# Vectors Applications

• Aug 21st 2013, 07:01 AM
DaItalianFish
Vectors Applications
A pilot wishes to fly from Bayfield to Kitchener, a distance of 100 km on a bearing of 105°. The speed of the plane in still air is 240 km/h. A 20 km/h wind is blowing on a bearing of 210°.
Remembering that she must fly on a bearing of 105° relative to the ground (ie the resultant must be on that bearing), find

1. the heading she should take to reach her destination
2. how long will the trip take
• Aug 21st 2013, 08:39 AM
HallsofIvy
Re: Vectors Applications
Start by drawing a picture. Draw a line at 105° and label it "100 km". Call that "D" (the "D"irect line to the "D"estination). Draw a second line,, starting at the same initial point, at 210° and label it "20t km" (t is the time in hours). Call that "W" ("W"ind). Finally, draw the line connecting the ends to form a triangle. That line represents a vector that, added to "W" gives "D". Call that "P" (for "P"lane).

Now, there are two ways to find length and direction of "P". One is to find "x" and "y" components for vector "D" and "W". You want P+ W= D (the motion of the plane plus the motion of the wind takes the plane along the correct route) so that P= D- W. Subtract corresponding components of D and W to get the components of P.

Or, my preference, "solve the triangle. You have two adjacent sides of lengths 100 and 20t with angle between them 210- 105= 105°. You can use the "cosine law" to find the length of the third side of the triangle (the length of vector P in terms of t so that the coefficient of t is the speed of the airplane) and then use, say, the sine law to determine the angle.
• Aug 21st 2013, 12:16 PM
DaItalianFish
Re: Vectors Applications
Quote:

Originally Posted by HallsofIvy
Start by drawing a picture. Draw a line at 105° and label it "100 km". Call that "D" (the "D"irect line to the "D"estination). Draw a second line,, starting at the same initial point, at 210° and label it "20t km" (t is the time in hours). Call that "W" ("W"ind). Finally, draw the line connecting the ends to form a triangle. That line represents a vector that, added to "W" gives "D". Call that "P" (for "P"lane).

Now, there are two ways to find length and direction of "P". One is to find "x" and "y" components for vector "D" and "W". You want P+ W= D (the motion of the plane plus the motion of the wind takes the plane along the correct route) so that P= D- W. Subtract corresponding components of D and W to get the components of P.

Or, my preference, "solve the triangle. You have two adjacent sides of lengths 100 and 20t with angle between them 210- 105= 105°. You can use the "cosine law" to find the length of the third side of the triangle (the length of vector P in terms of t so that the coefficient of t is the speed of the airplane) and then use, say, the sine law to determine the angle.

I attempted the question earlier, here is the link. I know it is wrong. If possible, can you point out where I made the mistakes and help me correct them?