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Q10.11.12要幫幫忙 圖片參考：http://tw.yimg.com/i/tw/ugc/rte/smiley_17.gif ，中英答都ｏｋ。唔該 圖片參考：http://i350.photobucket.com/albums/q437/mangorayyau/111213.jpg 更新: answers Q10. 43.9ms-1 Q11. N87.0E Q12. a. 7000s Q12. b. 58ms-1(S53.0E)
最佳解答:
(10) Let v be the speed of the raindrops relative to the earth. Since they are vertically falling relative to the passenger, their horizontal speed component should be 15 m/s and therefore: v sin 20 = 15 v = 43.9 m/s (11) Suppose that the wind is in the direction of N θ E, then the vector diagram consisting of the resultant velocity of the plane, the original velocity of the plane and the wind velocity will be an isos. triangle with equal sides 100 m/s and the base side 10 m/s Therefore by the cosine law: 1002 = 1002 + 102 - 2(100)(10)cos θ cos θ = 0.05 θ = 87 So the wind direction should be in N 87 E. (12a) The north-component of the plane's velocity will be 110 cos 25 and therefore the time of journey = 700000/(110 cos 25) = 7021 s which is about 117 minutes. (b) So to speak, the resultant velocity of the plane is 700000/(3 x 3600) = 64.8 m/s Therefore the wind's velocity has a southern component of 110 cos 25 - 64.8 = 34.9 m/s In addition to the eastern component of 110 sin 25 = 46.5 m/s, and by the Pyth. theorem, the wind's velocity has a magnitude of 58.1 m/s And for the direction, θ = tan-1 (34.9/64.8) = 53.0 Thus the direction is S53.0E.
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relative motion發問:
Q10.11.12要幫幫忙 圖片參考：http://tw.yimg.com/i/tw/ugc/rte/smiley_17.gif ，中英答都ｏｋ。唔該 圖片參考：http://i350.photobucket.com/albums/q437/mangorayyau/111213.jpg 更新: answers Q10. 43.9ms-1 Q11. N87.0E Q12. a. 7000s Q12. b. 58ms-1(S53.0E)
最佳解答:
(10) Let v be the speed of the raindrops relative to the earth. Since they are vertically falling relative to the passenger, their horizontal speed component should be 15 m/s and therefore: v sin 20 = 15 v = 43.9 m/s (11) Suppose that the wind is in the direction of N θ E, then the vector diagram consisting of the resultant velocity of the plane, the original velocity of the plane and the wind velocity will be an isos. triangle with equal sides 100 m/s and the base side 10 m/s Therefore by the cosine law: 1002 = 1002 + 102 - 2(100)(10)cos θ cos θ = 0.05 θ = 87 So the wind direction should be in N 87 E. (12a) The north-component of the plane's velocity will be 110 cos 25 and therefore the time of journey = 700000/(110 cos 25) = 7021 s which is about 117 minutes. (b) So to speak, the resultant velocity of the plane is 700000/(3 x 3600) = 64.8 m/s Therefore the wind's velocity has a southern component of 110 cos 25 - 64.8 = 34.9 m/s In addition to the eastern component of 110 sin 25 = 46.5 m/s, and by the Pyth. theorem, the wind's velocity has a magnitude of 58.1 m/s And for the direction, θ = tan-1 (34.9/64.8) = 53.0 Thus the direction is S53.0E.
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