windy word problem: "While on vacation, you decide to go hang-gliding...."

[emm1138]

Hi all,

i have went round and round on this word problem from my Algebra 2 class, what am I missing?:roll:

While on vacation, you decide to go hang-gliding. On the first day, you hang-glide to the water's edge with the wind at your back and then back to your start point with the wind in your face. The total 4 mile trip took 16 minutes. On the second day, you make an identical trip but there is no wind and it took you only 12 for an average speed of 20 mph. Assuming that your speed without wind is the same both days, what was the wind speed on day one?

My calculations:

Known 2nd day:
4 miles
12 minutes = .20 of an hour
20 mph
3 minutes per mile

Known 1st day:
4 miles
16 minutes = .2667 of an hour
14.99 mph avg
4 minutes per mile
ws= speed of wind in mph ?
ws=mpm= wind speed minutes per mile?

What is ws?

MY Formula:
2 miles*(3 mpm-wsmpm)+2miles*(3mpm+wsmpm) = 16 minutes
or 2 miles /(20mph-ws)+2miles /(20mph+ws)=.2667
either way you get 6 minutes -2wsmpm+6minutes+2wsmpm=16 minutes

What I don't understand is that in this perfect world, wouldn't the wind speed add and subtract the same amount of time on the way there and back? Why would it take 16 minutes instead of 12?

ODD ENOUGH: If I assume that the wind is at 10 mph, i use this formula

2/(20-10)+2/(20+10)=
6/30 +2/30=
8/30=0.2667
which is 16 minutes.

But how would I figure this problem out without guessing that the ws is 10mph?

 

[emm1138]

I think we all assumed that at one point; but AIN'T SO!!

Take this example (r=regular speed, w = wind speed):
distance one way = 60 miles
r = 10 mph, w = 5 mph

60/15 (with wind) + 60/5 (against wind) = 4 + 12 = 16 hours

60/10 (no wind) + 60/10 (no wind) = 6 + 6 = 12 hours

Your head spinning?

Thanks Denis.

So if I didn't guess that the WS = 10 how do I get the formula with ws as part of the denominator.

Thanks : ):cool: