Path displacement and reference frame

Mechanical motion can be described only together with a reference frame: a reference body, a coordinate system attached to it, and a device for measuring time.

Path $s$ is a scalar equal to the length of the segment of the trajectory covered during the time interval considered; path is never negative and never decreases with time.

Displacement is the vector $\Delta\vec r = \vec r_2 - \vec r_1$ joining the initial and final positions of the body; in coordinates on a plane

$$\Delta\vec r = (x_2 - x_1,\; y_2 - y_1), \qquad |\Delta\vec r| = \sqrt{(\Delta x)^2 + (\Delta y)^2}.$$

Because the shortest distance between two points is a straight segment, we always have $s \ge |\Delta\vec r|$, and equality holds only for motion along a straight line in one direction.

Path and displacement depend on the reference frame in different ways: changing the reference frame adds a constant vector to the displacement, while the path changes only when the trajectory itself changes (for example, because the reference body moves). In SI the quantities $s$, $|\Delta\vec r|$, $x$, $y$ are measured in metres (m).

A runner moves along a straight track: first from the point $x_1 = 20\ \text{m}$ to the point $x_2 = 140\ \text{m}$, then back to the point $x_3 = 80\ \text{m}$. Find the path $s$ and the displacement for the whole motion.

A tourist walks 30 m north, then 40 m east, then 30 m south. The reference frame is fixed to the ground: the $Ox$ axis points east, the $Oy$ axis points north. Find the path and the displacement (magnitude and direction).

A train moves east at $20\ \text{m/s}$ relative to the ground. A passenger walks from the rear to the front of the carriage at $2\ \text{m/s}$ relative to the train, covering $6\ \text{m}$ along the carriage in $3\ \text{s}$. Compare the passenger's path and displacement in the train frame and in the ground frame.

A safe self-study plan (no hazardous measurements).

1. Open the PhET simulation "The Moving Man" and switch on the distance-travelled graph.
2. Set the man to move forward to $+5\ \text{m}$, then back to $0\ \text{m}$, and read the path $s$ and the displacement $\Delta x$ from the simulation.
3. Compare $s$ with $|\Delta\vec r|$: check whether $s \ge |\Delta\vec r|$ holds.
4. Repeat the run with straight-line motion in one direction; verify that now $s = |\Delta\vec r|$.
5. Then shift the motion to another reference frame: imagine the reference body placed on a uniformly moving platform and explain which quantities changed and which stayed the same.

Record the results in a table: run number, $s$, $\Delta x$, $|\Delta\vec r|$, conclusion.

To show the difference between path length (the length of the trajectory) and displacement (the vector from the start point to the end point) and to stress that displacement does not depend on the shape of the trajectory.

A sketch: a curved trajectory from point $A(x_1, y_1)$ to point $B(x_2, y_2)$; the path length $s$ is labelled along the curve; a dashed arrow $\vec{AB} = \Delta\vec r$ drawn straight from $A$ to $B$ with its magnitude $|\Delta\vec r|$ labelled; next to it a separate drawing with coordinate axes $Ox$, $Oy$ and the legs $\Delta x$, $\Delta y$ of the displacement right triangle.

An athlete runs one stadium lap of length $400\,\mathrm{m}$ in $50\,\mathrm{s}$. The start and finish coincide. The reference frame is attached to the stadium, with the origin at the start line, so the start and finish coordinates are $(0,0)$.

Path is the scalar sum of travelled segments: $s=400\,\mathrm{m}$. Displacement is the vector $\Delta \vec r=\vec r_f-\vec r_s$; here $\vec r_f=\vec r_s$, so $\Delta \vec r=0$. The choice of reference frame fixes the coordinates $(0,0)$ but does not change the conclusion: when initial and final points coincide, displacement is zero while path is not.

A bus travels from home to school: $3.0\,\mathrm{km}$ east, then $4.0\,\mathrm{km}$ north, taking $12\,\mathrm{min}$. The reference frame is attached to home: the $x$-axis points east, the $y$-axis points north.

Path equals the sum of the segment magnitudes: $s=3.0+4.0=7.0\,\mathrm{km}$. Displacement is the vector $\Delta \vec r=(3.0\,\mathrm{km},4.0\,\mathrm{km})$; its magnitude is $|\Delta \vec r|=\sqrt{(3.0)^2+(4.0)^2}=5.0\,\mathrm{km}$. Thus the reference frame fixes the vector components, while path does not depend on the choice of axes.

A phone app records a bike ride: path $12.0\,\mathrm{km}$ in $40\,\mathrm{min}$. The cyclist starts and finishes at the same café. In the GPS map reference frame, the start and finish have the same coordinates, for example $(0,0)$ in a local plane grid.

The app builds path as the sum of small track segments: $s=12.0\,\mathrm{km}$. Displacement $\Delta \vec r=\vec r_f-\vec r_s=(0,0)$ when returning to the same point, although path is not zero. Here the reference frame is the Earth-fixed map; it lets the start and finish points be written as coordinates and the displacement vector be computed.