Trying to understand the logic behind adding vectors with an angle between them

[renobueno4153]

Homework Statement

Two aircrafts travel on the same plane with velocity v1 = 500 km/h and v2 = 800 km/h
respectively. The directions of motion generate an angle θ = 30◦ as the two aircrafts move
away from each other. Evaluate:

(a) The magnitude of the relative velocity of the second aircraft with respect to the first
one

Relevant Equations

law of cosine
vector addition

My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h.

So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since the planes just change direction but their speeds stay the same. With a lot of prompting with Chatgpt I got this sketch with the calculation out of it. And this just confuses me. Like why would I even want to know about the vector between them and then use the law of cosine. I mean I can see using the cosin in a sense of actually involving one of the given variables but intuitively I dont.

 

[PeroK]

Suppose you have a triangle with two sides that are 3 and 4 units respectively. Does the length of the third side depend on the angle between the first two sides?

 

[wrobel]

There is a regular theory which is good to know before solving such problems.
Banach Mech
You need formula (I) at page 57 of this book; and explanation of this formula above at page 56.

 

[kuruman]

Homework Statement: Two aircrafts travel on the same plane with velocity v1 = 500 km/h and v2 = 800 km/h
respectively.

Since the planes just change direction but their speeds stay the same.

Where does it say that the directions change?

 

[PeroK]

There is a regular theory which is good to know before solving such problems.
Banach Mech
You need formula (I) at page 57 of this book; and explanation of this formula above at page 56.

That source is perhaps a little advanced for the OP at this stage.

 

[kuruman]

Like why would I even want to know about the vector between them and then use the law of cosine.

Because, like, you are asked to find the relative velocity between the two planes. That is defined as the vector difference between the two velocities and is represented graphically as the arrow that you call $v_{21}$ in your drawing. Read about relative velocity here.

 

[gmax137]

Like why would I even want to know about the vector between them and then use the law of cosine.

@renobueno4153 I really hope you think about this until you understand it -- it holds the key idea in much of what follows in classical mechanics.

 

[Herman Trivilino]

My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h.

So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them.

In other words, you don't subtract the magnitudes of the vectors, you subtract the vectors.

This means you find the vector that you add to $\vec{v_1}$ to get $\vec{v_2}$.

 

[renobueno4153]

In other words, you don't subtract the magnitudes of the vectors, you subtract the vectors.

This means you find the vector that you add to $\vec{v_1}$ to get $\vec{v_2}$.

Ohhh I think I see where was confused. I think it was my mathematical reasoning was flawed and that was reason i didnt understand it!

 

[renobueno4153]

Thanks for all of your time!

 

[renobueno4153]

In other words, you don't subtract the magnitudes of the vectors, you subtract the vectors.

This means you find the vector that you add to $\vec{v_1}$ to get $\vec{v_2}$.

Wait but why do I know the 500km/h and 800km/h are their magnitude and not their vector components?

 

[wrobel]

Some questions merely disappear during systematically studying the subject.

 

[PeroK]

Wait but why do I know the 500km/h and 800km/h are their magnitude and not their vector components?

Those are speeds. Speed, by definition, is the magnitude of a velocity. The original statement of the problem is sloppy in using "velocity", when it means speed.

To specify a vector component, you would need to specify the direction or coordinate system involved.

The original problem gives the speeds and the angle between the velocities. This allows you to choose any useful coordinate system. You would then have vector components relating to your chosen coordinate system.

Alternatively, you can solve the problem geometrically without the need for any specific coordinate system. This is what you did in the original post.

 

[PeroK]

Wait but why do I know the 500km/h and 800km/h are their magnitude and not their vector components?

Let me show you the alternative way to solve this problem, using vector components.

We'll let $\vec v_1$ be along the positive x-axis. Note that this is different from your diagram. And, we let $\vec v_2$ be at an angle of $\theta = 30^{\circ}$ (anticlockwise from the x-axis). The vector components (in km/h) are:
$$\vec v_1 = (500, 0); \ \vec v_2 = (800\cos \theta, 800\sin \theta) = (693, 400)$$Now, relative to $\vec v_1$, the components of $\vec v_2$ are:
$$\vec v_{21} = (193, 400)$$Which has a magnitude of $444 km/h$.

We can also calculate the angle, using the chosen coordinate system:
$$\tan(\theta_{21}) = \frac{400}{193}$$$$\theta_{21} = 64^{\circ}$$To think about this physically, we can imagine that both planes travel at theses constant velocities for an hour. During that hour, the second plane has moved, relative to the first, by a displacement of $444km$ in the given direction. That is the definition of relative velocity.

 

[Herman Trivilino]

Wait but why do I know the 500km/h and 800km/h are their magnitude and not their vector components?

The typical notation used in a physics class is that, for example $\vec{v}$ is a vector and $v$ is the magnitude of that vector.

So as @PeroK said the problem is poorly worded but the meaning is clear from the context to an experienced problem solver.