Expressing Velocity in Vector Form

[opus]

Let me start off by stating a given problem:

A baseball is hit, and leaves the bat at a speed of 100 mph and at an angle of 20° from the horizontal. Express this velocity in vector form.

So we're given the velocity and the angle at which the ball is hit. The speed corresponds to the vector's magnitude, and the angle corresponds to the vector's direction. To get this into vector form we can use the fact that:
$$u=\left<\left\|u\right\|cos(θ), \left\|u\right\|sin(θ)\right>$$
$$=\left\|u\right\|\left<cos(θ),sin(θ)\right>$$

So in this case,

$$=\left\|u\right\|\left<cos(θ),sin(θ)\right>$$
$$=100\left<cos(20°),sin(20°)\right>$$
$$≈\left<93.97,34.20\right>$$

Now here's my question:
These components don't seem to tell nearly as much as the initial problem stated. In looking at the components, you can't tell the velocity of the ball, or the angle at which it was hit. We can just see the horizontal and vertical displacement. So what's the point of getting anything into vector form like this?

 

[Delta2]

Allow me to say that it all depends on how your brain is used to read and interpret information. If you are used to read and interpret the velocity with the magnitude and the angle with the horizontal x-axis then fine. But you might as well used to read velocity as the x and y component and in many problems we need to know the x and y components of the velocity and not the magnitude neither the angle.

And something else important (the way you stated your question seems that you are wrong on this) ,from the x and y component you can find both the magnitude of velocity and the angle with the horizontal. it will be $||v||=\sqrt{v_x^2+v_y^2}$ and $\tan\theta=\frac{v_y}{v_x}$, so the vector form and the (magnitude,angle) form are absolutely equivalent

 

[opus]

Interesting! I tried those formulae out with the vector components and they were extremely close to the ones in the given problem (probably a little off due to rounding).
So is the "magnitude of velocity" the same thing as speed then? Because in the given problem, the speed was 100 mph.
Also, why does it reverse the process like that? When I took $tan^{-1}θ=\frac{34.20}{93.97}≈20°$, how does it get the original angle back, considering we distributed 100 into $\left<cos(20°),sin(20°)\right>$? And I guess the same goes for finding the speed/magnitude of velocity as well.

 

[FactChecker]

It's very common to track motion in XY coordinates so that the current position is always available. Adding a series of velocity vectors in polar coordinates in not easy, but it is easy in XY coordinates. There are also several situations where the velocity is caused by forces that are most easily defined in Cartesian coordinates and only effect the X or Y velocity components (gravity, fixed springs, etc.). So it is very important to be able to switch back and forth between these two coordinate systems easily.

 

[opus]

Ah that makes a bit more sense. So say if it's vector components are $\left<5,4\right>$, and its velocity has a magnitude of say 5 feet/sec, then in another second it would be at $\left<10,9\right>$? Or is the velocity "built in" to the components, where after a second has passed it would actually be $\left<10,8\right>$?

 

[Delta2]

Interesting! I tried those formulae out with the vector components and they were extremely close to the ones in the given problem (probably a little off due to rounding).
So is the "magnitude of velocity" the same thing as speed then? Because in the given problem, the speed was 100 mph.
Also, why does it reverse the process like that? When I took $tan^{-1}θ=\frac{34.20}{93.97}≈20°$, how does it get the original angle back, considering we distributed 100 into $\left<cos(20°),sin(20°)\right>$? And I guess the same goes for finding the speed/magnitude of velocity as well.

Yes magnitude of velocity is commonly referred to as just "speed".
It is just some trigonometry math. In the OP you wrote the x and y components as $v_x=||v||\cos\theta$ , $v_y=||v||\sin\theta$ so taking
$\sqrt{v_x^2+v_y^2}=\sqrt{||v||^2(\cos^2\theta+sin^2\theta)}=\sqrt{||v||^2}=||v||$
and
$\frac{v_y}{v_x}=\frac{||v||\sin\theta}{||v||\cos\theta}=\frac{\sin\theta}{\cos\theta}=tan\theta$

 

[Delta2]

Ah that makes a bit more sense. So say if it's vector components are $\left<5,4\right>$, and its velocity has a magnitude of say 5 feet/sec, then in another second it would be at $\left<10,9\right>$? Or is the velocity "built in" to the components, where after a second has passed it would actually be $\left<10,8\right>$?

Not sure what you mean here, <5,4> is the position vector or the velocity vector? In my posts $v_x$ and $v_y$ are the x and y components of the velocity vector, they are not the same as the x and y components of the position vector (the latter are simply the x and y coordinates of the position of the particle).

 

[FactChecker]

Ah that makes a bit more sense. So say if it's vector components are $\left<5,4\right>$, and its velocity has a magnitude of say 5 feet/sec, then in another second it would be at $\left<10,9\right>$? Or is the velocity "built in" to the components

It's very important to keep everything well identified and organized using clear notation.
A velocity can be represented in $(x,y)_{XY}$ Cartesian XY coordinates or in $(r,\theta)_{polar}$ polar coordinates. Both methods represent the same velocity and do not represent the current position.
EXAMPLE:
Suppose at time 0 we have the velocity vector $v_0 = (5, 36.87^{\circ})_{polar}$ in polar coordinates. Then $v_0 = (4,3)_{XY}$ in XY cartesian coordinates. (I'm using the 3:4:5 right triangle as an easy velocity vector example.) The values of 3, 4, and 5 are in units of feet/second.
Suppose that the object with this velocity has a position at time 0 of $p_0 = (10,20)_{XY}$. Then after 2 seconds of motion $v_0$, its position will be at $p_2 = p_0 + 2*v_0 = (10,20)_{XY} + 2*(4,3)_{XY} = (18,26)_{XY}$
Suppose that motion is followed by one second of velocity $v_2 = (1,90^{\circ})_{polar} = (0,1)_{XY}$
Then the object's position at time 3 would be $p_3 = p_2 + v_2 = (18,26)_{XY} + (0,1)_{XY} = (18,27)_{XY}$
These types of absolute position calculations are easier to do in Cartesian coordinates. On the other hand, velocities are often directly measured in polar coordinates and final relative position answers are ofter desired in polar coordinates. So conversion between the two coordinate systems is common.

 

[opus]

Not sure what you mean here, <5,4> is the position vector or the velocity vector? In my posts $v_x$ and $v_y$ are the x and y components of the velocity vector, they are not the same as the x and y components of the position vector (the latter are simply the x and y coordinates of the position of the particle).

I didn't know that there was a difference. So we have separate vectors for position and velocity? Is this what is distinguished as $\left<x,y\right>$ vs $\left(x,y\right)$? Where the former is the displacement of the x and y coordinates (velocity), and the latter is the position of the coordinates (as a single point in space)?
To my understanding, velocity is displacement in a given direction. So in the case of $\left<x,y\right>$, this is saying that for every change in x, there is a change in y, which would be displacement in a certain direction if you took the magnitude of the hypotenuse, and the magnitude of that displacement would be speed? And I guess if you wanted to know the direction, you would need to just compare the two and see where the vector points?
I guess not having taken a Physics class yet isn't much help to me. If I'm jumping into the weeds in advance, please let me know.

It's very important to keep everything well identified and organized using clear notation.
A velocity can be represented in $(x,y)_{XY}$ Cartesian XY coordinates or in $(r,\theta)_{polar}$ polar coordinates. Both methods represent the same velocity and do not represent the current position.
EXAMPLE:
Suppose at time 0 we have the velocity vector $v_0 = (5, 36.87^{\circ})_{polar}$ in polar coordinates. Then $v_0 = (4,3)_{XY}$ in XY cartesian coordinates. (I'm using the 3:4:5 right triangle as an easy velocity vector example.) The values of 3, 4, and 5 are in units of feet/second.
Suppose that the object with this velocity has a position at time 0 of $p_0 = (10,20)_{XY}$. Then after 2 seconds of motion $v_0$, its position will be at $p_2 = p_0 + 2*v_0 = (10,20)_{XY} + 2*(4,3)_{XY} = (18,26)_{XY}$
Suppose that motion is followed by one second of velocity $v_2 = (1,90^{\circ})_{polar} = (0,1)_{XY}$
Then the object's position at time 3 would be $p_3 = p_2 + v_2 = (18,26)_{XY} + (0,1)_{XY} = (18,27)_{XY}$
These types of absolute position calculations are easier to do in Cartesian coordinates. On the other hand, velocities are often directly measured in polar coordinates and final relative position answers are ofter desired in polar coordinates. So conversion between the two coordinate systems is common.

So essentially, you have two vectors? You have a velocity vector and a position vector. At some initial starting position, $p_0$, you will take the position and add to it some multiple of the velocity (depending on the time) to get the new position?

 

[FactChecker]

Position, velocity, and acceleration are three different things, each of which can be represented by a vector. An arbitrary object can be in any position, $\vec p = (x_p,y_p)_{XY}$, and have any velocity, $\vec v = (x_v,y_v)_{XY}$. The values of $x_p$ and $x_v$ do not have to be the same at all. Likewise for $y_p$ and $y_v$. The same goes for the components of the acceleration vector, $\vec a = (x_a,y_a)_{XY}$. In fact, in any problem there can be many different vectors with different component values.

 

[Delta2]

We can use any of <x,y> or (x,y) notations, to denote a vector. We can also use the XY notation (as @FactChecker does) below and right of the parentheses (or of the <>) to denote that the vector is in cartesian coordinates. The exercise has to tell us which vector is which physical quantity like saying that the velocity vector is (10,9) and the position vector is (5,4).

Also velocity is displacement in a given direction per unit of time. That is we divide the displacement of a particle by the time it takes to do that displacement, in order to find the velocity.

 

[opus]

Thanks so much! It's starting to make more sense, I think I just need to use them in different problems to see how they work and how they can come in different forms. One example was a cat pushing a box off a table (stupid) but it seems that it had a similar flavor to it- a force vector for the gravity and a force vector for the cat's push. I just need to keep getting my hands dirty with this.

 

[opus]

I'd like to ask another question if that's alright. To keep the clutter minimized, I figured I'd just ask it in here rather than in another thread as it pertains to the same subject mostly.

If you'll look at the attached image, it is a representation of a 3-way tug-of-war, where three teams have ropes attached to a ring. I got this problem off of Khan Academy. The first picture is a representation of the word problem given.
Now according to the video, the team with the most magnitude in their direction wins. If you'd shift your attention to the lower picture, which I made up so you can visualize my question, this is where I'm a little confused.

In the diagram, it is clear the the green vector has the most magnitude out of the three. However, in looking at the blue and the pink, they are very close to the magnitude of green. Additionally, they are pretty much pulling in the same direction as each other. Now to me, it seems that although green has the highest magnitude, it wouldn't be able to overcome the somewhat combined magnitudes of the blue and the pink. So who wins here?

 

[fresh_42]

So who wins here?

The red line is the resulting force pointing upwards: green + violet + blue = red

 

[opus]

Apologies for the late response. Been juggling two classes so I've been jumping back and forth between the trig and algebra.

I'm not entirely sure what you mean here. So if the red is pointed upward, that would mean that green loses, even though it has the single highest magnitude?

 

[fresh_42]

All forces apply to the same point. To get the resulting force you have to add them. Vector addition is moving one at the end of the other and the result is from the old origin to the new end point. Here we have 9.8N up and 5 down, so a result of approximately 5N up shouldn't be a surprise. To calculate the exact value, you need some angles and perform a bit triangle trigonometry.

 

[CWatters]

The directions the teams are pulling in the two drawings don't seem to agree. Is the is the second one correct?

It's good practice to provide a link or post the whole problem word for word.

The solution is to add the three vectors together and work out the direction of the resultant. The direction of the resultant determines who wins, the magnitude determines how fast they win. The direction will depend on "which team has the most magnitude in their direction".

 

[CWatters]

Apologies for the late response. Been juggling two classes so I've been jumping back and forth between the trig and algebra.

I'm not entirely sure what you mean here. So if the red is pointed upward, that would mean that green loses, even though it has the single highest magnitude?

Yes.

 

[opus]

All forces apply to the same point.

By this, do you mean that all forces are being applied on the center ring?

To get the resulting force you have to add them. Vector addition is moving one at the end of the other and the result is from the old origin to the new end point. Here we have 9.8N up and 5 down, so a result of approximately 5N up shouldn't be a surprise. To calculate the exact value, you need some angles and perform a bit triangle trigonometry.

Ok I think I got this part. Because vectors are defined only by their magnitude and direction, so we can move them around in space as long as we don't change their magnitude and direction. So we move the force vectors around so that they are in a manner that can be added.

The directions the teams are pulling in the two drawings don't seem to agree. Is the is the second one correct?

So the first drawing is from the word problem. The second (lower) drawing is one I used to frame my general question.
In the first drawing, they are pulling in a fairly spaced apart manner. I made the second picture to show a situation where two vectors that are pulling in close directions are pulling against a vector of higher magnitude.
My train of thought was as such:

The green vector has magnitude of 5.0 kN. The other two vectors are pulling almost opposite of the green, but each have a lower magnitude of 4.9 kN. Since the green vector has the highest magnitude, it would win. But then I thought that the two smaller vectors are essentially pulling with each other, so their combined force would overcome green's.

It's good practice to provide a link or post the whole problem word for word.

The solution is to add the three vectors together and work out the direction of the resultant. The direction of the resultant determines who wins, the magnitude determines how fast they win. The direction will depend on "which team has the most magnitude in their direction".

Here is a link to the word problem, which I found on Khan Academy:


In the video, we solve for the magnitude of all three force vectors and vector b has the greatest and is said to win. I looked at this problem and asked "what if the vectors weren't so evenly spaced, and what if two were very close together? Would it still be true that the team with the highest magnitude wins?"

 

[fresh_42]

By this, do you mean that all forces are being applied on the center ring?

Yes. If they apply at different points, then it depends on the situation. A force applied on me and one on you won't add up at all. At different points on the same piece of steel there will be a force translation in the material, which e.g. could lead to bend it.

Ok I think I got this part. Because vectors are defined only by their magnitude and direction, so we can move them around in space as long as we don't change their magnitude and direction. So we move the force vectors around so that they are in a manner that can be added.

Correct.

So the first drawing is from the word problem. The second (lower) drawing is one I used to frame my general question.
In the first drawing, they are pulling in a fairly spaced apart manner. I made the second picture to show a situation where two vectors that are pulling in close directions are pulling against a vector of higher magnitude.
My train of thought was as such:

The green vector has magnitude of 5.0 kN. The other two vectors are pulling almost opposite of the green, but each have a lower magnitude of 4.9 kN. Since the green vector has the highest magnitude, it would win. But then I thought that the two smaller vectors are essentially pulling with each other, so their combined force would overcome green's.

Yes. Imagine three teams pulling on ropes. Which team would you place your money on?

Here is a link to the word problem, which I found on Khan Academy:


In the video, we solve for the magnitude of all three force vectors and vector b has the greatest and is said to win. I looked at this problem and asked "what if the vectors weren't so evenly spaced, and what if two were very close together? Would it still be true that the team with the highest magnitude wins?"

Well, do the trig computations. The highest magnitude will win, as long as you define the highest magnitude according to the direction properly. If you pull something heavy straight up, you will never move it horizontally, no matter how hard you pull it upwards: $\cos 90° = 0\,.$

 

[CWatters]

Ok I think I got this part. Because vectors are defined only by their magnitude and direction, so we can move them around in space as long as we don't change their magnitude and direction. So we move the force vectors around so that they are in a manner that can be added.

Sometimes referred to as the head to tail method of adding vectors.

So the first drawing is from the word problem. The second (lower) drawing is one I used to frame my general question.
In the first drawing, they are pulling in a fairly spaced apart manner. I made the second picture to show a situation where two vectors that are pulling in close directions are pulling against a vector of higher magnitude.
My train of thought was as such:

The green vector has magnitude of 5.0 kN. The other two vectors are pulling almost opposite of the green, but each have a lower magnitude of 4.9 kN. Since the green vector has the highest magnitude, it would win. But then I thought that the two smaller vectors are essentially pulling with each other, so their combined force would overcome green's.Here is a link to the word problem, which I found on Khan Academy:


In the video, we solve for the magnitude of all three force vectors and vector b has the greatest and is said to win. I looked at this problem and asked "what if the vectors weren't so evenly spaced, and what if two were very close together? Would it still be true that the team with the highest magnitude wins?"

It's only the team with the highest magnitude under the symetrical case. If you change the direction in which the teams are pulling but not the position of their winning zone then it's possible for a weak team to win. The general solution is to look at the direction of the resultant and see which win zone it points to.

 

[Chestermiller]

Were these homework problems?

 

[opus]

Were these homework problems?

No sir! Just questions I used as an example for my conceptual question.

Thanks everyone for the responses. Being very late on my responses due to two exams coming up and fighting off nasty infections from a bunch of spider bites I got. Still dissecting everyone’s posts to make better sense of it and will have my responses in not too long.

 

[opus]

Ok now that my finals are done, I've been able to come back to my side questions and I think I have a better understanding of this.
So if the three teams here are having their tug-of-war, and the case is symmetrical (which I will assume means that they have equal distance between each other), then the team with the highest magnitude wins. However, if the case is not symmetric, as shown in my drawing, then the angles are taken into account and it is possible for a team with a weaker pulling force to win. In this case, we'd want to add up all the force vectors (which include magnitude and direction), and this will tell us where the ring will go.
Is this correct?

 

[CWatters]

Yes. As I said above. The direction the ring takes is the vector sum (aka the resultant or net) of the three vectors applied by the teams.

This can be applied to both symmetric and non-symmetric cases - it's just that the symmetrical case is simpler to analyse.

Edit: Strictly speaking the resultant or net force give the direction in which the ring will accelerate. In this case the initial velocity is zero so its also the direction it will move, but in other problems that might not always be true.

 

[FactChecker]

In this case, we'd want to add up all the force vectors (which include magnitude and direction), and this will tell us where the ring will go.
Is this correct?

This is about the most fundamental fact in Newtonian physics. To determine the acceleration of the center of mass of an object, $\vec {A_{cm}}$, always do a vector addition of all the applied forces to get a total force vector, $\vec {F_{total}}$. Then $\vec {F_{total}}=m \vec {A_{cm}}$. Anything else (like symmetry) is secondary.