Math Destroyer question

[zammyd]

---

Members don't see this ad.

"Sam travels 40 miles from point A to B at 60mph. At what speed must he travel back from point B to A so that the average velocity for the round trip is 75 mph."

The answer in the book says 100mph, but,

Why is the answer not 90mph? if you go 60mph one way and then 90mph back, shouldn't that average out to 75mph?

 

[ns718]

No. The average speed is not the arithmetic average of the 2 speeds. the average speed is the total distance divided by the total time.

 

[zammyd]

But if the distance is the same for both trips, shouldn't you just be able to just average V1 and V2?

 

[FeralisExtremum]

The distance is the same for both trips. But the time spent during each leg of the trip is not. So even though Sam may move faster during the same distance, he spends less time travelling at that speed, so it contributes less to the overall average. This is why we calculate average velocity as total distance and total time, not simply averaging the two velocities.

To use an extreme example, let's say you have a 2 mile trip. You travel the first mile at 1 mph and you travel the second mile at 999 mph. Is your average velocity for the trip 500 mph? Of course not, because you spent an hour traveling the first half of the trip alone. If your average speed was actually 500 mph, the entire two mile trip would have taken you less than half a minute. Demonstrably, we can see that the arithmetic average of velocities is not valid in these situations.

 

[zammyd]

Thanks guys, got it now 🙂