Your child already knows how to find the arithmetic mean of two numbers: add them and divide by two. When a question supplies two speeds and asks for an average, that familiar procedure may arrive before the student thinks about the journey. The arithmetic can be perfectly executed while the mathematical model is wrong.

The issue is what each number represents. A speed describes distance travelled per unit time. If a traveller spends two hours at one speed and one hour at another, the two speed values do not describe equal amounts of time. Giving them equal weight discards information that determines how far the traveller went during the complete interval.

Use the outward-and-return example from the introduction. Travelling 60 km at 30 km/h takes 2 hours. Travelling 60 km at 60 km/h takes 1 hour. The complete distance is 120 km and the complete moving time is 3 hours. The average speed is 40 km/h because a constant speed of 40 km/h for 3 hours would cover that same 120 km.

This explanation gives the average a job: summarise the total distance over the specified time interval. It is not merely a middle number between the displayed speeds. A numerical answer between 30 and 60 can still be wrong, as 45 is here. Plausibility is useful, but the totals establish the actual value.

Parents can ask, “What journey would this average describe?” Let your child connect the proposed speed with the full time and distance. If 45 km/h for 3 hours gives 135 km rather than 120 km, the mismatch becomes visible. That check helps repair the meaning of average speed without dismissing the child’s correct arithmetic-mean calculation as a meaningless attempt.

For a stated journey interval, average speed preserves the relationship between total distance and total elapsed time. Write v_avg = D_total/T_total, with compatible distance and time units. The numerator measures how far the traveller actually moved. The denominator measures how long the requested interval lasted, including any part of that interval spent stationary.

A constant speed equal to this average would cover the same total distance in the same total time. For 90 km travelled in 2 hours, the average is 45 km/h. This does not claim the traveller maintained 45 km/h throughout. They may have changed speed or stopped; the average summarises the complete interval rather than describing every moment within it.

Parents can use an everyday mathematical question: “If we replaced the whole journey with one steady speed, what would cover the same distance in the same time?” That interpretation distinguishes average speed from the maximum speed, the final speed and a typical speed someone happens to remember. Each describes something different about the journey.

The interval matters. An average for the entire trip may differ from an average for only the moving portions. If the question asks for the complete time from departure to arrival, include a stated stop within that interval. If it explicitly asks for the average while moving, use the moving interval. Read the request before deciding what belongs in the denominator.

Begin a solution by naming the total distance and total time being measured. This can be one line rather than a lengthy explanation. The purpose is to stop individual speed values from replacing the quantities the definition actually needs. Once the totals are clear, the final division becomes an understandable calculation instead of a procedure chosen because the word “average” appeared.

A small table can prevent the student from combining quantities too early. Give each stage a row and record distance, time and speed with units. If two of the three are known, use distance = speed × time, time = distance ÷ speed, or speed = distance ÷ time to recover the missing quantity.

For a first stage of 60 km at 30 km/h, write 60 km, 2 h and 30 km/h. For a second stage of 60 km at 60 km/h, write 60 km, 1 h and 60 km/h. Only after the rows are complete should the totals be formed: distance 120 km and time 3 h.

Do not add the speed column to create a “total speed”. Speed is a rate, so a sum such as 30 + 60 does not describe how fast the complete journey occurred. Distances across successive stages add, and durations across those stages add. Their quotient supplies the overall average speed.

If a stop belongs to the interval, give it its own row. Its distance is zero, its time is the stated duration, and its speed while stationary is zero. This makes the stop visible in the denominator without adding an invented distance. It also prevents a pause from disappearing simply because no travelling speed was stated for it.

The table below models the equal-distance outward-and-return example. Parents can ask the child to explain how each time was found and why the final row has no “total speed” entry. The table is a reasoning aid, not a requirement to draw a large grid for every question. Once the relationships are secure, a concise pair of calculations can serve the same purpose.

The arithmetic mean of two constant stage speeds works when the traveller spends equal positive times at those speeds and the average concerns exactly those two stages. Suppose a traveller moves at 30 km/h for 1 hour and 60 km/h for 1 hour. The distances are 30 km and 60 km, so the average is 90 km divided by 2 hours, or 45 km/h.

This agrees with (30 + 60)/2 because the equal durations give the two speeds equal weight. The agreement follows from the totals: (30 × 1 + 60 × 1)/(1 + 1) = 45. The shortcut is justified by a condition, rather than by the mere presence of two speeds.

The same reasoning works for equal half-hour intervals. At 30 km/h for 0.5 h, the distance is 15 km. At 60 km/h for 0.5 h, it is 30 km. Total distance 45 km divided by total time 1 h again gives 45 km/h. Equal durations matter; they do not have to be one hour.

If a stop is added to the requested interval, the simple mean of the two moving speeds no longer describes the whole interval. For the two one-hour stages followed by a half-hour stop, the average over all 2.5 hours is 90/2.5 = 36 km/h. The extra time has zero distance attached to it.

Parents can ask, “What makes it fair to give these speeds equal weight?” A strong answer identifies equal time, not equal distance or two stages of a similar-looking diagram. This helps your child retain a useful shortcut while knowing when it applies. The goal is not to ban arithmetic means, but to connect their use to the journey conditions that justify them.

For a fixed positive distance, travelling more slowly takes longer. If two stages cover the same distance at different positive speeds, the slower stage occupies more of the total time. That makes an equal weighting of the speed numbers inappropriate for the average over the complete journey.

Take two 90 km stages at 45 km/h and 90 km/h. The first takes 90/45 = 2 hours; the second takes 90/90 = 1 hour. The total distance is 180 km and the total time is 3 hours, so the average is 60 km/h. The arithmetic mean of 45 and 90 is 67.5 km/h, which does not preserve those totals.

You can test that incorrect value against the journey. A constant speed of 67.5 km/h for 3 hours would cover 202.5 km, more than the stated 180 km. The result is not wrong because it is insufficiently close to the slower speed; it is wrong because it fails the distance–time relationship.

A visual explanation uses the time spent in each stage. The traveller spends two thirds of the moving time at 45 km/h and one third at 90 km/h. The time-weighted calculation is (2/3) × 45 + (1/3) × 90 = 60 km/h. This is the same calculation as total distance divided by total time, expressed through time proportions.

Parents can compare an equal-time and an equal-distance journey using the same two speed values. Ask which stage takes longer in each version. That controlled contrast makes the hidden assumption behind the shortcut visible. It is often more useful than a long set of exercises where the distances change but the student continues to average the displayed speeds automatically.

For two equal positive distances travelled at positive speeds u and v, the average moving speed is 2uv/(u + v). This can be derived from the definition rather than memorised separately. If each distance is d, total distance is 2d and total time is d/u + d/v.

Combining the time fractions gives d(u + v)/(uv). Dividing 2d by that expression gives 2uv/(u + v), with d cancelling because both stages have the same distance. The formula therefore depends on equal distances and positive speeds. It is not a replacement for reading the stage lengths or identifying the requested interval.

For u = 30 km/h and v = 60 km/h, the expression gives 2 × 30 × 60/(30 + 60) = 3,600/90 = 40 km/h. For u = 45 and v = 90, it gives 8,100/135 = 60 km/h. These agree with the explicit stage-time calculations already worked.

If the stage distances differ, this particular shortcut does not apply. A 30 km stage at 30 km/h followed by a 60 km stage at 60 km/h takes 1 hour each and has average 45 km/h. The equal-distance expression would give 40 km/h for the same speed pair, answering a different journey.

Use the shortcut only if it is useful and appropriate to the student’s current algebra. Parents need not introduce it before the totals are secure. Ask your child to state its conditions and explain where the time terms came from. A fast formula remembered without equal-distance conditions can recreate exactly the method-selection problem that the article is trying to repair.

When the stage distances differ, calculate each stage’s time separately. Suppose a traveller covers 30 km at 40 km/h and then 60 km at 80 km/h. The first time is 30/40 = 3/4 hour. The second is 60/80 = 3/4 hour. Total distance is 90 km and total time is 3/2 hours.

The average is 90 ÷ (3/2) = 60 km/h. In this example the arithmetic mean of 40 and 80 also gives 60, because the stage times happen to be equal. Unequal distances do not automatically make the simple mean wrong; equal time is the relevant condition. Calculate or establish the durations before deciding whether the shortcut fits.

Now change the second distance to 80 km while keeping its speed at 80 km/h. Its time becomes 1 hour, while the first remains 3/4 hour. The average is (30 + 80)/(3/4 + 1) = 110/(7/4) = 440/7 km/h, approximately 62.86 km/h to two decimal places.

The changed distance gives the faster stage a larger share of the total time, so the average rises from 60 to about 62.86. That interpretation is a useful check, while the exact fraction establishes the value. Keep 3/4 and 7/4 exact during the working rather than rounding durations unnecessarily.

Parents can ask what changed between the two versions and how it changed the time weights. This encourages the child to follow the journey relationships rather than declare that all equal-distance questions use one formula and all unequal-distance questions use another. The total-distance-over-total-time method remains valid across both, making it a dependable routine when a shortcut’s conditions are uncertain.

When stage times and speeds are given, find each distance using distance = speed × time. Suppose a traveller moves at 20 km/h for 1.5 hours and at 50 km/h for 0.5 hour. The distances are 30 km and 25 km. Total distance is 55 km and total time is 2 hours.

The average speed is 55/2 = 27.5 km/h. It is closer to 20 than to 50 because the traveller spends three quarters of the time at the slower speed. The time-weighted expression (20 × 1.5 + 50 × 0.5)/(1.5 + 0.5) shows that relationship directly.

If the child averages 20 and 50 to obtain 35 km/h, ask them to test the result over the full 2 hours. That speed would cover 70 km, not 55 km. The incorrect answer comes from treating the time contributions as equal even though the question explicitly gives unequal durations.

For three stages, the same definition continues to work. At 20 km/h for 1 hour, 40 km/h for 0.5 hour and 60 km/h for 0.5 hour, the distances are 20, 20 and 30 km. Total distance 70 km divided by total time 2 hours gives 35 km/h. Averaging the three speeds directly would give 40 km/h.

Parents can encourage a short record of each speed–time product before the totals. The student does not need a new rule for three stages or four stages: add the distances and add the durations across the stated interval. This makes the method scale naturally while retaining a clear meaning for every contribution.

Include a stop when it lies within the interval the question asks you to average over. Suppose a cyclist travels 12 km in 30 minutes, rests for 10 minutes, and then travels 8 km in 20 minutes. The total distance is 20 km. The complete elapsed time from departure to final arrival is 60 minutes.

The average over that complete interval is 20 km/h. If the question explicitly asks for the average while moving, the moving time is only 50 minutes, or 5/6 hour. The moving average is 20 ÷ (5/6) = 24 km/h. Both values can be correct for different defined intervals.

Do not decide whether to include a stop by asking which answer seems more impressive or more typical of the traveller. Read the wording and identify the time interval. If the request is unclear, state the interpretation or seek clarification. A stop does not add distance, but it can add time to the denominator.

A stationary stage can be represented as speed zero for its stated duration. In the full-interval calculation, its distance contribution is zero times that duration, while the duration still contributes to total time. This shows why adding an unweighted zero to a list of moving speeds is not a general way to account for a stop.

Parents can ask, “When does our clock start, and when does it stop?” That question often makes the relevant interval easier to understand. Then list any pause that occurs between those endpoints. The next useful step is a consistent numerator and denominator, rather than a universal instruction to include or exclude every stop regardless of what the question asks.

Match the time unit to the speed unit before multiplying or dividing. A speed of 60 km/h multiplied by 30 minutes cannot be calculated as 60 × 30 km without a conversion. Thirty minutes is 1/2 hour, so the distance is 60 × 1/2 = 30 km.

Likewise, 15 minutes is 1/4 hour, 45 minutes is 3/4 hour and 90 minutes is 3/2 hours. Time written in clock-style notation is not a decimal fraction of an hour: 1 hour 30 minutes is 1.5 hours, not 1.30 hours. The conversion uses 60 minutes per hour, not 100.

Suppose a 15 km stage occurs at 30 km/h and another 15 km stage at 45 km/h. Their times are 1/2 hour and 1/3 hour. Total time is 5/6 hour, or 50 minutes. The average is 30 ÷ (5/6) = 36 km/h. Dividing 30 km by 50 minutes would instead give 0.6 km/min, which is equivalent but has a different unit.

For metres per second and kilometres per hour, 1 m/s = 3.6 km/h. This follows from 1 metre = 0.001 kilometre and 1 hour = 3,600 seconds. A speed of 5 m/s is 18 km/h. Use the conversion only when needed, and keep each stage’s units visible.

Parents can ask the child to predict the unit of the final division: kilometres divided by hours gives km/h. If the working contains kilometres and minutes, either report km/min or convert the time appropriately. Units will not detect every modelling error, but they can expose a mismatch before an otherwise tidy calculation produces a misleading numerical result.

A journey described by departure and arrival times needs an elapsed duration before calculating average speed. From 09:20 to 10:05 is 45 minutes, or 3/4 hour. It is not 0.85 hour obtained by subtracting the displayed digits as decimals. Clock readings use hours and minutes, so calculate the duration in compatible time units.

For 36 km covered from 09:20 to 10:05, the average speed is 36 ÷ (3/4) = 48 km/h. You can check the duration by moving from 09:20 to 10:00 for 40 minutes, then to 10:05 for another 5 minutes. The complete interval is 45 minutes.

If a stated stop occurs within that departure-to-arrival interval, do not add it again to an elapsed duration that already includes it. For example, a ten-minute rest between those clock times is part of the 45 minutes. Adding another ten minutes would count the same stop twice. Subtract it only if the question asks for moving time.

For an overnight interval, account for the date boundary explicitly. From 23:40 to 00:10 the following day is 30 minutes. The word “following” or equivalent date information matters. If dates or the interval are ambiguous, do not assume an unsupported travel duration simply to obtain a numerical answer.

Parents can encourage a small timeline when the clock calculation is the difficulty. It need not be drawn to scale; marked start, stop and finish times can show what belongs in the interval. Then return to total distance divided by that duration. This separates a time-reading issue from an average-speed modelling issue and gives the child a clearer next repair.

On a graph of cumulative distance travelled against elapsed time, the average speed over an interval is the total increase in distance divided by the elapsed time. It can be represented by the slope of the line joining the interval’s endpoints. Read the axis values and units, rather than counting grid squares without checking their scales.

Suppose the graph begins at distance 0 km at time 0 h and ends at cumulative distance 90 km at time 3 h. The average over the complete interval is 90/3 = 30 km/h. The graph may contain several different slopes and a horizontal rest section; the endpoint calculation still uses the complete distance and time.

A horizontal section on a cumulative-distance graph indicates no added distance during that interval. If the average covers that section, its duration belongs in the denominator. A steep section represents a higher speed than a shallower rising section, provided the same axes and scales are used. Do not average the segment slopes equally unless their time intervals justify equal weighting.

Be careful about what the vertical axis measures. A position or displacement graph can decrease during a return journey, unlike cumulative distance travelled. The final change in position does not give total distance travelled when direction changes. Identify the graph’s quantity before using its endpoint difference to calculate an average speed.

Parents can ask, “Does this vertical value record how far has been travelled altogether, or where the traveller is relative to a reference point?” That distinction prevents a return journey from being treated as though its travelled distance disappeared. Once the axis meaning is clear, the graph becomes another representation of the same distance-and-time totals, rather than a separate set of unexplained rules.

For a speed–time graph with compatible units, the area under the graph over an interval gives distance travelled. Total elapsed time comes from the horizontal axis. Average speed is therefore the total area under the speed graph divided by the interval length. The graph’s height alone gives a speed at a particular time, not the whole-interval average.

For a constant 4 m/s over 10 seconds followed by 8 m/s over 5 seconds, the areas are 4 × 10 = 40 m and 8 × 5 = 40 m. Total distance is 80 m and total time is 15 s, giving average 80/15 = 16/3 m/s, approximately 5.33 m/s.

A straight line rising from 0 to 10 m/s over 4 seconds forms a triangular area of (1/2) × 4 × 10 = 20 m. The average over those 4 seconds is 20/4 = 5 m/s. Here the mean of the endpoint speeds works because the graph is linear across the interval, not merely because two endpoint values are visible.

If a speed graph is not linear, averaging its endpoint heights is not generally sufficient. Two journeys can begin and end at the same speeds while covering different distances between those times. The area, or equivalent complete information about the speed changes, determines the distance. Do not invent a straight segment when the problem does not establish one.

Use this graph connection when it belongs to the child’s current learning. Parents can ask what each area represents and check the units: metres per second multiplied by seconds gives metres. This connects graph geometry with the journey table and helps the student see the same total-distance-over-total-time relationship in another form.

Average speed uses total distance travelled, so an outward-and-return journey has positive average speed if positive distance is covered over positive time. Returning to the starting point does not erase the distance travelled. A 5 km outward trip and a 5 km return trip cover 10 km in total.

If the complete trip takes 2 hours, the average speed is 10/2 = 5 km/h. The net displacement is zero because the traveller finishes where they began. In a context studying velocity, average velocity over that interval is net displacement divided by time, giving zero. Speed and velocity summarise different quantities.

This distinction matters when a graph records position relative to the starting point. Its endpoint change may be zero after a return, while the path length travelled is not. For one-dimensional motion with a clearly stated route, add the distances along each moving stage to find total distance. Do not substitute the net change in position.

Parents need not introduce vector terminology early if the student’s current mathematics only requires speed. A simple prompt is enough: “Did the traveller move ten kilometres, or no kilometres?” The start and finish locations answer where the traveller ended; the stages answer how far they travelled. Keep those questions separate.

When the work includes displacement or velocity, label the requested quantity before choosing the numerator. This is the same habit used throughout the article: identify what the calculation must preserve. A correct time interval cannot rescue a numerator that describes a different physical quantity, even if the final unit still looks like distance per time.

Working backwards begins with the overall distance and target average. Suppose a journey consists of two 60 km stages and must have average speed 40 km/h over the moving interval. The total distance is 120 km, so the required total time is 120/40 = 3 hours.

If the first stage is travelled at 30 km/h, it takes 60/30 = 2 hours. That leaves 1 hour for the second 60 km stage. Its required speed is 60/1 = 60 km/h. The solution uses the complete time budget rather than treating the target average as the arithmetic mean of an unknown speed and 30.

Check the answer by recombining the stages: 2 h + 1 h = 3 h and 60 km + 60 km = 120 km. Then 120/3 = 40 km/h, matching the target. This final check confirms both the missing-stage calculation and the interpretation of the requested interval.

If a stop is included in the required interval, subtract its duration from the available time as well. For example, if the same three-hour total includes a half-hour stop after the two-hour first stage, the second stage has only half an hour available. The mathematical required speed would be 120 km/h. This is an illustrative calculation, not travel advice or a recommendation to meet such a target.

Parents can ask the child to label the total time budget and each time already used. The remaining time is often the bridge between an overall average and an unknown rate. If it is zero or negative, pause before dividing: that may indicate the target cannot be achieved under the stated conditions, rather than an arithmetic problem to force through.

A target average can imply a total time that has already been used before the journey is complete. Suppose the two stages are 60 km each and the first is travelled at 30 km/h, taking 2 hours. A desired overall average of 60 km/h for 120 km would require total time 120/60 = 2 hours.

That leaves no time for the second positive-distance stage. No finite positive speed can cover 60 km in zero time. The target is therefore impossible under the stated conditions. Reaching a very large speed later can reduce the remaining time, but cannot make it exactly zero.

For an overall target above 60 km/h, such as 80 km/h, the required total time is 120/80 = 1.5 hours. The first stage has already taken 2 hours, leaving a negative remaining time. That is another clear sign of impossibility. Do not report a negative “required speed” as a valid solution to this physical journey.

This reasoning can be understood without formal limits. Any finite completion of the second stage adds positive time to the first stage’s 2 hours. The total time must therefore exceed 2 hours, so the average for 120 km must be below 60 km/h. The upper boundary comes from the time already spent.

Parents can invite the child to check feasibility before rearranging a long equation. “How much time is available, and how much has already been used?” is a concrete question. It teaches that a numerical method must respect the context and that recognising an impossible target can be a complete mathematical conclusion, rather than a failure to find the right calculator input.

A useful Secondary 3 Mathematics tuition lesson can separate several possible difficulties: interpreting the requested interval, recalling distance–time relationships, converting units, weighting stage speeds and executing arithmetic. A student may manage most of these and still get an average-speed question wrong. Begin with a small contrast that makes the uncertain decision visible.

Give the same speeds in two journeys: equal time at each speed, then equal distance at each speed. Ask the student to find the durations and distances before calculating the average. If they apply the arithmetic mean to both, focus on why equal time matters. If they choose totals correctly but convert 30 minutes to 0.30 hour, repair the time conversion.

A tutor can then add a stop, change a distance or ask for an unknown stage speed. Each controlled change checks whether the original relationship survives. Worked examples should include the reason for the numerator and denominator, rather than show only a final formula. The student can explain what the complete interval contains before using it.

Useful evidence of progress includes identifying unequal time weights, including a stop consistently, distinguishing travelled distance from displacement, and checking a proposed average against the original totals. These decisions are observable. A correct answer copied from a familiar pattern gives less information about whether the method will transfer to changed wording.

Parents exploring a Punggol Secondary 3 Maths tutor can bring one example where averaging the speed values caused the error. Ask how equal-time and equal-distance cases will be contrasted and how corrected work will be revisited. Enquire directly about available lesson arrangements. A helpful discussion clarifies the child’s next learning need without assuming an article guarantees a schedule or a particular result.

Choose one completed question and ask for the total distance and total time. Your child can point to the journey stages, write two sums or make a small table. You do not need to recite a special formula for each story. The aim is to see whether the numerator and denominator describe the same requested interval.

If the child has averaged two speeds, ask whether the time at each speed was equal. Let them calculate the stage times if distances are given. A slower stage taking longer is often the missing observation. Once that is visible, return to the total-distance-over-total-time definition rather than introduce several new shortcuts.

Keep feedback specific. “You calculated each stage time correctly; now add the times before dividing” preserves a successful step. If the totals are correct but the unit is wrong, say that separately. This helps the child identify what needs repair instead of interpreting every mistaken final answer as evidence that they understand none of the topic.

One paired example can be enough for an evening check. Compare one hour at each of two speeds with equal distances at those speeds. Ask what stayed the same and what changed. The same speed values can produce different averages because their time contributions differ. That is a useful explanation to carry into the next lesson.

Stop when a clear next action emerges: practise time conversions, define the averaging interval, or check equal-time conditions before using a mean. Record one representative example if support is needed. A calm note about the exact first decision that went wrong is more useful than a long list of corrected answers with no explanation of the pattern.

Before calculating, identify the interval and keep distance and time units compatible. Question A: travel 40 km at 40 km/h and then 40 km at 80 km/h, with no stop. Find the average speed. Question B: travel at those same speeds for 1 hour each. Find the average over those two hours.

Question C: travel 12 km in 30 minutes, stop for 15 minutes, and travel 18 km in 45 minutes. Find the average from initial departure to final arrival. Question D: travel at 6 m/s for 10 seconds and 3 m/s for 20 seconds. Find the average over all 30 seconds.

Question E: a 100 km journey must average 50 km/h. Its first 40 km take 1 hour, with no stops included. Find the required speed for the remaining 60 km. Question F: two 50 km stages must average 50 km/h, but the first is travelled at 25 km/h. Decide whether a finite positive speed can achieve the target on the second stage.

For A, total time is 1 + 1/2 = 3/2 hours and distance is 80 km, giving 160/3 km/h, approximately 53.33 km/h. For B, distances are 40 and 80 km over 2 hours, giving 60 km/h. For C, distance is 30 km over 90 minutes, or 1.5 hours, giving 20 km/h.

For D, distance is 6 × 10 + 3 × 20 = 120 m over 30 s, giving 4 m/s. For E, total allowed time is 100/50 = 2 hours; one hour remains, so the required speed is 60 km/h. For F, the first stage already uses the entire two-hour allowance. No finite positive speed can complete the second 50 km in zero remaining time.

Use the same sequence each time: define the interval, identify the total travelled distance, find the total elapsed time, divide and check. If stages are given separately, calculate missing distances or times before combining them. This routine works for equal-distance, unequal-distance and equal-time cases, and does not depend on remembering which story normally goes with a shortcut.

Use the arithmetic mean of two constant stage speeds when their durations are equal and the requested interval consists of those stages. Use an equal-distance shortcut only when its equal-distance conditions hold. If the conditions are uncertain, return to the definition. A slower but understandable starting method is useful while the student learns which faster routes are justified.

Keep stops and unit conversions visible. A stated departure-to-arrival interval already includes time spent resting between those endpoints. A moving-only interval excludes it. Kilometres and hours produce km/h; metres and seconds produce m/s. The same labels should remain consistent from the stage calculations through the final division.

Check the result against the totals. Multiplying the proposed average by total time should return total distance, allowing for any stated final rounding. A moving average with positive time weights lies between the stage speeds; stops included as zero-speed intervals can lower the complete average. Plausibility helps catch errors, while the actual totals establish the answer.

For parents, progress may sound simple: “The slower part took longer, so I cannot give both speeds equal weight,” or “The stop belongs to the interval, so its time stays in the denominator.” If you are considering Secondary 3 Mathematics tuition in Punggol, bring the example that exposed the confusion. The next useful step is a clearer account of distance and time, making average speed a relationship your child can explain and reuse.