You are standing in a railway carriage, holding a handrail. Your distance from the door stays the same, but platform posts glide past the window. Are you moving or standing still? Both answers can be correct, provided you say relative to what.
Motion is a change in an object’s position over time in a chosen reference frame. Speed describes how fast it moves; velocity also includes the direction of the change in position. A number such as “5 metres per second” therefore needs another detail: relative to the platform, the carriage or another object? OpenStax: motion and reference frames.
Standing still needs a reference too
To describe motion, first agree where to measure position from, which direction counts as positive and how to measure time. For example, fix the coordinate origin at a post, make rightwards along the track positive and start a stopwatch when the carriage passes the post. A reference object, a coordinate system attached to it and a way to keep time give us a reference frame.
If a passenger’s position relative to the carriage does not change, the passenger is at rest in the carriage frame. In the platform frame, the same passenger moves with the carriage. Saying “really standing still” without naming a frame adds nothing here: both descriptions concern the same person.
That does not make measurements arbitrary. Observers must agree which events and coordinates they are comparing. We will start with straight-line motion at constant speeds far below the speed of light. Acceleration, turns and the delay in receiving light lie outside this example. OpenStax: limits of the classical description of relative motion.
One metre per second in the carriage, six on the platform
Imagine a carriage moving to the right at 5 m/s relative to the platform. A passenger walks forward at 1 m/s relative to the carriage. In one second the carriage carries the passenger five metres, and walking adds another metre. The passenger therefore moves at 6 m/s relative to the platform. This addition applies to the same straight direction and our low speeds.
Now move the coordinate origin into the carriage. The carriage itself has zero velocity, the passenger has plus one metre per second, and the platform post has minus five: it moves away to the left. The minus sign identifies a direction along our chosen axis, rather than “less movement”.
Before switching the diagram, predict where the post will end up. Then compare descriptions of the same two seconds. Every object is represented by a point, with their initial coordinates chosen to coincide. This does not draw the carriage’s dimensions.
The same two seconds in two reference frames
Predict where the post will move in the carriage frame. Switch frames and compare the coordinates. The keyboard arrows change the selection.
After two seconds in the platform frame, the passenger’s coordinate is 12 m and the post’s is 0 m. In the carriage frame they are 2 m and −10 m. The passenger–post distance is 12 m in both descriptions. Switching frames has not moved the passenger: it has changed how we record the position. OpenStax: combining relative velocities.
Branch · With equationsOptional · you can skip this
Subtracting the carriage’s motion
Let x be an object’s coordinate relative to the platform in metres, t the time in seconds and u the carriage’s constant velocity relative to the platform in m/s. If the two origins coincide at t = 0, the carriage origin shifts by ut during time t. The coordinate in that frame is:
(x)^(′) = x − ut.
For the passenger after 2 s, (x)^(′) = 12 − 5 ⋅ 2 = 2 m. Similarly, the velocity along this axis is (v)^(′) = v − u: subtract the carriage’s 5 m/s from the platform value of 6 m/s to obtain 1 m/s in the carriage. For the post, 0 − 5 = − 5 m/s. Multiplying m/s by seconds gives metres, as a coordinate requires.
This is a classical, nonrelativistic approximation for uniformly moving frames with parallel axes and a shared measure of time. Near the speed of light, this simple subtraction fails; light cannot be treated as another pedestrian in the carriage.
Returning to the start is different from going nowhere
Leave the carriage and consider a walk from a fixed marker. You go six metres right, then six metres back, stopping where you began. A counter of metres travelled records a distance of 12 m. Distance travelled is the length of the path; returning adds metres rather than erasing them.
Displacement answers another question: how does the final position differ from the initial one? In this walk they coincide, so displacement is zero. If you stopped two metres to the right of the start, displacement along the axis would be +2 m, regardless of extra steps along the way. OpenStax: distance and displacement.
This distinction leads to two averages. Average speed is the entire distance travelled divided by the entire chosen time interval. Average velocity is displacement divided by that same interval. The first tells us how quickly travelled metres accumulated; the second retains information about the overall change in position and its direction.
Suppose the six metres out took three seconds, the return took another three, and you then stood still for two seconds. Over the entire eight-second interval, average speed is 12 m / 8 s = 1.5 m/s. Average velocity is 0 m/s, because the final displacement is zero. That zero does not deny that you walked.
The two stationary seconds belong to this interval. Excluding them gives a different question: the average over only the six moving seconds, 2 m/s. Before comparing average speeds, check the beginning and end of the interval as well as the units. OpenStax: average speed and velocity.
Branch · With unitsOptional · you can skip this
Why one metre per second equals 3.6 km/h
An hour contains 3,600 seconds and a kilometre contains 1,000 metres. At a constant speed of 1 m/s, an object travels 3,600 m, or 3.6 km, in an hour. Therefore:
1 m/s = 3.6 km/h.
Multiply by 3.6 to go from m/s to km/h; divide for the reverse conversion. For example, 36 km/h = 10 m/s. Converting changes the unit, not the motion or the reference frame.
The average for a journey and the speed now describe different things
Knowing only 12 m and 8 s cannot recover the details of the walk. You could have walked at different speeds on different sections or arranged the stops differently. A single average compresses a whole history into one ratio.
Instantaneous speed describes motion at a particular moment. Picture measuring position changes over shorter and shorter intervals around that moment. A speedometer reading concerns speed now, rather than the average since the journey began. OpenStax: instantaneous velocity and speed.
When direction matters, we represent velocity by an arrow: a vector. Two people can walk equally fast in opposite directions: their speeds are equal, but their velocity vectors differ. Likewise, motion in a circle can maintain a constant speed while the velocity’s direction keeps changing. In equations and translations, check whether a quantity includes direction or only magnitude.
The reference frame matters even more for a spacecraft. Earth itself moves, so a probe’s velocity relative to Earth and relative to the Sun are different quantities. Astronomical measurements record the velocity frame along with the data; it is part of what the number means. NRAO: velocity reference frames.
Return to our passenger at the handrail. The passenger is stationary relative to the carriage and moving relative to the platform. To understand a speed or velocity, establish three things: relative to what, over which interval and whether direction is included. And when someone returns to the start, do not confuse zero displacement with zero distance travelled.
Try describing a familiar object’s motion first relative to the room, then relative to something you move alongside it. For a larger scale, return to Earth’s rotation and the cycle of day and night. Next in the course, we will ask why the motion of objects changes.