DIY Solar Panels

4 Responses

1. hayharbr Says:
   January 28th, 2010 at 5:34 am

   x and y are not miles; they are speeds, so you can’t add them up to get 1725 miles.

   They give you the speed in still air so don’t use a variable for that. The formula is distance = speed times time, or
   time = distance divided by speed

   Let w = the speed of the wind. Then 520 + w = plane’s speed with wind and 520 – w = plane’s speed against the wind.

   Since the times are the same,
   1725 / (520 + w) = 1395 / (520 – w)

   So cross multiply and solve for w.
   References :

2. Ivans Says:
   January 28th, 2010 at 6:22 am

   You are mostly right, however what it means with 1725 miles with the wind and 1395 miles against the wind in the same amount of time. is that after time t we have these equations:

   t(520+y) = 1725 (time * rate = distance)
   t(520-y) = 1395
   now t=1725/(520+y) from the first equation
   and t=1395/(520-y) from the second, I can equate these to get:
   1725/(520+y)=1395/(520-y)

   Now we can cross multiply to get
   1725(520-y)=1395(520+y)
   897000 – 1725y = 725400 + 1395y
   171600=3120y
   y=55

   hope that’s clear.
   References :

3. Brock Sampson Says:
   January 28th, 2010 at 6:55 am

   Assuming that the wind is pushing the airplane at the same speed the wind is blowing, than
   1725=speed with wind
   13395=speed against wind
   520=speed of plane in still air
   x=speed of wind
   y=amount of time plane is flying
   (520+x)y=1725
   (520-x)y=1395
   1040y=3120
   y=3
   (520-x)3=1395
   520-x=465
   x=55 mph

   Remeber this is only right if the plane is being pushed by the wind exactly as fast as the wind is blowing.
   References :

4. whabtbob Says:
   January 28th, 2010 at 7:16 am

   Since they already give you this info:
   "the speed of the plane in still air is 520 mph"
   then it is unnecessary to set:
   "x = speed of plane in still air" (because you already know it is 520 mph).

   You have to use the formula
   rate * time = distance
   to solve the problem. Notice they give you that both trips are completed in the same amount of time. Solving the formula for time give us:

   time = distance/rate

   Let’s set w= speed of the wind. Then the speed of the plane with the wind is 520+w, and the speed of the plane against the wind is 520-w. Plug this in the formula:

   Let w = the speed of the wind. Then 520 + w = plane’s speed with wind and 520 – w = plane’s speed against the wind.

   1725 / (520 + w) = 1395 / (520 – w)
   Let w = the speed of the wind. Then 520 + w = plane’s speed with wind and 520 – w = plane’s speed against the wind.

   Since the times are the same,
   1725 / (520 + w) = 1395 / (520 – w)
   Let w = the speed of the wind. Then 520 + w = plane’s speed with wind and 520 – w = plane’s speed against the wind.

   Since the times are the same,
   1725 / (520 + w) = 1395 / (520 – w)
   Let w = the speed of the wind. Then 520 + w = plane’s speed with wind and 520 – w = plane’s speed against the wind.

   Since the times are the same,
   1725 / (520 + w) = 1395 / (520 – w)
   Let w = the speed of the wind. Then 520 + w = plane’s speed with wind and 520 – w = plane’s speed against the wind.

   Since the times are the same,
   1725 / (520 + w) = 1395 / (520 – w)
   1725 * (520 – w) = 1395 * (520 + w)
   897000 – 1725w = 725400 + 1395w
   171600 = 3120w
   55 = w

   This is much more logical an answer, and that is not rounded; it is exactly 55, which makes me suspect it is the correct answer.

   Hope this helps.
   References :

Leave a Comment

Name

Mail (will not be published)

Website

Please note: Comment moderation is enabled and may delay your comment. There is no need to resubmit your comment.