Respuesta :
Answer:
(a) 1.18 h
(b) 1.07 h
(c) 1.124 h
Solution:
As per the question:
Speed of the airplane, [tex]v_{a} = 645\ km/h[/tex]
Speed of the wind, [tex]v_{w} = 32.1\ km/h[/tex]
Distance, d = 724 km
Now,
(a) To calculate time taken for the trip when the wind blows towards south:
The plane flies towards North and the wind blows towards South.
Thus the velocity of the airplane is given by:
[tex]v = v_{a} - v_{w} = 645 - 32.1 = 612.9\ km/h[/tex]
Thus the time taken, t = [tex]\frac{d}{v} = \frac{724}{612.9} = 1.18\ s[/tex]
t = 1.18 h
(b) To calculate the time interval in case of the tailwind:
Now, the direction of the wind and the plane is same:
Velocity of the plane, [tex]v' = v_{a} + v_{w} = 645 + 32.1 = 677.1\ km/h[/tex]
Time interval, t' = [tex]\frac{d}{v'} = \frac{724}{677.1} = 1.07\ h[/tex]
(c) To calculate the time interval in case of a crosswind:
[tex]v" = \sqrt{v_{a}^{2} - v_{w}^{2}}[/tex]
[tex]v" = \sqrt{645^{2} - 32.1^{2}} = 644.2\ km/h[/tex]
Time taken, t" = [tex]\frac{d}{v"} = \frac{724}{644.2} = 1.124\ h[/tex]
