Answer:
[tex]209[/tex] mph
Explanation:
[tex]V[/tex] = Speed of bird in still air
[tex]v[/tex] = Speed of wind = 44 mph
Consider the motion of the bird with the wind
[tex]D_{1}[/tex] = distance traveled with the wind = 9292 mi
[tex]t_{1}[/tex] = time taken to travel the distance with wind
Time taken to travel the distance with wind is given as
[tex]t_{1} = \frac{D_{1}}{V + v}[/tex]
[tex]t_{1} = \frac{9292}{V + 44}[/tex] eq-1
Consider the motion of the bird with the wind
[tex]D_{2}[/tex] = distance traveled against the wind = 6060 mi
[tex]t_{2}[/tex] = time taken to travel the distance against wind
Time taken to travel the distance against wind is given as
[tex]t_{2} = \frac{D_{2}}{V + v}[/tex]
[tex]t_{2} = \frac{6060}{V - 44}[/tex] eq-2
As per the question,
Time taken with the wind = Time taken against the wind
[tex]t_{1} = t_{2}[/tex]
[tex]\frac{9292}{V + 44} = \frac{6060}{V - 44}[/tex]
[tex](9292) (V - 44) = (6060) (V + 44)[/tex]
[tex]9292V - 408848 = 6060V + 266640[/tex]
[tex]3232V = 675488[/tex]
[tex]V = 209[/tex] mph
