(a) To determine: Thevelocity (magnitude and direction) of the plane, relative to the ground. Solution: 174 .33 m/s Explanation: Given: The velocity of the airplane due south V p = 688 k m / h = ( 688 × 10 3 ) 3600 m/s =191 .11 m/s The velocity of the wind from the southwest V w = 90.0 k m / h = 90.0 × 10 3 × 1 3600 = 25 m/s The velocity of the plane relative to the ground V =? Formula used: The direction south coincides with y-axis. The direction east coincides with x- axis. The unit vector along the East is i and the unit vector along the South is j . The velocity of the airplane with respect to ground, V → = V → p + V → w ... ( 1 ) The velocity of the airplane, V p = 0 i ^ - 191 .11 j ^ The velocity of the wind, V w = 25 sin 45 o i ^ + 25 cos 45 o j ^ Calculation: Velocity of the airplane with respect to the ground, V → = ( 0 + 25 sin 45 o ) i ^ + ( − 191.11 + 25 cos 45 o ) j ^ ... ( 2 ) V → = 17 .68 i ^ − 173.43 j ^ T h e magnitude of V → is V= 17.68 2 + ( − 173.43 ) 2 V = 312.58 +30077 .96 V = 30390.54 V = 174.33 m/s Conclusion: The velocity of the plane relative to the ground is 174 .33 m/s . (b) To determine: The distance from its intended position it will be after 11 min if the pilot takes no corrective action. Solution: The distance between the intended and the real position of the flight is x = 11.08 km Explanation: Given: The velocity of the airplane due south, V p =191 .11 m/s The velocity of the wind from the southwest, V w = 25 m/s Time, t = 11 min t = 11 × 60 t = 660 s s Formula used: v e l o c i t y = d i s tan c e t i m e Calculation: The position of the plane after 660s is: The velocity of the plane V → p = 0 i → -191 .11 j → The magnitude, V p = 191 .11 m/s Distance that the plane moves: d = V p × t = 191 .11 × 660 = 126132.6 m Taking the velocity of the plane with respect to air, V → = 17 .68 i → − 173.43 j → The real position of the flight after 11 min, = | V → | × t = 174 .33 × 660 = 115057 .8 The distance between the intended and the real position of the flight is x = 126132 .6 - 115057 .04 x = 11075 .56 m x = 11 .08 km Conclusion: The distance between the intended and the real position of the flight is x = 11 .08 km .

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