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1.3.2)

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Destination distance = 600 km in west direction.

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Destination time t_d= 55 minutes

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t_d= 55/60 = 0.917 hr

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Wind  blowing speed v_w=   75 m/s in north-west direction.

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v_w=   75 * 3.6 = 270 km/hr

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∠COA = Relative angle

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Speed of flight = distance / time = 600 / 0.917 = 654 km/hr

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OA = 645 km/hr

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OB = 270 km/hr

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Resultant speed DE² = OA² + OB² - 2 OA OB cos45

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DE² = 645² + 270² - 2 (270)(645) cos45

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DE² = 645² + 270² - 2 (270)(645) cos45

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DE = 492.58 km/hr

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For parallelogram

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(diagonal1)² + (diagonal2)² = (side1 + side2)²

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(DE)² + (OC)² = (OE + OD)²

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(OC)² = ( 645 + 270 )² - 492.58²

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OC = 771.5 km/hr.

\ Solution : \

the velocity of the aircraft with respect to the ground is 771.5 km/hr

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