Frame of reference and uniform motion

The train starts moving, but the passenger sitting opposite stays in the same place. How can both observations describe the same person correctly? First we must agree on what we measure position relative to. Then we can try to predict that position a few seconds later.

Imagine two observers: one stands at the station and the other sits on the train. The passenger is moving relative to the first observer and at rest relative to the second. Neither is mistaken: they compare positions with different objects. To repeat a measurement, we need a reference object, a coordinate system fixed to it and a clock. Together these form a frame of reference. Without this choice, saying that an object moves is not yet a complete description.

Suppose we want the travel time between two cities. Would tracking every wheel of the car change the answer? Hardly: following one point is enough. A point particle is a model of an object whose size and shape can be neglected for the problem being solved. Now try parking in a narrow garage: the dimensions become essential. We choose a model to suit the question, not simply because the object itself is large or small.

How would you tell someone where you are on a straight road? Choose a visible post as the origin and one direction as positive. The coordinate x gives your position on this axis, in metres. Writing x = −3 м means three metres from the origin in the negative direction. Moving the origin or reversing the axis changes the number even if you stand still. The coordinate depends on the convention we choose.

Walk from home to a shop and back. Your step counter records a nonzero distance, but you return to your starting position. We therefore need two quantities. Distance travelled is the total length of the path followed; it is nonnegative. Displacement is a vector: an arrow from the initial position to the final one. It is zero after you return. Displacement tells us where you ended up relative to the start; distance tells us how far you walked.

For motion along a line, it is convenient to use the displacement's component: sₓ = x − x₀, where x₀ is the initial coordinate. Why subtract? We want the change in position, not the distance from the chosen origin. If the coordinate increases, sₓ is positive; if it decreases, sₓ is negative. Shifting the origin changes both coordinates by the same amount, so their difference stays the same. A negative component does not make the distance travelled negative.

Mark a toy cart's position every second. If it makes equal displacements along a straight line in any equal time intervals, it has uniform rectilinear motion. The velocity component vₓ tells us how much the coordinate changes each second; its unit is m/s. If vₓ = −2 м/с, the coordinate decreases by 2 m every second. The speed is 2 m/s: the minus sign specifies direction, not slowness.

In one second the coordinate changes by vₓ metres; in two seconds it changes twice as much. With constant velocity, these equal changes add up, giving sₓ = vₓt. Here t is the elapsed time in seconds; multiplying m/s by s gives metres. Doubling the time doubles the displacement; if vₓ = 0, the displacement is zero. If the cart accelerates, its initial velocity alone no longer describes all the following seconds.

Displacement alone does not tell us where an object is: we also need its starting position. Add the change to the initial coordinate: x = x₀ + vₓt. At t = 0 we obtain x₀, as expected. A more negative velocity makes the coordinate decrease faster; a positive velocity makes it increase. The starting point need not be the origin. This equation predicts position for as long as the velocity component stays constant.

Plot time t horizontally and coordinate x vertically. Equal changes in coordinate over equal times give a straight line. The ratio Δx/Δt is its slope and also equals vₓ. A horizontal line means rest, an upward slope means motion in the positive direction, and a downward slope means motion in the negative direction. With the same axis scales, a steeper line means greater speed. A downward-sloping graph does not by itself mean that the object is slowing down.

A walker starts at x₀ = 6 м and moves with vₓ = −2 м/с. Where are they after 4 s?

The initial position is to the right of the origin: x₀ = 6 м. The velocity vₓ = −2 м/с means that the coordinate decreases by 2 m every second.

In 4 s the change is sₓ = vₓt = −2 · 4 = −8 м. Therefore x = x₀ + sₓ = 6 − 8 = −2 м.

Check the motion: the origin is 6 m away, so the walker reaches it in 3 s. During the final second they travel another 2 m in the negative direction. They never turn around, so the distance travelled is 8 m.

After 4 s the walker is at x = −2 м, two metres from the origin in the negative direction. The distance travelled is 8 m; the displacement component is −8 m.

Describing motion starts with an observer, an axis and a clock: different coordinates can describe the same event.

Distance adds up the whole route; displacement compares the final position with the initial one.

With constant velocity, equal changes in position add up to x = x₀ + vₓt. The sign of the velocity component specifies direction.

Next we will describe changes in velocity and learn to read an acceleration graph.