Uniform motion

You are cruising on the freeway, the speedometer holds one value, and the kilometer markers pass in a steady rhythm. You are not pulling away from lights and you are not braking. Each second adds the same distance. That is uniform motion: speed does not change with time, so acceleration a = 0.

In everyday estimates and in many problems you work with distance s and with v as a positive number. If direction is also fixed, the velocity vector is constant. On a straight line without turning back, distance and displacement match as numbers. If you turn around or close a loop, distance keeps growing while displacement may end near zero. This article sticks to a one-way stretch.

The core relation is one formula:

s = v·t

Know any two of v, s, t and the third follows: v = s/t or t = s/v. That is not three different physics stories, only one definition at v = const. Phrases like “steady speed”, “constant speed”, or “no acceleration” keep you with this block.

First numeric intuition: v = 20 m/s for t = 90 s gives s = 1800 m, about 1.8 km. That is a calm freeway pace, or a strong cycling pace on flat ground. The same numbers appear in the problem below: work it out by hand, then check in the calculator.

Second intuition: at constant v = 25 m/s (about 90 km/h), a stretch s = 5000 m takes t = 200 s, or 3 min 20 s. If your head says “five kilometers at ninety is a few minutes”, you are in the right order of magnitude. If you get half an hour or a few seconds, hunt units or the wrong formula.

On an s(t) graph you get a straight line with slope v. Larger v means a steeper rise in distance. On a v(t) graph you get a horizontal line: speed does not depend on time. Keep both pictures before you move on to acceleration, where both lines change and ½at² appears.

Units matter. Use distance in meters, time in seconds, speed in m/s. If you are given km/h, divide by 3.6. For example 90 km/h is 25 m/s. If you are given minutes, multiply time by 60 before you put it into s = v·t. Mixed units are the most common reason for a weird result: someone types 90 “as m/s” and in 90 s gets a distance as if they crossed half a continent.

The model assumes that on the segment you study you may treat v as constant, or that you knowingly use the idealization “steady the whole way”. A car in traffic is not uniform motion for the whole trip, even if it rolls steadily for a moment between two braking stretches. Then either split the trip into segments or switch to average speed.

That is why the tools split. The uniform-motion calculator lets you fill any two of v, s, t. The distance-only page always computes s = v·t from two inputs. Separately sits average speed: v̄ = s/t from total distance and total time, stops included.

Trip average is not the same as the speedometer reading. Do not drop it in as “constant v” in the uniform model without reading the problem. Phrases like “at constant speed” or “steady” keep you with s = v·t. Phrases like “on average”, “including stops”, or “in traffic” switch you to average speed. That trap is common when the problem first says “60 on average” and then asks for travel time with no stops.

A dramatic trap with two equal lengths looks like this. You cover s at v₁ and the same s at v₂, so you take (v₁+v₂)/2 and feel safe. The answer is rarely right. You spend more time on the slower stretch, so the mean sits closer to the smaller speed: 2v₁v₂/(v₁+v₂), the harmonic mean. The average-speed calculator does not guess your arithmetic. It uses total s and t: sum the segments first, then divide.

Other common mistakes: using s = v·t while accelerating from rest; feeding km/h straight into a formula with seconds; mixing the distance-only tool with full v-s-t; computing arrival time with bare s = v·t and forgetting a coffee stop. The formula alone will not add breaks. Either fold stops into total t and use average speed, or split the trip into driving segments and sum the times.

When v is not constant but a is, move to accelerated motion. There s = v₀t + ½at². Uniform motion is the special case a = 0 of the same language. You do not erase the old formulas. At a = 0 the t² term simply vanishes.

Uniform motion also returns as a segment inside longer stories. Driver reaction before braking is often s = v·t_r at nearly constant v. Horizontal flight at constant v_x in a horizontal throw is the same block sideways. A cruise between accelerations is again s = v·t. You learn one tool, then slot it into a larger puzzle.

Sanity-check the magnitude. 20 m/s is about 72 km/h. In 90 s (1.5 min) at that speed, about 1800 m looks sensible: under two kilometers. If you get 180 km in a minute and a half, you almost certainly typed km/h as if it were m/s. If you get 18 m, you probably divided instead of multiplying, or used the wrong t.

The model is not enough when speed clearly rises or falls on the stretch you care about, when the problem wants stops folded into one “constant v”, or when the s(t) graph bends. Then you need accelerated motion, average speed, or another tool. Do not rescue the situation with a made-up “average v” unless the problem gives an average outright.

In dynamics, straight-line uniform motion pairs with zero net force: drive balances drag, or nothing really pulls you along. This article stays kinematic. You compute s, v, t. Forces and masses enter only when you ask “why is v constant”, not “what is s”.

Before you start, check these points in order: (1) whether the problem guarantees constant v on that stretch, (2) whether the units match the choice in the page header, (3) which of the three tools you need, (4) whether the answer matches everyday experience. Then check the arithmetic in the calculator, and only then trust the number.

If you come back here from the acceleration article, compare the graphs. At constant v, distance grows evenly. At constant a, later seconds add more distance than earlier ones. Uniform motion is the simplest block in the series, but it returns in braking, throws, and average speed. Worth having it in hand before you move on.