SOLUTION: An aircraft flies downwind from A to B with speed of 400mph for 2 hours and returns in absence of wind in 2.5 hours. What is the wind speed?

Question 890887: An aircraft flies downwind from A to B with speed of 400mph for 2 hours and returns in absence of wind in 2.5 hours. What is the wind speed?

Found 4 solutions by JulietG, gargaditya, Alan3354, greenestamps:
Answer by JulietG(1812)   (Show Source): You can put this solution on YOUR website!
With the wind, the airplane travels 800 miles in 2 hours.
Without the wind, the airplane travels 800 miles in 2.5 hours.
800 / 2.5 = 320 mph, the speed of the plane.
Therefore, the wind speed is 80 mph. (400 - 320)
.
That's quite a tailwind!
Answer by gargaditya(1)   (Show Source): You can put this solution on YOUR website!
There are 2 approaches to solve the same.
Ok after a lot of reading online I could not get a single page to highlight these points clearly:
1.When question says downwind or upwind it refers to Net or Vag(velocity of aeroplane with respect to ground)
2.Vaw(Velocity of plane with respect to air) aka speed in still air←-YES they mean same thing (still air means assuming air to be stationary and hence its velocity of plane relative to air):
this is a fixed value ie only groundspeed(Vag)aka net value of speed in a direction and windpeed keep changing as per condition/windspeed may add or subtract
3.distance calculations always use ‘net speed’ aka ‘groundspeed’
D=Snet *time
OR
D=Vag *time
——————-
Approach 1 :
Let speed of plane in still air be a (ie Vaw)
Let speed of wind be w
given, “NET” aka groundspeed(Vag)= a+w=400
Remember D=Snet X T
therefore distance =net speed Xtime = 400X 2 =800 mi
Returning back implies distance is same=800 mi
D=S’ XT’
using no wind condns ie net speed =a
800 =a X2.5
giving a=320
hence a+w=400 implies w=80

Approach 2:
According to relative velocity concept,
Vaw =Vag - Vwg (1)
(P.S. Here qty on left side signifies wind is the frame of reference AND also note that these are vectors ie take care of sign)
given 400mph is the GROUNDSPEED(OR NET)
**let speed of wind be w
ie Vaw= 400-(w) =400-w (2) (downwind means sign of w is positive like Vag and Vaw)
D=”Vag”*time =400*2=800 (3)
In second scenario,
Vaw= V’ag -V’wg **Always note Vaw remains same as first scenario,only vel of plane wrt ground and windspeed change
ie Vaw =V’ag -0 (4)
distance to go=distance to return back
and distance=V’agXtime‘ (5)
800= V’ a,g X t’
800=V’agX2.5 giving V’ag=320
therefore using eq 4,Vaw =V’ag=320 (6)
using eq 6 and eq 2
320=400-w
w=80
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
An aircraft flies downwind from A to B with speed of 400mph
=========
Is that its airspeed or groundspeed?
======================================
A "trick question."
Airspeed does not vary with windspeed, so groundspeed must be assumed.

Answer by greenestamps(13206)   (Show Source): You can put this solution on YOUR website!

Let's forget about the technical definitions of the terms air speed and ground speed. This is a problem for learning how to do math; it is clear what the intended meaning is.

Flying with the wind, the plane flew for 2 hours at 400mph; that is 2*400 = 800 miles.

Flying the same distance of 800 miles back, the trip took 2.5 hours. The speed is 800/2.5 = 320mph.

The difference in the two speeds is the speed of the wind: 400-320 = 80mph.

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