- #1
physics4ever25
- 35
- 0
Moved from a technical forum, so homework template missing
Hello guys,
I have a vectors word problem and I found 2 different ways to solve the same problem but I'm getting different answers. Apparently, both answers are correct since I've looked for the answer online and I found both answers from different sources, so I'm really confused now.
_______________________________________________________________________________________
Question: An airplane is flying at 550 km/h on a heading of 080 degrees. The wind is blowing at 60 km/h from a bearing of 120 degrees. Find the ground velocity (resultant vector) of the airplane.
________________________________________________________________________________________
Method 1: First you would draw the vector diagram. You would draw a y-axis and a x-axis. Then from the origin you would draw the plane vector at an angle (from north) of 80 degrees and this vector would be 550 km/h in magnitude. Then from the origin you would draw the wind vector at an angle (from north) of 120 degrees and this vector would be 60 km/h in magnitude. Then you would connect the outside ends (away from the origin) of each vector together. This side would be the resultant. This will form a triangle. Then using cosine and sine laws you can find the magnitude and angle of the resultant vector.
________________________________________________________________________________________
Method 2: I found this method online and I don't even quite understand it myself but apparently it works. All values used in this method are from the question.
Vr=Vp+Vw
=[550(cos80+i*sin80)] + [60(cos120+i*sin120)]
=[95.50+541.64i] + [-30+51.96i]
=65.50+593.60i
Since this is in a+bi form you will now convert it to polar form. You will do this using the Pythagorean theorem and tan.
r=square root (65.50^2+593.60^2)
=597.2
TanTheta=593.60/65.50
Theta=83.70
Therefore, the magnitude of the resultant would be 597.2 km/h and the angle would be 83.70 degrees (from north).
________________________________________________________________________________________
Anyways, these are the two methods. If you solve this using method 1 you will get a different answer than method 2. However, both answers are apparently correct since I checked for the answer to this question online and I found both answers. I'm confused now about which answer would be the correct one (using method 1 or method 2).
I have a vectors word problem and I found 2 different ways to solve the same problem but I'm getting different answers. Apparently, both answers are correct since I've looked for the answer online and I found both answers from different sources, so I'm really confused now.
_______________________________________________________________________________________
Question: An airplane is flying at 550 km/h on a heading of 080 degrees. The wind is blowing at 60 km/h from a bearing of 120 degrees. Find the ground velocity (resultant vector) of the airplane.
________________________________________________________________________________________
Method 1: First you would draw the vector diagram. You would draw a y-axis and a x-axis. Then from the origin you would draw the plane vector at an angle (from north) of 80 degrees and this vector would be 550 km/h in magnitude. Then from the origin you would draw the wind vector at an angle (from north) of 120 degrees and this vector would be 60 km/h in magnitude. Then you would connect the outside ends (away from the origin) of each vector together. This side would be the resultant. This will form a triangle. Then using cosine and sine laws you can find the magnitude and angle of the resultant vector.
________________________________________________________________________________________
Method 2: I found this method online and I don't even quite understand it myself but apparently it works. All values used in this method are from the question.
Vr=Vp+Vw
=[550(cos80+i*sin80)] + [60(cos120+i*sin120)]
=[95.50+541.64i] + [-30+51.96i]
=65.50+593.60i
Since this is in a+bi form you will now convert it to polar form. You will do this using the Pythagorean theorem and tan.
r=square root (65.50^2+593.60^2)
=597.2
TanTheta=593.60/65.50
Theta=83.70
Therefore, the magnitude of the resultant would be 597.2 km/h and the angle would be 83.70 degrees (from north).
________________________________________________________________________________________
Anyways, these are the two methods. If you solve this using method 1 you will get a different answer than method 2. However, both answers are apparently correct since I checked for the answer to this question online and I found both answers. I'm confused now about which answer would be the correct one (using method 1 or method 2).