Introduction
A route can involve several changes of position before arriving at its destination. The individual legs describe how the route proceeds, but a different question asks for the overall change from its beginning to its end. That overall change is the net displacement.
Net displacement is a vector: it records both direction and amount of change in position. Total distance traveled is a length: it measures how much ground the route covers. These quantities are related, but they are not generally equal. Keeping them separate is essential when interpreting a route.
Displacements Along a Route
Suppose a route visits positions \(\mathbf{p}_0,\mathbf{p}_1,\ldots,\mathbf{p}_n\) in three-dimensional space, in that order. The displacement vector for leg \(i\) points from the position at the start of that leg to the position at its end.
Vector subtraction and addition are performed component by component. Thus, if the legs are given directly as vectors, their sum gives the net displacement. The order in which we write the vectors does not change their sum, but the order of the route still matters when identifying which displacement belongs to which leg and how to find the intermediate positions.
This sum gives the vector from the starting point to the ending point, regardless of how many intermediate positions the route visits. It does not describe the route’s detailed shape.
Why the Leg Vectors Add to the Net Displacement
The addition rule follows directly from how each leg is defined. The displacement of a leg is the difference between its ending and starting positions. When we add those differences in sequence, intermediate positions appear once with a positive sign and once with a negative sign, so they cancel.
For example, \(\mathbf{p}_1\) appears first as \(+\mathbf{p}_1\), then as \(-\mathbf{p}_1\) in the next leg; these terms cancel. The same cancellation occurs for every intermediate position \(\mathbf{p}_2,\ldots,\mathbf{p}_{n-1}\). Only the final position and the negative of the initial position remain. The result is exactly the definition of net displacement.
Worked Example: A Four-Position Route in Three Dimensions
Worked Example: A Four-Position Route in Three Dimensions
A route begins at \(\mathbf{p}_0=(1,-2,0)\) and has three successive displacement vectors \(\mathbf{d}_1=(3,1,2)\), \(\mathbf{d}_2=(-2,4,-1)\), and \(\mathbf{d}_3=(0,-3,2)\). Find each intermediate position, the net displacement, and the total distance if each leg is a straight segment.
For the first leg, add \(\mathbf{d}_1\) to the initial position:
For the second leg:
For the third leg:
Add the three displacements to find the net displacement:
Check this result using the endpoints: \(\mathbf{p}_3-\mathbf{p}_0=(2,0,3)-(1,-2,0)=(2-1,0-(-2),3-0)=(1,2,3)\), as required.
For straight legs, the length of each leg is the magnitude of its displacement vector. The three lengths are:
Therefore the total distance traveled along these straight segments is \(\sqrt{14}+\sqrt{21}+\sqrt{13}\) units. The magnitude of the net displacement is \(\sqrt{1^2+2^2+3^2}=\sqrt{14}\) units. These are different measurements: one adds the lengths of the route’s three legs, while the other measures the direct change from start to finish.
Total Distance and the Shape of Each Leg
For a straight-segment leg, the displacement runs directly from one endpoint to the other. Its magnitude is the length of that segment. Consequently, if every leg of a route is straight, the total distance traveled is the sum of the magnitudes of the leg displacement vectors.
This distinction means that a list of displacement vectors determines the route’s net displacement, but does not generally determine the distance traveled. If a leg is curved, its endpoint displacement does not reveal how much the path bends or how long it is.
Worked Example: A Route That Returns to Its Start
Worked Example: A Route That Returns to Its Start
A route consists of three straight legs with displacements \(\mathbf{d}_1=(4,0,0)\), \(\mathbf{d}_2=(-1,3,0)\), and \(\mathbf{d}_3=(-3,-3,0)\). Find its net displacement and total distance.
Add the vectors component by component:
The route’s net displacement is the zero vector, so it ends at the same position where it began. For example, starting at \((0,0,0)\), the intermediate positions are \((4,0,0)\), then \((3,3,0)\), then \((0,0,0)\).
Because all three legs are straight, their lengths are the magnitudes of their displacement vectors:
The total distance is \(4+\sqrt{10}+3\sqrt{2}\) units. The magnitude of the net displacement is \(\|(0,0,0)\|=\sqrt{0^2+0^2+0^2}=0\) units. The zero net displacement does not mean that no movement took place: it means only that the route ends where it started.
As a check, the endpoint calculation gives \((0,0,0)-(0,0,0)=(0,0,0)\), agreeing with the vector sum. The positive leg lengths also confirm that the route covered a positive distance despite returning to its initial position.
Worked Example: Identical Displacements, Different Distances
Worked Example: Identical Displacements, Different Distances
Consider a two-leg route that starts at \(A=(0,0,0)\), reaches \(B=(4,0,0)\), and then reaches \(C=(4,0,3)\). The first leg could be the straight segment from \(A\) to \(B\), or it could be the upper semicircle in the \(xy\)-plane with the same endpoints. The second leg is straight. Compare the net displacement and total distance in these two cases.
The displacement of the first leg is \(B-A=(4-0,0-0,0-0)=(4,0,0)\). The second-leg displacement is \(C-B=(4-4,0-0,3-0)=(0,0,3)\). Thus the net displacement is:
Checking with the endpoints gives \(C-A=(4-0,0-0,3-0)=(4,0,3)\). Its magnitude is \(\sqrt{4^2+0^2+3^2}=\sqrt{16+9}=5\) units. This net displacement is the same whether the first leg is straight or curved, because its endpoints have not changed.
If the first leg is straight, its length is \(\sqrt{4^2+0^2+0^2}=4\) units. The second leg has length \(\sqrt{0^2+0^2+3^2}=3\) units. The total distance in this case is \(4+3=7\) units.
For the curved option, parameterize the upper semicircle by \(\mathbf{r}(\theta)=(2+2\cos\theta,2\sin\theta,0)\), with \(\theta\) running from \(\pi\) down to \(0\). At \(\theta=\pi\), this gives \((2-2,0,0)=(0,0,0)=A\); at \(\theta=0\), it gives \((2+2,0,0)=(4,0,0)=B\). Its derivative is \(\mathbf{r}'(\theta)=(-2\sin\theta,2\cos\theta,0)\), whose magnitude is \(\sqrt{4\sin^2\theta+4\cos^2\theta}=2\). The curved-leg length is therefore \(\int_0^\pi 2\,d\theta=2\pi\) units.
The second leg still has length \(3\) units, so the total distance for the curved route is \(2\pi+3\) units, while its net displacement remains \((4,0,3)\), of magnitude \(5\) units. The two routes have the same leg displacement vectors but different total distances. This confirms that displacement vectors alone do not determine the length of a curved leg.
A Common Mistake: Confusing Net Displacement with Distance
A common error is to take the magnitude of the net displacement as the total distance traveled. In the return-to-start example, that incorrect method would calculate:
and conclude that the route’s total distance is zero. That conclusion is wrong. The vector sum answers the question “What is the overall change in position?” It does not add up the length of the route. Since those three legs are straight, the correct distance calculation is:
The correction is to identify which quantity the problem asks for. For net displacement, add the vectors first. For distance along a route made entirely of straight segments, find each leg’s magnitude and add those lengths. For curved legs, use the actual path lengths; do not assume a displacement vector provides them.
A Reliable Procedure
Read them in route order, or calculate each as ending position minus starting position.
Add all \(x\)-components, all \(y\)-components, and all \(z\)-components to find the net displacement.
Subtract the initial position from the final position and confirm that it equals the vector sum.
For straight legs, add the magnitudes of their displacement vectors. For curved legs, use their actual path lengths.
Why the Distinction Matters
Net displacement summarizes the start-to-finish change in position. It is useful when the overall destination relative to the starting point is the quantity of interest. Total distance describes how much ground was covered along the route. It can depend on every turn or curve, including motion that eventually brings the route back toward its starting point.
These measurements answer different questions. If a route begins and ends at the same place, its net displacement is zero, but its distance may be substantial. If route legs are straight, their lengths can be calculated from their displacement vectors. If a leg curves, the endpoint vector still describes that leg’s displacement, but more information about the path is needed to find its length.
Check Your Understanding
Use the distinction between net displacement and distance to answer the following questions.
- A route has displacement vectors \((2,1,0)\) and \((-2,-1,0)\). What is its net displacement?
- If both legs in Question 1 are straight, what expression gives the total distance traveled?
- Why can a route have zero net displacement but positive total distance?
- Two routes have the same starting and ending points, but one has a curved leg. What can be concluded about their net displacements, and what cannot be determined from that fact alone?
- When does the magnitude of a leg’s displacement equal the distance traveled along that leg?