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anonymous
 3 years ago
i really need help with vectors! can someone please explain to me how to go about solving this problem? The pilot of an airplane flies at a constant speed of 200 km/h at a bearing of N 25degrees S. There is a 40 km/h crosswind blowing southeast (S 45degrees E). What are the plane's actual speed and direction? ***will draw picture***
anonymous
 3 years ago
i really need help with vectors! can someone please explain to me how to go about solving this problem? The pilot of an airplane flies at a constant speed of 200 km/h at a bearing of N 25degrees S. There is a 40 km/h crosswind blowing southeast (S 45degrees E). What are the plane's actual speed and direction? ***will draw picture***

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360291818800:dw

stamp
 3 years ago
Best ResponseYou've already chosen the best response.1Your picture is a bit all over the place. Let us try again, using vector notation.\[V_{plane}=200\frac{km}{h}<cos25°,sin25°>\]Can you find the other vector?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0would it be 40km/hr <cos45,sin45> ?? i dont understand how you got that equation

stamp
 3 years ago
Best ResponseYou've already chosen the best response.1The velocity doesn't have direction, it is a scalar that varies. The direction is determined by the angle it forms with the axis. Observe:dw:1360292754973:dw

stamp
 3 years ago
Best ResponseYou've already chosen the best response.1Your vector is ok except southeast means that your y vector is negative\[V_{wind}=40\frac{km}{h}<cos45°, sin45°>dw:1360292917561:dw\]

stamp
 3 years ago
Best ResponseYou've already chosen the best response.1Please note that my vector drawing should have arrows on the tails, and my axis should be labeled x and y appropriately

stamp
 3 years ago
Best ResponseYou've already chosen the best response.1\[V_{wind}=40\frac{km}{h}<cos45°, sin45°>\]

stamp
 3 years ago
Best ResponseYou've already chosen the best response.1The planes actual speed and direction will be\[V_{actual}=V_{plane}+V_{wind}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so i would do 200km/h <cos25, sin25> + 40km/h <cos45,sin45> in order to get the magnitude of the vector i am looking for?

stamp
 3 years ago
Best ResponseYou've already chosen the best response.1Not to get the magnitude of the vector you are looking for, but to get the resultant vector. The resultant vector is how fast the plane is actually going, because even though the airplane travels at 200km/h in a direction, the wind offsets it by blowing against. The planes true course is determined by the added vectors. To find the magnitude of the vector, you would use pythagoream theorem.
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