Question

City A lies directly west of city B. When there is no wind, an airliner makes the round trip flight of distance s between them in a flight time of t1 while traveling at the same speed in both directions. When a strong, steady wind is blowing from west to east and the airliner has the same air speed as before, the round trip takes a flight time of t2.
How fast the wind blowing ??

Answer 1

Answer:

Speed of the wind is $= \sqrt{(\frac{s}{t_1})^2-(\frac{s}{t_1t})^2 }$

Explanation:

Given that,

airliner cover distance s, in time t1  between the cities faces opposite A and B

Let the speed of the wind be v

and the speed of the plain be $v_p$

The speed of the plain is

$v_p = \frac{s}{t}$

Time taken from one side is

= half distance travelled / (vp +v)

Time taken during A and B is

$t_A_B = \frac{s/2}{s/t_1 +v}$

Time taken during B to A

$t_B_A = \frac{s/2}{s/t_1 -v}$

Total timetaken by plane at the presence of wind

$t = t_A_B + t_B_A$

$t = \frac{s/2}{\frac{s}{t_1} + v} + \frac{s/2}{\frac{s}{t_1} -v}$

rearrange the equation to get v

$v = \sqrt{(\frac{s}{t_1})^2-(\frac{s}{t_1t})^2 }$

Speed of the wind is $= \sqrt{(\frac{s}{t_1})^2-(\frac{s}{t_1t})^2 }$