Maths Solver

Home › Word problems › Distance, speed, time

Distance, speed, and time

By Sara Lin · Last updated April 2026 · 8 min read

For motion at a constant speed, distance equals speed times time:

$$d = v \cdot t.$$

That is the entire content of this topic. Everything else — rearranging for $v$ or $t$, the unit conversions, the “there and back” problems — is bookkeeping. The trouble is that the bookkeeping is exactly where students make all the mistakes, so it is worth being careful about.

The three forms

Rearranging $d = vt$ for the other variables:

$$v = \frac{d}{t}, \qquad t = \frac{d}{v}.$$

Some textbooks present these as a triangle to memorise. I prefer just remembering $d = vt$ and rearranging on the spot — the same operation works for any two-variable formula and you do not have to keep separate triangles in your head for $V = IR$, $F = ma$, and so on.

Units: where almost everything goes wrong

The formula $d = vt$ is an equation between physical quantities, and the units have to match. If $v$ is in kilometres per hour and $t$ is in hours, then $d$ comes out in kilometres. If $v$ is in metres per second and $t$ is in minutes, then... well, then you have to convert.

Three conversions worth knowing by heart:

Always write the units down next to your numbers as you work. “$v = 60$” is ambiguous; “$v = 60$ km/h” is not, and lets you spot a unit mismatch before it becomes a wrong answer.

The “there and back” trap

“A car drives 60 km from A to B at 60 km/h, then returns at 40 km/h. What is the average speed for the round trip?”

The instinctive answer is the arithmetic mean: $(60 + 40)/2 = 50$ km/h. That is wrong.

Average speed is total distance divided by total time, not the average of the two speeds. The first leg takes $60/60 = 1$ hour; the second takes $60/40 = 1.5$ hours. Total distance is $120$ km, total time is $2.5$ hours, so the average speed is $120/2.5 = 48$ km/h. The slower leg pulls the average down because more time is spent at the slower speed.

The general formula for two equal-length legs at speeds $v_1$ and $v_2$ is the harmonic mean:

$$\bar v = \frac{2 v_1 v_2}{v_1 + v_2}.$$

For our example: $2 \cdot 60 \cdot 40 / (60 + 40) = 4800/100 = 48$ km/h. Same answer. If you do enough of these, the harmonic mean is worth memorising; otherwise just compute distance and time separately.

The two-vehicle problems

“Two trains leave the same station at the same time, in opposite directions. One travels at 60 mph, the other at 80 mph. After how long are they 350 miles apart?”

The two distances add: at time $t$, the first has gone $60t$ miles and the second $80t$, for a total of $140t$. Set this equal to $350$ and solve: $t = 350/140 = 2.5$ hours.

“Two trains leave at the same time, travelling toward each other from stations 350 miles apart, at 60 mph and 80 mph. When do they meet?”

Same algebra: their distances still add, this time to a total of 350 miles. Same answer: $t = 2.5$ hours.

“A train leaves at noon at 60 mph. A second train leaves the same station at 1 PM in the same direction at 80 mph. When does the second train catch up?”

At time $t$ hours after noon, the first train is $60t$ miles out, the second is $80(t - 1)$ miles out. Set equal: $60t = 80(t - 1)$, so $60t = 80t - 80$, $20t = 80$, $t = 4$ hours past noon — that is, 4 PM.

The mistakes I see most often

1. Mixing units in the same equation

$v = 60$ km/h, $t = 30$ min → $d = vt = 60 \cdot 30 = 1800$ km. That is obviously wrong; the answer should be $30$ km. The fix is to convert one of them so the units cancel. Either $t = 0.5$ h (giving $d = 30$ km) or $v = 1$ km/min (also $d = 30$ km). Always convert.

2. Averaging speeds arithmetically when they should be averaged harmonically

Covered above. Average speed = total distance / total time. Always.

3. Forgetting that the second vehicle started later

In the catch-up problem above, if you wrote $80t$ instead of $80(t - 1)$ for the second train, you would conclude they meet at $t = 0$, which makes no sense. Read the problem twice and identify the offsets.

4. Speed cannot be negative, but velocity can

At school the words are often used interchangeably; in physics they are not. Speed is the magnitude (always ≥ 0); velocity is a vector with direction (can be positive or negative on a 1D problem). If a question gives you negative numbers for “speed,” it almost certainly means velocity in opposite directions.

Where this comes up later

Constant-speed motion is the simplest case in mechanics. In physics the relation generalises to $\Delta x = \int v\,dt$ when $v$ varies in time, which is the integral version of the same statement. Distance-rate-time problems also appear identically in flow problems (“a tap fills a tank in $X$ minutes; with a second tap...”) and in work-rate problems — same algebra, different units.

References