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Suppose that during the whole journey there is a constant wind blowing from A to B. If the speed of the aircraft in still air is 165 km/h, find the speed of the wind.

2007-01-17 01:46:54 · 5 answers · asked by Zin Mar 1 in Science & Mathematics Physics

5 answers

d=vt
t=d/v
5.5 = 450/(165+x) + 450/(165-x)

Solve for x
x=15km/h

2007-01-17 01:54:15 · answer #1 · answered by catarthur 6 · 1 0

Set up an equation. Remember that distance divided by speed equals time. X is the wind speed. The first term is the time for the trip with the wind, the second term is the time for the trip against the wind. The sum of the two times should equal the round-trip time:

450/(165+x) + 450/(165-x) = 5.5

Then multiply to get a common denominator:

450(165-x)/(165+x)(165-x) + 450(165+x)/(165+x)(165-x) = 5.5

[450(165-x)+450(165+x)]/(165^2-x^2) = 5.5

450(330) = 5.5(165^2-x^2)

27000=27225-x^2

x=15 km/h

2007-01-17 10:04:07 · answer #2 · answered by Gerfried 2 · 0 0

The aircraft measures its speed relative to the surrounding air and this is V = 165 km/h. When there is wind (this is u and is required), the velocity of the aircraft is found by a vector sum of the two velocities (composition of velocities in different reference frames).
When the aircraft is moving from A to B, its speed is (wind in the tail)::

v1 = V + u

When it is moving from B to A, its speed is (wind in the head):

v2 = V - u

The time from A to B is:

t1 = s/v1 = s/(V-u)

and from B to A is:

t2 = s/v2 = s/(V+u)

The total time is:

t = t1 + t2 = s/(V-u) + s/(V+u)

This is an equation with one unknow and its solution is:
t = s/(V-u) + s/(V+u) /.(V-u)(V+u)
t*V^2 - t*u^2 = s*V + s*u + s*V - s*u

V*(V*t - 2*s) = t*u^2

u^2 = V^2*(1 - 2*s/(V*t))

u = V*Sqrt(1 - 2*s/(V*t))

u = 165*Sqrt(1 - 2*450/(165*5.5)) km/h

u = 15 km/h

2007-01-17 09:49:40 · answer #3 · answered by Bushido The WaY of DA WaRRiOr 2 · 0 0

My math may be off but he cant get there in 5.5 hours with a constant wind speed and if the wind is 166km/h he would be flying backwards on the return leg and never get back to A.

2007-01-17 10:14:29 · answer #4 · answered by Tony N 3 · 0 0

Let the wind peed be w km/hr
we then have the eqn.
450/[165+w]+450/[165-w]=5.5
450[1/[165+w]+1/[165-w]=5.5
450*2*165/[165+w][165-w]=5.5
[165^2-w^2]=900*165/5.5
w^2=165^2-900*165/5.5
=165[165-163.64]
=165*1.36=224.4
w=15km/hr approx

2007-01-17 10:29:07 · answer #5 · answered by openpsychy 6 · 0 0

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