Those stupid aeroplane questions

Everyone who does Spec will know what I'm talking about. "This aeroplane wants to fly from airfield A to airfield B, but there's a wind blowing from x direction at x velocity. What direction should the plane fly in?" I hate these questions for two reasons. 1) They're annoying. 2) I like planes and I abhor their usage in annoying questions. Unfortunately, if you do Spec you'll have no other choice than to encounter and solve them.

Let's have a look at one of these questions- this time using a helicopter, not an aeroplane. Good riddance to helicopters.

A helicopter can fly at 75m/s in still air. The pilot wishes to fly from airport A to a second airport B, 300km due North of A. If i is a unit vector due East and j a unit vector due North find (in the form ai + bj) the velocity vector that the pilot should set and the time the journey will take if
a) there is no wind blowing,
b) there is a wind of (21i + 10j) m/s blowing.

Part a) is easy. Since there is no wind to affect the movement of the helicopter, the helicopter can just fly directly North for 300km (i.e. with a velocity vector of 75j). You can then use distance / speed to get the time required. You do need to be careful in this one- you have to convert m/s into km/h, or you can convert km into m.

75*3.6 = 270 km/h
300/270 = 1.11 hr = 1 hr 7 minutes (nearest minute)

Alternatively:
300*1000 = 300 000 metres
300000/75 = 4000 seconds = 66.7 minutes (1 d. p.) = 1 hr 7 minutes (nearest minute)

Now part B... that's the part I hate simply because it requires more work. Drawing a diagram can help to visualise the situation.

I have to admit that my diagram is a bit misleading (especially because I've only just realised that I wrote "aeroplane" instead of "helicopter" and "airfield" instead of "airport" out of sheer habit). Unless the journey is complete in exactly 1 second or 1 of whatever unit of time you're using, the resultant of the wind and aircraft vectors will not be equal to the distance to the airport or wherever the question's asking you to go to. The resultant is a scalar multiple of the position vector of the second airport relative to the first- that is, it's in the same direction, but shorter or longer.

Now, the question said that the helicopter can go at 75m/s in still air. This means that the magnitude of the aeroplane's velocity should be 75m/s. If we let the helicopter's velocity be xi + yj m/s, then we can create this equation for x and y:

sqrt(x^2 + y^2) = 75 m/s

Now, as you probably already know, when we have 2 variables, we need at least 2 equations to solve for the variables. We can get another equation by adding together the wind and helicopter vectors- as previously mentioned, the resultant should be a scalar multiple of the position vector of the second airport relative to the first. Sounds like a lot of garble, but it's actually not that complex.

21 i + 10 j + xi + yj = k(300 000j) where k is a constant.

Then just equate the i components and the j components separately.

21 + x = 0
therefore x = -21

10 + y = 300000k

Since this example is quite easy we can get a result for x straight away. We can then substitute it into that equation we had earlier to work out what y is.

sqrt((-21)^2 + y^2) = 75 m/s

y = 72 (reject -72 as we can see from the above diagram that y has to be positive)

Therefore, the velocity vector of the plane is -21i + 72j m/s.

To work out resultant velocity, add wind and helicopter velocities together. You can then work out the magnitude from there and use that to complete the question.

21i + 10j - 21i + 72j = 82j

|82j| = 82m/s = 295.2km/h
300/295.2 = 1.02 hours = 61 minutes (nearest minute)

And that's how you solve an aeroplane question!

That was a relatively simple question, though, since the velocity vector is directly north. Here I'll give a very brief explanation on what to do if the velocity vector isn't directly north. I might expand on it later if I can be bothered.

Write out the position vector of the second location relative to the first in component form. For example, if the airport was actually 300km north-east, you could write it as 300 cos 45 i + 300 sin 45 j. It doesn't matter here whether you use the same units (kilometres or metres) as your velocity vectors, since these numbers are going to be cancelled out soon anyway.

When you equate the velocity vectors to a scalar multiple of the position vector, you'll end up with this:

21 i + 10 j + xi + yj = k(300 cos 45 i + 300 sin 45 j)

Equating i and j components:

21 + x = 300 cos 45 k
10 + y = 300 sin 45 k

Divide the first equation by the second and rearrange it to get an equation for x or y.

(21 + x)/(10 + y) = (300 cos 45 k)/(300 sin 45 k)
(21 + x) / (10 + y) = cos 45 / sin 45
21 + x = (cos 45 / sin 45)(10 + y)
x = (cos 45 / sin 45)(10 + y) - 21

You could then substitute this into the sqrt((-21)^2 + y^2) = 75 m/s equation.

By the way, the equation x = (cos 45 / sin 45)(10 + y) - 21 could be further simplified to x = 10 + y - 21 and then x = y - 11 because cos 45 happens to be equal to sin 45. The only reason I left them unsimplified is because you won't always get nice numbers that can simplify down like that.