Relative Motion of a plane in a storm

[retracell]

Homework Statement

The navigator of an airplane plans a flight from one airport to another 1200km away, in a direction 30 degrees east of north. The weather office informs him of a prevailing wind from the west, of 80km/h. The pilot wants to maintain an air speed of 300km/h.

What heading should the navigator give the pilot?

Homework Equations

Basic kinematic and relative motion equations. eg. v=dt

The Attempt at a Solution

I tried drawing many different triangles regarding the heading of the plane and the direction of travel. I got to two equations, something like:

tan30 = (Vx+80)/Vy
Vx^2 + Vy^2 = 300^2

If this is right, I was wondering if there was a simpler way.

 

[SeanGillespie]

It is solved by using vectors. The best way to simplify it is by drawing a triangle. If the pilot flew straight towards the airport at 300km per hour they would end up east of the airport due to the eastward wind.

The pilot would need to fly at an angle towards the northwest to overcome the cross wind. You need a right angle triangle where the air speed is the adjacent to the angle of travel and the speed of the wind is the side opposite the plane's heading. You should be able to find the angle using tan.

These questions can be confusing though, so if you have the answer at hand try changing the side which has the airspeed to the hypotenuse and recalculate the angle.

I got in a muddle with a similar question a number of weeks ago, and to be honest, I've forgot the correct solution already.

 

[kerimek]

I would approach this problem by first understanding the concept of relative motion. The relative motion of an object is its motion with respect to a particular frame of reference. In this case, the frame of reference would be the ground, as the plane is moving relative to the ground.

To solve this problem, we can use vector addition to find the resultant velocity of the plane. The plane's velocity can be represented by a vector in the direction of its heading, with a magnitude of 300 km/h. The wind's velocity can also be represented by a vector, in the opposite direction of its motion, with a magnitude of 80 km/h.

Using basic trigonometry, we can find the components of the plane's velocity in the x and y directions. The x component would be 300*cos(30) = 260.6 km/h and the y component would be 300*sin(30) = 150 km/h.

Now, we can use vector addition to find the resultant velocity of the plane. This can be done by adding the x and y components of the plane's velocity to the x and y components of the wind's velocity. The resultant velocity would have a magnitude of 300 km/h and would be at an angle of 30 degrees east of north.

Therefore, the heading that the navigator should give the pilot is 30 degrees east of north. This would ensure that the plane maintains an air speed of 300 km/h and reaches its destination 1200 km away.