I did a there-and-back bike trip to a bakery that’s 8 km away. Zoomed there with a friendly tailwind at 24 km/h, crawled back into a grumpy headwind at 12 km/h. I figured the overall average speed would just be (24 + 12)/2, but when I try that and then compare it to adding the time for each leg and doing total distance divided by total time, I get different results. My calculator is judging me, and my croissant is losing its flakiness while I stare at the numbers.
What’s the right way to set this up so I get the correct total time and average speed for the whole trip? Should I add the times for each leg first and only then do distance/time? Any tips for not tripping over hours vs minutes (and fractions vs decimals)? Also, if I took a 2 km detour on the way back, would the method change, or is there a neat general way to handle different speeds and different distances?
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5 Responses
Average speed isn’t the simple average of speeds because the slow stretch hogs more time: add times first t = 8/24 + 8/12 = 1 h, so v_avg = total distance/total time = 16/1 = 16 km/h; in general v_avg = (d1+d2)/(d1/v1 + d2/v2) (equal distances gives the harmonic mean 2/(1/v1+1/v2)), and a 2 km detour just means use d2 = 10 km in the same formula. Stick to one unit (all hours or all minutes) or use fractions to dodge decimal traps-nice refresher here: https://www.khanacademy.org/science/physics/one-dimensional-motion/displacement-velocity-time/a/what-is-velocity Hope this helps!
Short answer: average speed is total distance divided by total time. Do not average the speeds unless each leg takes the same amount of time.
Why the arithmetic mean fails here
– The fast leg takes less time than the slow leg, so the slow speed “counts” more. Averaging 24 and 12 equally gives too much weight to the fast leg.
– With equal distances, the correct average speed is the harmonic mean of the speeds, not the arithmetic mean.
Your trip (8 km out at 24 km/h, 8 km back at 12 km/h)
– Outward time: 8/24 hours = 1/3 h = 20 min
– Return time: 8/12 hours = 2/3 h = 40 min
– Total time: 1/3 + 2/3 = 1 h
– Total distance: 8 + 8 = 16 km
– Average speed: 16 km / 1 h = 16 km/h
Check against the harmonic mean:
– Harmonic mean of 24 and 12 is 2 / (1/24 + 1/12) = 2 / (3/24) = 16 km/h.
General method (works for any distances and speeds)
– For each leg i with distance d_i and speed v_i, time is t_i = d_i / v_i.
– Total distance: D = Σ d_i.
– Total time: T = Σ (d_i / v_i).
– Average speed: v_avg = D / T.
This automatically handles different distances, detours, and varying speeds.
Your 2 km detour on the way back
– Out: 8 km at 24 km/h → time = 8/24 = 1/3 h.
– Back: 10 km at 12 km/h → time = 10/12 = 5/6 h.
– Total distance: 8 + 10 = 18 km.
– Total time: 1/3 + 5/6 = 7/6 h.
– Average speed: 18 ÷ (7/6) = 18 × 6/7 = 108/7 ≈ 15.43 km/h.
Simple worked example to keep the idea clear
– 1 km out at 30 km/h, 1 km back at 10 km/h.
– Times: 1/30 h and 1/10 h → total time = 1/30 + 1/10 = 4/30 = 2/15 h.
– Total distance: 2 km.
– Average speed: 2 ÷ (2/15) = 15 km/h (not (30 + 10)/2 = 20).
– Again, harmonic mean of 30 and 10 is 15.
Practical tips
– Always sum times first (or equivalently compute D / Σ(d_i/v_i)).
– Keep units consistent. Convert minutes to hours by dividing by 60 (40 minutes = 40/60 = 2/3 h, not 0.40 h).
– Fractions help avoid rounding errors; convert to decimals at the end if you want.
Your calculator isn’t judging you, it’s just very loyal to time! You can only average speeds directly when you spend equal time at each speed; here you spent more time at 12 km/h than at 24 km/h, so the slow stretch “wins.” The right method is: average speed = total distance ÷ total time. For your bakery run: time out = 8/24 h = 1/3 h (20 min), time back = 8/12 h = 2/3 h (40 min), so total time = 1 h and total distance = 16 km, giving an average speed of 16 km/h (not the simple average 18). Handy fact: when the distances are equal, the average speed is the harmonic mean, 2v1v2/(v1+v2) = 16. In general, for any mix of distances and speeds, use v_avg = (d1 + d2 + …)/(d1/v1 + d2/v2 + …). If you took a 2 km detour on the way back, the return is 10 km at 12 km/h, so time back = 10/12 h = 5/6 h (50 min), total time = 1/3 + 5/6 = 7/6 h, total distance = 18 km, and average speed = 18 ÷ (7/6) = 108/7 ≈ 15.43 km/h. Unit tip so your croissant stays flaky: keep everything in hours (fractions are great) or everything in minutes-don’t mix decimals of hours with minutes (remember 0.2 h is 12 minutes, not 20). Does that match your calculator now, and how would you set it up if we added a third segment or a coffee-stop pause?
Don’t average the speeds unless the distances are equal (hmm, I might be mixing that up with equal times), so the reliable method is to keep units consistent and do total distance ÷ total time. Example: 8 km at 24 km/h and 8 km at 12 km/h gives time = 8/24 + 8/12 = 1 h, so average = 16/1 = 16 km/h; with a 2 km detour back (10 km at 12 km/h), time = 8/24 + 10/12 and average = 18 ÷ that time; more here: https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-rates-and-percent/cc-6th-rates-word-problems/v/average-speed-word-problem
You’re bumping into a classic trap: averaging speeds only works if you spend the same amount of time at each speed, but on an out-and-back with equal distances, you spend more time at the slower speed, so it “counts” more. The right method is total distance divided by total time. For your bakery run: out is 8 km at 24 km/h, so time out = 8/24 = 1/3 hour = 20 minutes; back is 8 km at 12 km/h, so time back = 8/12 = 2/3 hour = 40 minutes. Total time = 1 hour, total distance = 16 km, so the overall average speed is 16 km/h (not the simple average 18). A neat shortcut when the distances are the same is that the average speed is the harmonic mean: 2/(1/24 + 1/12) = 16 km/h; more generally, always do average speed = (sum of distances) ÷ (sum of time for each leg) = (Σ d_i) ÷ (Σ d_i / v_i). If you take a 2 km detour on the way back, nothing about the method changes: out is still 8/24 = 1/3 hour, back is now 10/12 = 5/6 hour (50 minutes), so total time = 1/3 + 5/6 = 7/6 hours, total distance = 18 km, and average speed = 18 ÷ (7/6) = 108/7 ≈ 15.43 km/h. To avoid unit stumbles, keep everything in hours (or everything in minutes) until the end, and use fractions if decimals start to get messy-convert to minutes only after you’ve got the time. I learned this the hard way planning a run in college: I “averaged” two paces and promised a friend I’d be back in 40 minutes; I returned late and sweaty with a sheepish grin and a new respect for time-weighted averages (and their effect on snack freshness!).