different paths that have the same displacement but very different path lengths.

Fix your start and end points. Now draw a straight line between them, then draw a wildly winding route between the same two points. The displacement is identical. The path lengths are completely different.

What pattern is emerging? Displacement only cares about two things: where you started and where you ended. Path length cares about every step in between.

Concept Reveal

Let us now give these ideas precise definitions.

Trajectory

The trajectory is the complete geometric path an object follows through space. It is the shape of the motion --- the curve you would see if the object left a trail of paint behind it. The trajectory tells you the route but says nothing about when the object was at each point.

In one dimension, a trajectory is just a segment (or collection of segments) of the number line. In two or three dimensions, it is a curve through space.

Displacement

Displacement is the vector pointing from the initial position to the final position. It answers the question: where did you end up, relative to where you started?

For motion in one dimension, if an object starts at position $x(t_1)$ and ends at position $x(t_2)$, the displacement is:

$$\Delta x = x(t_2) - x(t_1)$$

In two or three dimensions, displacement is a vector:

$$\Delta \vec{r} = \vec{r}(t_2) - \vec{r}(t_1)$$

Key properties of displacement: - It is a vector: it has both magnitude and direction. - It depends only on the starting and ending positions, not on the route taken. - It can be zero even if the object moved a great deal (the runner on the track). - It can be positive or negative in 1D, indicating direction along the axis.

Path length

Path length (also called distance traveled) is the total length of the trajectory. It answers the question: how much ground did you cover?

Key properties of path length: - It is a scalar: it has magnitude only, no direction. - It depends on every detail of the route taken between start and finish. - It is always non-negative. You cannot travel a negative distance. - It is always greater than or equal to the magnitude of the displacement.

That last point deserves emphasis. The shortest distance between two points is a straight line. The magnitude of the displacement is the straight-line distance between start and end. Any deviation from a straight path adds to the path length without changing the displacement. So:

$$\text{path length} \geq |\Delta \vec{r}|$$

with equality only when the motion is in a straight line without reversing direction.

A contrast table

Feature	Displacement	Path length
Type	Vector	Scalar
Depends on	Start and end positions only	Entire route
Can be zero for real motion?	Yes (closed loop)	Only if the object never moved
Can be negative?	Yes (in 1D)	Never
Question it answers	"Where did you end up?"	"How far did you travel?"

Worked Example

A car drives east along a straight road for 5 km, then turns around and drives west for 2 km.

[Animation: A number line with a dot at the origin. The dot moves right to the 5 km mark, pauses, then moves left to the 3 km mark. Two quantities are tracked below: displacement updates from 0 to +5 km then down to +3 km; path length increases from 0 to 5 km and continues upward to 7 km.]

Displacement: The car started at $x = 0$ and ended at $x = 3$ km. So $\Delta x = 3 \text{ km}$ (eastward).

Path length: The car traveled 5 km east and then 2 km west. Total path length $= 5 + 2 = 7 \text{ km}$.

Notice how the path length kept accumulating regardless of direction, while the displacement reflects only the net result.

Variation Theory: What Changes, What Stays the Same?

Same displacement, different path lengths

Consider three different journeys, all starting at the origin and ending at $x = 4$ m:

1. Walk directly from $x = 0$ to $x = 4$ m. Path length = 4 m.
2. Walk from $x = 0$ to $x = 8$ m, then back to $x = 4$ m. Path length = 12 m.
3. Walk from $x = 0$ to $x = -3$ m, then to $x = 10$ m, then back to $x = 4$ m. Path length = 19 m.

[Animation: Three panels showing a dot on a number line executing each journey. All three end at the same position. The displacement arrow (from 0 to 4) is identical in all three panels. The path length, shown as a growing counter, ends at very different values: 4, 12, and 19.]

What changed? The route. What stayed the same? The displacement. What does this tell you? Displacement contains no information about what happened between the start and end.

Same path length, different displacements

Now consider three journeys where each person walks exactly 10 m total:

1. Walk 10 m to the right. Displacement = 10 m.
2. Walk 5 m to the right, then 5 m to the left. Displacement = 0 m.
3. Walk 8 m to the right, then 2 m to the left. Displacement = 6 m.

What changed? The displacement. What stayed the same? The path length. What does this tell you? Knowing how far someone traveled does not tell you where they ended up.

Connection

These ideas connect directly to what you have already learned. In Section 1.2, you saw that the position function $x(t)$ gives a complete description of motion. Displacement falls out of that function immediately:

$$\Delta x = x(t_2) - x(t_1)$$

You just evaluate the position at two times and subtract. You do not need to know anything about what happened between $t_1$ and $t_2$.

Path length is different. To compute it, you need to track every bit of motion, including backtracking. You need to accumulate the total amount of movement along the way. In one dimension, this requires knowing how $x(t)$ changes direction, and in multiple dimensions, you need to integrate the speed along the curve:

$$\text{path length} = \int_{t_1}^{t_2} |\vec{v}(t)|\, dt$$

We have not introduced velocity or integrals yet --- those are coming in Sections 1.5 and later. For now, the takeaway is that displacement is easy to compute from the position function (just subtract), while path length requires more information about the full journey. This is a preview of why calculus will become essential.

And recall Section 1.3: your choice of coordinate system affects the numbers you assign to positions, but it does not change the displacement or the path length. Both are real, physical quantities. The runner's zero displacement and 400 m path length are facts about her motion, not artifacts of how you set up your axes.

Practice Layers

Layer 1: Concrete computation

Problem 1. A particle moves along the $x$-axis. Starting at $x = 2$ m, it moves to $x = 7$ m, then to $x = 3$ m.

(a) What is the displacement?

(b) What is the path length?

Solution. (a) Displacement $= x_f - x_i = 3 - 2 = 1$ m. (b) Path length $= |7 - 2| + |3 - 7| = 5 + 4 = 9$ m.

Problem 2. A hiker walks 3 km north, then 4 km east.

(a) What is the path length?

(b) What is the magnitude of the displacement?

(c) Why are these different?

Solution. (a) Path length $= 3 + 4 = 7$ km. (b) Displacement magnitude $= \sqrt{3^2 + 4^2} = 5$ km. (c) The path was not a straight line; it had a right-angle turn. The displacement is the hypotenuse of the right triangle, which is shorter than the sum of the two legs.

Layer 2: Pattern --- what can you conclude?

Problem 3. You are told that an object's displacement over some time interval is 6 m to the right.

(a) What is the minimum possible path length?

(b) What is the maximum possible path length?

(c) Can you determine the exact path length from the displacement alone?

Think about this before reading the answer.

Solution. (a) The minimum path length is 6 m (straight-line motion, no backtracking). (b) There is no maximum --- the object could have zigzagged arbitrarily many times. (c) No. Displacement alone does not determine path length.

Problem 4. You are told that an object's path length is 10 m.

(a) What is the maximum possible magnitude of the displacement?

(b) What is the minimum possible magnitude of the displacement?

(c) Can you determine the displacement from the path length alone?

Solution. (a) The maximum displacement magnitude is 10 m (straight-line motion). (b) The minimum displacement magnitude is 0 m (a closed loop). (c) No. Path length alone does not determine displacement.

Layer 3: Structure --- why it works this way

Problem 5.

Can the magnitude of displacement ever exceed the path length? Explain why or why not.

Solution. No, never. The displacement is the straight-line distance from start to end. The path length is the total distance traveled along the actual route. A straight line is the shortest distance between two points, so any actual path must be at least as long as the displacement. The inequality $\text{path length} \geq |\Delta \vec{r}|$ always holds, and it is an equality only when the motion is along a straight line with no reversal of direction.

This is not just a physics fact --- it is a geometric fact, an instance of the triangle inequality.

Layer 4: Creation

Problem 6.

Design a path in two dimensions where the displacement is zero but the path length is exactly 10 m. Describe or sketch your path.

There are infinitely many correct answers. Some possibilities: - A circle of circumference 10 m (radius $= \frac{10}{2\pi} \approx 1.59$ m). - A square with side length 2.5 m (perimeter $= 4 \times 2.5 = 10$ m). - Walk 5 m east, then 5 m west (back to the start). - Any closed loop with total perimeter 10 m.

The key insight: you need a closed path (displacement = 0) whose total length is 10 m. The shape does not matter.

Problem 7.

A friend says, "I walked 1 km today." Another friend says, "I ended up 1 km from where I started." Could they be describing the same walk? Must they be? Give an example where both statements are true simultaneously and another where only one is true.

Solution. They could be describing the same walk: a straight-line walk of 1 km. But they need not be. The first friend might have walked 1 km total along a winding path and ended up only 0.3 km from the start. The second friend might have taken a winding 5 km walk that ended 1 km from the start. "Walked 1 km" describes path length; "ended up 1 km from the start" describes displacement magnitude.

Reflection

Question 1: When would a physicist (or engineer, or navigator) care specifically about displacement? When would they care specifically about path length? Think of a concrete scenario for each.

Displacement matters when you need to know the net change in position --- for example, a pilot calculating how far off course an aircraft is from its destination. Path length matters when you care about the total effort or resource expenditure --- for example, calculating fuel consumption, wear on tires, or the total energy a runner expends.

Question 2: After this section, what is your one-sentence summary of the difference between displacement and path length?

There is no single right answer, but here is one way to put it: Displacement tells you where you ended up; path length tells you how you got there.

Question 3: Look back at the position function $x(t)$ from Section 1.2. Which is easier to compute from $x(t)$: displacement or path length? Why?

Displacement is easier --- you only need $x$ at two times. Path length requires knowing everything that happened in between, which is why it will eventually require integration.

Summary

• The trajectory is the geometric shape of the path through space.
• Displacement ($\Delta \vec{r} = \vec{r}(t_2) - \vec{r}(t_1)$) is a vector from start to finish. It depends only on the endpoints.
• Path length is a scalar measuring the total distance traveled along the trajectory. It depends on every detail of the route.
• Path length is always greater than or equal to the magnitude of displacement: $\text{path length} \geq |\Delta \vec{r}|$.
• These two quantities answer fundamentally different questions about the same motion. Treating them as interchangeable is one of the most common errors in introductory physics.