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ihatealgebrasomuch
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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***
 one year ago
 one year ago
ihatealgebrasomuch Group Title
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***
 one year ago
 one year ago

This Question is Closed

ryan123345 Group TitleBest ResponseYou've already chosen the best response.0
i love algebra <3
 one year ago

ihatealgebrasomuch Group TitleBest ResponseYou've already chosen the best response.0
dw:1360291818800:dw
 one year ago

stamp Group TitleBest ResponseYou've already chosen the best response.1
Your 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?
 one year ago

ihatealgebrasomuch Group TitleBest ResponseYou've already chosen the best response.0
would it be 40km/hr <cos45,sin45> ?? i dont understand how you got that equation
 one year ago

stamp Group TitleBest ResponseYou've already chosen the best response.1
The 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
 one year ago

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

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

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

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

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

stamp Group TitleBest ResponseYou've already chosen the best response.1
Not 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.
 one year ago
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