I biked 6 km to the library at 24 km/h, then pedaled back the same 6 km against a grumpy headwind at 12 km/h. My brain cheerfully said, “Average the speeds: (24 + 12) / 2 = 18 km/h!” But when I look at how long the whole round trip actually took me, that neat average seems to be lying through its mathematical teeth. I know average speed is supposed to be total distance divided by total time, but I get tangled when the speeds are different in each direction.
What really trips me up is when the problem changes from equal distances to equal times. Like, if I run for 10 minutes at one speed and then 10 minutes at another, my instinct is to average the speeds again, but I suspect there’s a sneaky catch. How do I set these up so I don’t fall into the wrong-average trap?
Could someone show me a clean way to think about these speed–distance–time puzzles-especially how to decide what to add and what to divide-and maybe a tiny checklist for unit conversions (minutes vs hours) so I don’t turn my bike ride into a unit soup?
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3 Responses
The trap is that “average of speeds” only works when you spend equal time at each speed; otherwise the slower part counts more because you’re stuck in it longer. Ask first: what’s equal here-distance or time? On your bike ride the distances are equal (6 km out and 6 km back), so you should average the times, not the speeds: time out = 6/24 = 0.25 h, time back = 6/12 = 0.5 h, total time = 0.75 h, total distance = 12 km, so average speed = 12/0.75 = 16 km/h. That matches the harmonic mean of the two speeds, 2v1v2/(v1+v2) = 2·24·12/(24+12) = 16, because equal distances lead to a harmonic-mean situation. If instead you run 10 minutes at v1 and then 10 minutes at v2 (equal times), the average speed is the ordinary arithmetic mean, (v1+v2)/2, because each speed gets the same time weight. A small analogy: think of half your trip on smooth pavement and half in sticky mud-since you’re in the mud longer per kilometer, the mud sets the overall pace. Tiny checklist: (1) Decide what’s fixed: distances or times; (2) Make units consistent-minutes to hours (÷60), km/h to m/s (×5/18) or back (×3.6); (3) Compute each leg’s time t = d/v; (4) Add times; (5) Average speed = total distance ÷ total time. For a deeper dive on why equal-distance problems use the harmonic mean, see a clear walkthrough here: https://brilliant.org/wiki/harmonic-mean/
Ooh, classic trap! When the distances are equal, the slower leg “weighs more” because you spend more time at that speed. For your bike ride: out 6 km at 24 km/h takes 6/24 = 0.25 h, back 6 km at 12 km/h takes 6/12 = 0.5 h. Total distance = 12 km, total time = 0.75 h, so average speed = 12 / 0.75 = 16 km/h, not 18. That 16 pops out of the harmonic mean of the two speeds: 2ab/(a+b). Neat pattern: equal distance → harmonic mean, because time is what varies and the slow segment dominates time.
Flip it to equal time and you get the opposite vibe: if you run 10 minutes at v1 and 10 minutes at v2, the distances scale with speed, so the average speed is the ordinary (arithmetic) mean: (v1 + v2)/2. Clean recipe: always compute total distance and total time in the same units, then average speed = total distance / total time. Tiny checklist: pick a time unit (usually hours) and convert minutes by dividing by 60; keep all distances in the same unit; use t = d/v or d = v·t to build totals; then divide distance by time once at the end. A nice walkthrough is here: https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-ratios-prop-topic/cc-6th-rates/v/average-speed-example
Hope this helps!
Your instinct is friendly but sneaky-wrong because you spend more time at the slower speed. For your ride: out time = 6/24 = 0.25 h, back time = 6/12 = 0.5 h, so total time = 0.75 h and total distance = 12 km. Average speed = total distance ÷ total time = 12 ÷ 0.75 = 16 km/h, not 18. When the two legs cover equal distances at different speeds v1 and v2, the correct “overall speed” is the harmonic mean: 2/(1/v1 + 1/v2). Plugging in 24 and 12 gives 2/(1/24 + 1/12) = 16. Intuition check: you linger longer in the slow leg, so the overall average must be pulled below the simple average.
If, instead, the times are equal-say 10 minutes at one speed and 10 minutes at another-then the average speed is the ordinary (arithmetic) mean: (v1 + v2)/2. That’s because total distance is v1 t + v2 t and total time is 2t, so the t’s cancel. A tidy way to keep your footing: always start from “average speed = total distance ÷ total time.” Then decide what’s equal. Equal distances? Use the harmonic mean. Equal times? Use the arithmetic mean. Anything else? Just add all distances, add all times (time for each leg is distance/speed), then divide. I once “proved” I was an Olympic sprinter because I averaged my speeds over two stretches without noticing I’d jogged the slow part twice as long-my watch was very impressed; reality, less so.
Tiny unit sanity-check so your bike doesn’t become a blender: keep one time unit everywhere (minutes to hours: divide by 60), keep one distance unit everywhere (meters vs kilometers), make sure your speed units match those choices (km/h pairs with km and hours; m/s with meters and seconds), convert before you compute, and only round at the very end. If the units line up, the averages will, too.