Pilot Must Maintain Velocity of 353.11km/h for London-Rome Flight

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The pilot must maintain a velocity of 353.11 km/h at an angle of 34º W of S to fly from London to Rome in 3.5 hours, covering a distance of 1400 km. The ground velocity required is calculated as 400 km/h [43º E of S]. Adjusting for a 75 km/h eastward wind, the necessary airspeed was derived through vector analysis. Despite discrepancies with the textbook answer of 550 km/h [15º S of W], the calculations presented appear accurate. The discussion confirms the pilot's required airspeed to successfully complete the flight.
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The displacement from London, Uk, to Rome is 1400km [43 degrees E of S]. A wind is blowing with a velocity of 75km/h [E]. The pilot wants to fly directly from London to Rome in 3.5h. What velocity relative to the air must the pilot maintain?

[p = plane, a = air, g = ground]

Vpa + Vag = Vpg

Vpg = d/t
= 1400km [43º E of S] / 3.5hr
= 400km/h [43º E of S]

Vpgx = 400sin43 = 272.79km/h
Vpgy = 400cos43 = 292.54km/h

Vagx = 75km/h
Vagy = 0

Vx = 272.79 - 75 = 197.79km/h
Vy = 292.52 + 0 = 292.52km/h

V = sqrt 197.79² + 292.52²
V = 353.11km/h

tan-1(197.79/295.52) = 34º

therefore the velocity the pilot must maintain is 353.11km/h [34º W of S] in order to get to Rome from London.

Can someone check over my work, my answer did not match the answer key in the back of my textbook (although it is filled with mistakes)

The answer in my textbook was 550km/h [15º S of W]
 
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Your answer seems fine to me.
 
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