An airplane is flying at 500 km/hr at an angle of 310 degrees. The wind is blowing at 45 km/hr at an angle of 225 degrees. What is the resultant speed and direction of the plane?

Guest Dec 14, 2014

#1**+5 **

We can set this up as two vectors, the first for the plane and the second for the wind...so we have

u = < 500cos310, 500sin310 > v = < 45cos225, 45sin225 >

Adding the x components, we have 500cos310 + 45cos225 = 289.57

Adding the y components, we have 500sin310 + 45sin225 = -414.84

The resultant is given by √(289.57^{2} + (-414.84)^{2} ) = about 505.9 km'hr

The direction is given by tan^{-1}(-414.84/289.57) = about (- 55.08) degrees = about 304.91 degrees

This makes sense......the wind is helping to increase the speed of the plane but it is also blowing it toward a more "southerly" direction.....

CPhill
Dec 14, 2014

#1**+5 **

Best Answer

We can set this up as two vectors, the first for the plane and the second for the wind...so we have

u = < 500cos310, 500sin310 > v = < 45cos225, 45sin225 >

Adding the x components, we have 500cos310 + 45cos225 = 289.57

Adding the y components, we have 500sin310 + 45sin225 = -414.84

The resultant is given by √(289.57^{2} + (-414.84)^{2} ) = about 505.9 km'hr

The direction is given by tan^{-1}(-414.84/289.57) = about (- 55.08) degrees = about 304.91 degrees

This makes sense......the wind is helping to increase the speed of the plane but it is also blowing it toward a more "southerly" direction.....

CPhill
Dec 14, 2014