Speed and average speed

Ask "how fast is it going?" about a train, a runner, or a falling stone, and physics answers with one clean idea: speed, the distance covered per unit time. For steady motion that single number tells the whole story. But most real journeys are not steady — a bus crawls through traffic, then speeds up on a highway — so physics also needs average speed, a summary number built from the total distance and the total time of the whole trip, not from averaging the individual speeds. This article, part of the CBSE Class 9 Motion chapter, builds both ideas from the definition upward, derives the correct way to combine a multi-stage journey, and works through a fully solved two-part bus trip so the formula becomes second nature rather than a rule to memorise.

The core insight

Speed is distance divided by time, and average speed for any journey — however many stages, however uneven — is always total distance divided by total time, never a simple average of the individual speeds.

The core idea

Speed is the distance an object travels per unit time. It is a scalar quantity — it has magnitude but no direction, unlike velocity.

\text{Speed} = \frac{\text{Distance travelled}}{\text{Time taken}}

where: distance is measured in metres ( $\text{m}$ ) and time in seconds ( $\text{s}$ ), so the SI unit of speed is metre per second ( $\text{m/s}$ ). A common practical unit is kilometre per hour ( $\text{km/h}$ ), and the two convert as $1\ \text{km/h} = \dfrac{5}{18}\ \text{m/s}$ . For an object in uniform motion, the speed never changes, so the formula above gives that one constant value at every instant — plug in any distance and the matching time and you get the same answer. For non-uniform motion, the speed keeps changing from moment to moment, and no single instantaneous value describes the whole trip. Physics instead uses average speed, built from the entire journey rather than any one moment:

\text{Average speed} = \frac{\text{Total distance travelled}}{\text{Total time taken}}

where: total distance is the sum of every distance covered in every part of the journey, and total time is the sum of every time interval, including any stops. Average speed does not mean the object actually moved at that value throughout — it means a constant speed of that size would have covered the same total distance in the same total time. Two journeys that look completely different moment to moment (one smooth and unhurried, another full of stop-start bursts) can still end up with identical average speeds, as long as their total distance and total time match. That is exactly why average speed is a summary, not a description of every instant — a distance-time graph is what actually shows how the speed varied along the way. The figure below makes this concrete for a journey covered in two stages at two different steady speeds: each stage is a straight line on a distance-time graph, and the slope of each line is that stage's speed.

Figure

Distance time graph of a two stage journey with a steeper first segment and a shallower second segment

Common core relations

A common problem type gives a journey in two or more parts, each at a different uniform speed. The average speed for the whole trip is never the plain average of the two speeds — it must be found by adding all the distances and dividing by the sum of all the times.

Derivation AAverage speed for a two-stage journey

The goal is to show, using nothing beyond the basic speed formula rearranged for time, exactly why the average speed of a two-part journey equals total distance over total time, and why this is generally not the same as simply averaging the two individual speeds — a distinction that trips up most students the first time they meet it.

Step 1. Set up the two stages. Let the first stage cover distance $d_1$ at uniform speed $v_1$ , and the second stage cover distance $d_2$ at uniform speed $v_2$ . Since speed is distance over time, the time for each stage is time equals distance over speed: $t_1 = \dfrac{d_1}{v_1}$ and $t_2 = \dfrac{d_2}{v_2}$ .

Step 2. Apply the definition of average speed to the whole journey. Average speed is total distance divided by total time, so $v_{avg} = \dfrac{d_1 + d_2}{t_1 + t_2} = \dfrac{d_1 + d_2}{\dfrac{d_1}{v_1} + \dfrac{d_2}{v_2}}$ .

Step 3. See why this is not the simple average $\dfrac{v_1 + v_2}{2}$ . The simple average would only be correct if the times $t_1$ and $t_2$ were equal, because average speed is genuinely a time-weighted combination of $v_1$ and $v_2$ , not a distance-weighted one. When the two stages take equal time, the formula in Step 2 reduces algebraically to $\dfrac{v_1 + v_2}{2}$ ; whenever the times differ, it does not, and using the plain average gives a wrong answer.

Step 4. Read off the working recipe. For any multi-stage journey: find the time for each stage separately using $t = d/v$ , add all the distances for the total distance, add all the times for the total time, and divide. This recipe works for any number of stages, at any speeds, with no shortcut needed.

Worked example

Question: A bus travels the first $60\ \text{km}$ of a journey at a uniform speed of $30\ \text{km/h}$ , and the remaining $40\ \text{km}$ at a uniform speed of $20\ \text{km/h}$ . Find the average speed of the bus for the entire journey.

Step 1. Find the time for stage 1 using $t = d/v$ : $t_1 = \dfrac{60\ \text{km}}{30\ \text{km/h}} = 2\ \text{h}$ .

Step 2. Find the time for stage 2 the same way: $t_2 = \dfrac{40\ \text{km}}{20\ \text{km/h}} = 2\ \text{h}$ .

Step 3. Add the distances and the times separately: total distance $= 60 + 40 = 100\ \text{km}$ ; total time $= 2 + 2 = 4\ \text{h}$ .

Step 4. Divide total distance by total time: $\text{Average speed} = \dfrac{100\ \text{km}}{4\ \text{h}} = 25\ \text{km/h}$ . The bus's average speed for the whole journey is $25\ \text{km/h}$ .

Competitive stretch

Convert that answer to SI units: $25\ \text{km/h} \times \dfrac{5}{18} \approx 6.9\ \text{m/s}$ . Note also that $25\ \text{km/h}$ happens to equal the simple average of $30$ and $20\ \text{km/h}$ only because $t_1 = t_2 = 2\ \text{h}$ in this problem; had the two distances been different, the simple average would have given the wrong number, even though the two speeds stayed the same.

Common traps

Averaging the two speeds directly, like $\dfrac{30+20}{2}$ , as a shortcut — correct only when the time spent at each speed is equal, wrong otherwise.
Mixing units — combining kilometres and hours with metres and seconds without converting everything to one consistent system first.
Treating average speed as the actual speed at every instant of the trip, rather than a single summary value for the whole journey.
Confusing speed (always positive, scalar) with velocity, and later applying the velocity-averaging formula $v_{avg} = \dfrac{u+v}{2}$ to speed problems where it does not apply.

Ask the tutor: Stuck on which distances and times to add up in a multi-stage speed problem? Ask the AI tutor to walk through your specific journey stage by stage.

If a bus covers equal distances at two different speeds is the average speed the same as covering equal times at those speeds

No, these give different answers. Equal-distance stages spend more time at the slower speed, so the average speed is pulled toward the slower value — it is not the plain average of the two speeds. Equal-time stages, by contrast, do average out to exactly $\dfrac{v_1+v_2}{2}$ , because the time-weighting in the average speed formula becomes a simple 50-50 split when the times themselves are equal. Always check what is equal — distance or time — before choosing a shortcut.

Can average speed ever be zero even if the object was moving the whole time

Average speed, as defined by total distance over total time, can only be zero if the total distance travelled is zero, which is impossible for an object that keeps moving forward without ever retracing its path. It is average velocity that can be zero for a round trip, because velocity depends on displacement (straight-line distance from start to end), and a trip that returns to its starting point has zero displacement even though the total distance covered, and hence the average speed, is clearly not zero.

Why do multi-stage problems always want time found first instead of just adding the speeds

Because speed is not something you can add across stages — only distances add to give total distance, and only times add to give total time. Speeds do not combine that simply; a speed carries information about both a distance and a time bundled together, so pulling two speeds apart and adding them loses that bundled information. Finding $t = d/v$ for each stage first, separately, is what correctly unbundles distance and time so they can be summed and then redivided into one honest average speed.