I’m stuck on these speed–distance–time problems where a trip is split into parts with different speeds. Example: drive the first 40 km at 80 km/h and the next 40 km at 40 km/h – what’s the average speed for the whole trip? My brain keeps doing (80 + 40) / 2 because it feels obvious, but I know that’s not right and I can’t seem to unlearn it. I get mixed up switching between “same distance” vs “same time,” and I’m not sure which one matters for average speed. Is there a quick mental way to handle this without writing a bunch of equations? Also, if the distances aren’t equal (like 30 km at 60 km/h and 50 km at 90 km/h), what’s the simplest way to keep the units straight when the times come out in awkward fractions of an hour?
Ready to make maths more enjoyable, accessible, and fun? Join a friendly community where you can explore puzzles, ask questions, track your progress, and learn at your own pace.
By becoming a member, you unlock:
3 Responses
I feel you-our brains want to average 80 and 40, but average speed is total distance divided by total time, so the slow part counts more because you spend more time there. It’s like walking to the store slowly and jogging back: even if the distances match, you spend longer in the slow phase, so the overall pace leans slow. For your 40 km at 80 km/h and 40 km at 40 km/h: time is 40/80 = 0.5 h plus 40/40 = 1 h, total 1.5 h; distance is 80 km; average speed = 80/1.5 ≈ 53.3 km/h, not 60. Quick mental trick when the distances are equal: use the harmonic mean, 2v1v2/(v1+v2). Here that’s 2·80·40/(80+40) = 6400/120 ≈ 53.3. When distances aren’t equal, just keep times as clean fractions in hours and add: for 30 km at 60 and 50 km at 90, times are 30/60 = 1/2 h and 50/90 = 5/9 h, so total time is 1/2 + 5/9 = 19/18 h; distance is 80 km; average speed = 80 ÷ (19/18) = 80·18/19 ≈ 75.8 km/h. Unit tip: since speeds are in km/h, keep time in hours as fractions (like 5/9) and only round at the end. A nice walkthrough is here: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:rate-word-problems/avg-speed/v/average-speed
I fall into that trap too-my brain wants to average the two speeds because it feels so tidy. But average speed is total distance divided by total time, so it “weights” each speed by how long you spent going at it. That means the slower part counts more if the distances are equal (you’re stuck at that speed longer). A neat mental trick: think in “pace” (minutes per km) instead of speed. Convert each speed v to pace 60/v minutes per km, average those by distance, then flip back. Simple average of speeds only works when the times are equal, not when the distances are equal.
Worked example (equal distances): 40 km at 80 km/h, then 40 km at 40 km/h. Times: 40/80 = 0.5 h and 40/40 = 1 h, so total time = 1.5 h for 80 km. Average speed = 80 / 1.5 = 53.33 km/h. If you like the pace trick: 80 km/h → 0.75 min/km, 40 km/h → 1.5 min/km; equal distances mean average pace = (0.75 + 1.5)/2 = 1.125 min/km, so average speed = 60/1.125 = 53.33 km/h. (That’s also the harmonic mean: 2v1v2/(v1+v2).)
Unequal distances: say 30 km at 60 km/h and 50 km at 90 km/h. I keep units clean by moving to minutes: time1 = (30/60) h = 30 min; time2 = (50/90) h ≈ 33.33 min; total ≈ 63.33 min = 1.0556 h. Total distance = 80 km, so average speed ≈ 80 / 1.0556 ≈ 75.8 km/h. Mental tip: converting to minutes is friendly because of the “60,” and the rule never changes-average speed = total distance / total time. For a nice visual explainer, this page is great: https://www.mathsisfun.com/average-speed.html
The key is that average speed is total distance divided by total time, so it only equals the simple average of the two speeds when you spend the same amount of time at each speed (not when the distances are the same). In your example, 40 km at 80 km/h takes 0.5 h and 40 km at 40 km/h takes 1 h, so total time is 1.5 h over 80 km, giving 80/1.5 ≈ 53.3 km/h; that’s lower than 60 because the slow part “lasts longer” and drags the average down, like a hike where the uphill section takes more of your day than the downhill. A quick mental trick when the distances are equal: use the harmonic mean of the speeds, 2ab/(a+b), which here is 2·80·40/(80+40) = 53.3 km/h (I sometimes think “geometric,” but it’s the harmonic one for equal distances). For unequal distances, I wouldn’t average the speeds; I compute each time as distance/speed, add them, then do total distance ÷ total time-for 30 km at 60 and 50 km at 90, times are 0.5 h and 50/90 ≈ 0.56 h, so total time ≈ 1.06 h and average speed ≈ 80/1.06 ≈ 75.8 km/h. To keep units clean, stick to km and hours throughout: write times as exact fractions (e.g., 50/90 h) and only round at the end. You can also think of it as a distance-weighted average of the speeds, though really what’s being weighted in the calculation is the time each speed is used, which is why the slower stretch counts more.