Use the relationship between distance, speed and time to solve problems, rearranging the formula to find any of the three quantities and converting between km/h and m/s

What this dot point is asking

NESA wants you to use the one relationship that links distance, speed and time. From any two of them, you should be able to find the third. That relationship is $D = S \times T$ (distance equals speed times time). The skill has two parts. First, rearranging the formula to make speed or time the subject. Second, keeping the units consistent so the answer lands in the unit the question wants. Almost every lost mark here is a units mistake (mixing minutes with hours, or kilometres with metres) or an "average speed" question answered by averaging the speeds instead of dividing total distance by total time. The deeper idea is that speed is a rate: the amount of distance covered in each unit of time. A rate behaves like a fraction, so the units on top and bottom have to match the units in the rate.

The answer

The one formula, three ways

Everything in this topic comes from a single equation:

Rearranged to make each quantity the subject, it becomes:

• $D = S \times T$ (distance, when you know speed and time),
• $S = \dfrac{D}{T}$ (speed, when you know distance and time),
• $T = \dfrac{D}{S}$ (time, when you know distance and speed).

The triangle above is the memory aid: write $D$ on top with $S$ and $T$ side by side underneath, then cover the quantity you want. Cover $D$ and you see $S$ next to $T$, which means $S \times T$. Cover $S$ and you see $D$ above $T$, which means $D \div T$. Cover $T$ and you see $D$ above $S$, which means $D \div S$. The triangle is just a picture of the algebra; if you prefer, rearrange $D = ST$ directly by dividing both sides by $T$ (or by $S$), which is the same skill you use in changing the subject of a formula.

Speed is a rate, so the units must match

A speed of $60$ km/h literally means $60$ kilometres travelled for every $1$ hour. The "per" is a division, so the unit is kilometres divided by hours. That is why the units have to be consistent before you compute:

• if the speed is in km/h, the time must be in hours and the distance in kilometres;
• if the speed is in m/s, the time must be in seconds and the distance in metres.

When the question gives a time in minutes but the speed in km/h, convert the minutes to hours first (divide by $60$). When it gives a distance in metres but the speed in km/h, convert the metres to kilometres first (divide by $1000$). Get the units agreeing before you substitute and the arithmetic looks after itself. Substitute mismatched units and the answer is wrong by a factor of $60$ or $1000$, even though every button you pressed was correct.

Converting between km/h and m/s

Road speeds are quoted in km/h, but many physics-style and sport contexts use m/s, so you must be able to swap between them. The conversion factor is $3.6$:

So to go from km/h to m/s you divide by $3.6$, and from m/s to km/h you multiply by $3.6$. A quick sanity check stops you doing it backwards: a metres-per-second figure is always the smaller number (a car at $100$ km/h is only about $28$ m/s), so if your converted answer to m/s came out bigger than the km/h figure, you multiplied when you should have divided. For example $100$ km/h $= 100 \div 3.6 \approx 27.78$ m/s, and $25$ m/s $= 25 \times 3.6 = 90$ km/h.

Average speed over a whole journey

The biggest conceptual trap in this dot point is the phrase average speed. Average speed is not the average of the speeds on each leg. It is always

Say a trip is driven partly fast and partly slow. You find the time for each leg (each part of the trip) with $T = \dfrac{D}{S}$, add the distances, add the times, and divide. Averaging the two speeds gives the wrong answer. The car spends more time on the slow leg, so the slow speed should count for more. The worked examples below show this clearly: two legs at $100$ and $80$ km/h give an average of about $92.31$ km/h, not $90$ km/h. A freight run at $70$ and $90$ km/h averages about $77.27$ km/h while moving, not $80$ km/h.

How exam questions ask about distance, speed and time

The wording tells you which form of the formula to reach for:

• "How far / what distance does it travel..." Distance is the unknown, so use $D = S \times T$. Check the time is in the unit that matches the speed.
• "How long / what time does it take..." Time is the unknown, so use $T = \dfrac{D}{S}$. The answer is often a decimal of an hour, which you may need to turn into hours and minutes (multiply the decimal part by $60$).
• "Find the average speed / what is its speed..." Speed is the unknown, so use $S = \dfrac{D}{T}$. If the trip has several legs, this means total distance over total time, not an average of speeds.
• "...in km/h" or "...in m/s" is a units instruction: convert with $3.6$ (divide km/h to m/s, multiply m/s to km/h), or convert the given time/distance first so it matches.
• "At what time does it arrive..." Find the duration with $T = \dfrac{D}{S}$, convert it to hours and minutes, then add it to the departure time.
• A trip with a rest stop or two different speeds is an average-speed question in disguise: handle each moving leg separately, then decide whether the question wants the average over the moving time only or over the whole elapsed time including the stop.

Reading a distance-time graph

A journey at a steady speed graphs as a straight line of distance against time, and the gradient of that line is the speed. The graph below shows the caravan's steady $95$ km/h drive: distance is directly proportional to time, so the line goes through the origin and rises by $95$ km for every $1$ hour across. Reading off at $2$ hours gives $190$ km, exactly $95 \times 2$, and the steeper the line, the faster the speed. A horizontal section (no gain in distance) would mean the vehicle is stopped, which is how a rest stop shows up on such a graph.

This is the same idea you meet again in linear relationships: the line is $D = 95T$, a direct variation through the origin with gradient $95$. Reading a value off the line (go up from a time to the line, then across to the distance) is exactly substituting into $D = ST$, just done graphically.