The Student Room Group
Let distance travelled (one way) be s. The time taken to get there was s/60. Time taken to get back was s/30. Average time = (s/60 + s/30)/2 = (3s/60)/2 = s/40. Average speed = distance / time = s / (s/40) = 40mph.
Oh, and, please, stop offering rep. Your rep power is zero. :biggrin:
Reply 3
I know generalebriety answered it, but I think a verbal (well, I suppose it's actually typed) explanation could give a bit more understanding. Although equal distances are travelled each way it takes twice as long when you go at half the speed, so 2/3 of the time is spent at 30mph, so the average will be closer to 30 than 60.
Yeah. It's not simply a case of averaging the two speeds - average speed = average distance / average time. Plus what harr said. :smile:
harr
I know generalebriety answered it, but I think a verbal (well, I suppose it's actually typed) explanation could give a bit more understanding. Although equal distances are travelled each way it takes twice as long when you go at half the speed, so 2/3 of the time is spent at 30mph, so the average will be closer to 30 than 60.

Or, more quantitatively, we can say:

Unparseable latex formula:

\tex \huge Average\: Speed = \frac{2}{3} \times 30 + \frac{1}{3} \times 60 = 20+20 = 40



precisely because of your above reasoning.