Speed and uniform motion

A passenger walks towards the rear of a train. To a fellow passenger they move backwards, but to someone on the platform they move forwards. Can both observers be right? First we must ask what the motion is measured relative to.

Mark a point on the carriage floor. A seated passenger stays the same distance from it but gets farther from a post on the platform. Motion is a change of position over time relative to a chosen reference object. To make the description testable, we also need a coordinate axis and a clock; together they form a reference frame. Before asking how fast something moves, state what its positions are compared with.

Imagine photographs taken every second. If the body covers equal distances between successive positions, and this holds for any equal time intervals, its motion is uniform: its speed stays constant. On a circular track, however, its direction keeps changing. The velocity vector is constant in uniform straight-line motion without reversals. Turning and speeding up are two different ways to change velocity.

A trolley travels uniformly for 12 m in 4 s. Each second accounts for 3 m: this is the meaning of v=S/t. Here S is distance in metres, t is a nonzero time interval in seconds, and v is speed in m/s. Over the same time, twice the distance requires twice the speed; for the same distance, twice the time means half the speed. If S=0 over a finite time, the average speed is zero.

Where does the conversion factor 3.6 come from? In an hour, or 3600 s, motion at 1 m/s covers 3600 m, or 3.6 km. Thus 1\,\text{м/с}=3{,}6\,\text{км/ч}. Both the distance unit and the time unit have changed. To convert km/h to m/s, divide by 3.6; for example, 36 km/h is 10 m/s. In one second, a car at that speed travels roughly the combined length of two passenger cars.

Walk 3 m forwards and 3 m back. A distance counter records 6 m, but your final position is your starting position. Displacement, the directed segment from the initial position to the final one, is zero. Distance adds the lengths of all sections and cannot be negative. The magnitude of displacement cannot exceed the distance; they are equal for straight-line motion without a change of direction.

How can we compare journeys with changes of speed and stops? Imagine uniform motion covering the same distance in the total time. Its speed v_{\text{ср}}=S/t is called the average speed. Time spent stopped belongs in the denominator even though it adds no distance. A longer stop over the same distance gives a smaller result. The arithmetic mean of two speeds works when the times spent at them are equal; equal distances are not enough.

In one second the carriage shifts relative to the platform, and the passenger also shifts relative to the carriage. An outside observer must include both displacements to find the passenger's new position. Let \vec v_1 be the passenger's velocity relative to the carriage, \vec v_2 the carriage's velocity relative to the platform, and \vec v the passenger's velocity relative to the platform. These details matter: the first two velocities answer different questions.

Dividing the sum of displacements over the same time by that time gives the classical rule \vec v=\vec v_1+\vec v_2. If right is positive, a carriage velocity of 10 m/s and a passenger velocity of −1 m/s relative to it give 9 m/s to the right. When \vec v_1=0, the passenger moves with the carriage. This rule applies to frames in translational motion at speeds much smaller than the speed of light; velocities add as vectors.

What would successive photographs of an accelerating trolley show? The gaps between its positions grow. Starting from rest with constant acceleration, the distances in successive equal time intervals have the ratio 1:3:5:7:\ldots. The trolley begins each new interval with a greater speed. This is a prediction of the model for now; in the next lesson we will find the area under the velocity graph and derive this sequence.

A pedestrian walks 300 m in 5 minutes, then stops for 1 minute. Find the average speed over the whole interval.

We use the whole interval, so the stopping time is included.

Distance S = 300 м; time t = (5 + 1) × 60 = 360 с.

Use the definition of average speed: vср = S/t = 300/360 ≈ 0,83 м/с.

Check: the speed during walking was 1 m/s; the stop reduced the average speed.

Average speed ≈ 0.83 m/s. Walking uniformly at about 83 cm each second for all 6 minutes would cover the same 300 m.

Motion is described relative to a chosen object; observers can measure different velocities without contradiction.

Distance remembers the whole route; displacement only the start and finish. Returning makes displacement zero, but not distance.

Average speed includes all the time, including stops; adding velocities requires both reference frames and directions to be stated.

Next we introduce acceleration and explain how distance grows as an object speeds up.