Ground speed of an airplane

Question

I'm really in big trouble with this homework question. I can't seem to figure out how to tackle this problem. Can somebody help out? Thanks!

A plane is heading due south. If its airspeed is 530 km/hr and there is a wind blowing 90 km/hr to the northwest, what is the ground speed of the plane? ground speed = ______

Answer 1

You have a triangle with two sides given, and the angle between them can be quickly figured out. Therefore, the Law of Cosines can be used to solve for the third side. The two sides of the triangle we know of are the plane's airspeed and the wind's speed. The angle between them is simply $45°$ because that is the angle between any vector in the northwest direction and any vector in the northward direction.

The Law of Cosines states that

$$c^2 = a^2 + b^2 - 2ab \cos C$$

We know $a$, $b$, and $C$, so we can plug into this formula to get

$$ c^2 = 530^2 + 90^2 - 2(530)(90)\cos 45°$$

Solving for $c$ yields the answer $c = \sqrt{289000 - 47700 \sqrt 2} $ $\frac{km}{hr}$, where $c ≈ 470.6824971$$\frac{km}{hr}$, which makes sense given the information in the problem.

Answer 2

I'm new to stackexchange, and unsure of whether or not homework help is allowed, but I'll answer this question anyway and assume the mods will remove it if it's not allowed.

The key to problems like these is unpacking the information given them and putting them into a form you can work with (usually vectors). Let's go through the problem and see what it tells us.

A plane is heading due south. If its airspeed is 530 km/hr

So the plane is moving due south. Assuming north is the positive y direction, we know the plane's velocity is going to be negative in the y direction. Because the plane is moving due south, we know it's not moving in the x direction. Thus, we can write the velocity of the plane as the vector {0, -530} in units of km/hr.

there is a wind blowing 90 km/hr to the northwest

There are two pieces of information here- magnitude and direction. Northwest is an angle of 135 degrees. The magnitude is 90 km/hr. We can break this magnitude up into its x- and y-parts by multiplying the magnitude by cos(135deg) and sin(135deg), respectively. The cosine of 135 degrees is -sqrt(2)/2, and the sine of the angle is sqrt(2)/2. Multiplying these values by the magnitude of the wind, we get a velocity vector of [-45sqrt(2), 45sqrt(2)].

So now we've successfully decoded the givens of the problem. But what is the problem asking for? The ground speed velocity, or the net velocity after wind speed has been accounted for. To calculate this, we simply add the vector of the plane's velocity and the vector of the wind's velocity (think about it this way- both the plane and the wind affect the speed, so we add them to see what the final result is).

[0,-530] + [-45sqrt(2), 45sqrt(2)] = [-45sqrt(2), 45sqrt(2)-530]

I'll leave you to convert that to decimal, but we've now found the velocity of the plane with respect to the ground. To find pure speed, we take the magnitude of the vector. To find the angle at which the plane is travelling, take the arctangent of the y-value divided by the x-value. I'll leave you to decide the best form to put the answer in.

Does that make sense how we solved that problem? If you have any questions, don't be afraid to ask.