An airplane is heading north at an airspeed of 600 kilometres per hour, but there is a wind blowing from the southwest at 80 kilometres per hour. How many degrees off course will the plane end up flying and what is the plane’s speed relative to the ground? Give your answers correct to one decimal place.
So the airplane is heading north at an airspeed of 600 kilometres per hour, so we have a magnitude and a direction. So this is a vector quantity, namely, velocity. We can represent this vector quantity on a diagram, so I’m taking north to be up, and we have this arrow, which represents the velocity vector with a magnitude of 600 kilometres per hour and a direction of north.
But while the airspeed of the plane is 600 kilometres per hour, that is, the speed of the airplane relative to the air is 600 kilometres per hour north, the air itself is moving because there is a wind blowing from the southwest at 80 kilometres per hour.
Again, this is a velocity because we have both a magnitude, 80 kilometres per hour, and the direction, from the southwest. And we can represent this vector quantity, velocity, on the diagram too. There are two velocity vectors coming from the blue dot, which represents our airplane, and so the net velocity will be the sum of those two vectors, which we’ll find using vector addition.
We can perform this vector addition using the parallelogram rule, so we put the vectors nose to tail, and then the sum we’re looking for is the diagonal of this parallelogram. The magnitude of this vector will be the plane’s speed relative to the ground, and the measure of the angle 𝜃 will be the number of degrees off course the plane will end up flying, assuming that the plane was meant to be flying due north.
Let 𝑖 be the unit vector pointing east and 𝑗 be the unit vector pointing north. Then we can rewrite our velocity vectors in terms of these 𝑖 and 𝑗. Our airspeed is 600 kilometres per hour north, which is 600𝑗 kilometres per hour. And with a bit more effort, we can see that the wind speed is 80 over square root two 𝑖 plus 80 over square root two 𝑗 kilometres per hour.
You can either show this using trigonometry or noticing that the components must be equal. And for the magnitudes to be 80 kilometres per hour, they must be therefore equal to 80 over root two.
The net velocity is the sum of these two, 80 over root two 𝑖 plus 600 plus 80 over root two 𝑗 kilometres per hour. We can draw a diagram to help us, and we can see that the angle we’re looking for, 𝜃, using what we know about angles in parallel lines is equal to this angle here.
And so using trigonometry, tan 𝜃 is equal to the magnitude of the opposite side, 80 over root two kilometres per hour, over the magnitude of the adjacent side, 600 plus 80 over root two kilometres per hour.
Using a calculator, we find that 𝜃 is equal to 4.924 and so on degrees, which to one decimal place is 4.9 degrees. We also need to find the plane’s speed relative to the ground; this is the magnitude of the velocity.
The speed is therefore the square root of the first component, 80 over root two squared, plus the second component, 600 plus 80 over root two squared. And putting this into our calculator, we get a speed of 659.0 kilometres per hour correct to one decimal place.
So there we have it. The airplane will be blown 4.9 degrees off course, assuming it was meaning to be [heading] north, and its ground speed will be 659 kilometres per hour, so notice it actually it’s been sped up by the wind as its air speed is only 600 kilometres per hour.