How Should a Plane Adjust Its Course in Crosswinds to Fly Due North?

[greyradio]

A plane is flying at a speed of 206 km/h in still air. There is a wind blowing at a speed of 76.6 km/h at 48.7 degrees to the east of north, and the pilot wishes to fly due north.

What angle should the plane fly? (assume the angle is measured between the plane and north.)
What speed does the plane fly relative to the ground with the wind blowing?I've tried drawing the vector diagram but I'm having trouble with it. I initially assumed the plane would be flying in a straight line from west to east. However, quickly realized that this created a right triangle where the hypotenuse is 76.6 km/h which is not possible as the largest vector is 206 km/h. My other attempts to solve it have failed. I was hoping someone could help me out. Thanks.

 

[Nate810]

The angle that the plane should fly is 48.7 degrees east of north. The speed of the plane relative to the ground with the wind blowing is 166 km/h.To solve this problem, you can draw a vector diagram. The vector for the plane's speed is 206 km/h at 0 degrees (due north). The vector for the wind is 76.6 km/h at 48.7 degrees (east of north). To find the total vector, add these two vectors together to get 166 km/h at 48.7 degrees. This is the direction and speed the plane should fly relative to the ground with the wind blowing. Therefore, the angle the plane should fly is 48.7 degrees east of north.

 

[Moetasim]

I would approach this problem using vector addition to determine the resultant velocity of the plane in the presence of the wind. This can be achieved by breaking down the velocities into their respective components and then adding them together.

First, we can break down the velocity of the wind into its north and east components using basic trigonometry. The north component would be 76.6 km/h * cos(48.7 degrees) = 50.1 km/h, and the east component would be 76.6 km/h * sin(48.7 degrees) = 58.5 km/h.

Next, we can add the wind's north component to the plane's velocity in still air, which would result in a northward velocity of 206 km/h + 50.1 km/h = 256.1 km/h. This means that the plane would have a northward component of velocity of 256.1 km/h relative to the ground.

To determine the angle at which the plane should fly, we can use inverse trigonometry to find the angle between the resultant velocity and the north direction. This would be arctan(58.5 km/h / 256.1 km/h) = 13.1 degrees. Therefore, the plane should fly at an angle of 13.1 degrees to the west of north in order to counteract the wind and maintain a northward direction.

Finally, to determine the speed of the plane relative to the ground, we can calculate the magnitude of the resultant velocity using the Pythagorean theorem. This would be √(206 km/h)^2 + (50.1 km/h)^2 = 211.2 km/h. Therefore, the plane would be flying at a speed of 211.2 km/h relative to the ground with the wind blowing.