Why is the velocity (vector) the same value as the speed (scalar)?

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  • #1
ggandy
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Why is the velocity(vector) the same value with the speed(scalar).png
 

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  • #2
anuttarasammyak
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Summary:: Why is the velocity(vector) the same value with the speed(scalar) in a parabolic motion?
I think the displacement is smaller than the distance, so the velocity must be smaller than the speed.

In OP's case of
[tex]x=v_xt[/tex]
[tex]y=-\frac{1}{2}gt^2[/tex]
Distance =
[tex]\int_0^T \sqrt{dx^2+dy^2} = \int_0^T \sqrt{v_x^2+g^2t^2} dt[/tex]
Displacement =
[tex](v_xT, -\frac{1}{2}gT^2)[/tex]
From your sketch it is obvious Distance > Displacement that
[tex]\int_0^T \sqrt{v_x^2+g^2t^2} dt > \sqrt{v_x^2T^2+\frac{1}{4}g^2T^4}[/tex]

speed = d Distance / dT =
[tex]\sqrt{v_x^2+g^2T^2}[/tex]
velocity = (d Dispacement_x/dT, d Dispacement_y /dT)=
[tex](v_x, -gT)[/tex]

We see speed equals to magnitude of velocity.

In general, infinitesimal Distance = magnitude of infinitesimal Displacement
[tex] dl = \sqrt{dx^2+dy^2}[/tex]
[tex] \frac{dl}{dt} = \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}[/tex]
 
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  • #3
Doc Al
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Be careful not to confuse average velocity with instantaneous velocity.
 
  • #4
ggandy
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Be careful not to confuse average velocity with instantaneous velocity.
Then in my sketch, does the velocity means instantaneous velocity?
 
  • #5
Delta2
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Then in my sketch, does the velocity means instantaneous velocity?
The calculation for ##\vec{v}## (velocity) is for the instantaneous velocity yes. The magnitude of the instantaneous velocity is always equal to the instantaneous speed.

However your assessment that "The displacement is smaller than the distance, hence the velocity should be smaller than the speed" is correct only if you replace the word "velocity" with "magnitude of the average velocity" and the word speed with "average speed". It isn't correct for the magnitude of the instantaneous velocity and the instantaneous speed which are always equal as i said in the previous paragraph.
 
  • #6
ggandy
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The calculation for ##\vec{v}## (velocity) is for the instantaneous velocity yes. The magnitude of the instantaneous velocity is always equal to the instantaneous speed.

However your assessment that "The displacement is smaller than the distance, hence the velocity should be smaller than the speed" is correct only if you replace the word "velocity" with "magnitude of the average velocity" and the word speed with "average speed". It isn't correct for the magnitude of the instantaneous velocity and the instantaneous speed which are always equal as i said in the previous paragraph.
Then in my sketch, the value of the velocity from the blue dotted line means the magnitude of the average velocity?
And the value of the speed from the red dotted line means the average speed?
Hence the magnitude of the average velocity is smaller than the average speed?
 
  • #7
Doc Al
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Then in my sketch, the value of the velocity from the blue dotted line means the magnitude of the average velocity?
The blue line shows the displacement, which you can use to calculate the average velocity.

And the value of the speed from the red dotted line means the average speed?
Not sure what you mean. The value that you calculated is the instantaneous velocity after one second.

Hence the magnitude of the average velocity is smaller than the average speed?
To calculate the average speed of the projectile you'd have to find the total distance traveled (distance, not displacement). That's what you are looking to compare with the average velocity.
 
  • #8
ggandy
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The blue line shows the displacement, which you can use to calculate the average velocity.


Not sure what you mean. The value that you calculated is the instantaneous velocity after one second.


To calculate the average speed of the projectile you'd have to find the total distance traveled (distance, not displacement). That's what you are looking to compare with the average velocity.
Then if I calculate the average speed from the total distance traveled, would it be bigger than the average velocity?
 
  • #9
Delta2
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Then if I calculate the average speed from the total distance traveled, would it be bigger than the average velocity?
Yes. In your sketch you calculate the instantaneous velocity and instantaneous speed after 1 second. If you calculate the average velocity and the average speed for the duration of this 1second you ll find that the average speed is greater than the magnitude of the average velocity.
 
  • #10
anuttarasammyak
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Then if I calculate the average speed from the total distance traveled, would it be bigger than the average velocity?

As a clear and extreme case, say a body starts from the origin at time 0 and go around a path and come back to the origin at time T.
There is no displacement so Displacement / T = 0 < Length of path / T

Then let us do Galilean (or Lorentz for SR) transformation so that
starting point x=0, time=0
goal point x=X, time =T
Displacement is a straight line meaning inertial, constant velocity motion. Path of the body are curves. The line is the shortest one among the curves. So in this new frame also
Displacement < Length of path (=Distance though I myself am not accustomed to saying so)
Displacement /T < Length of path/T (=average speed)


About the terminology, I confess my coarse usage was :
Path length is the length that a body moved in some way = Distance here
Distance is the length of line connecting the two points. The shortest distance.= Displacement here
ref.
"A circle is a shape consisting of all points in a plane that are a given distance from a given point, the centre;" In this Wiki definition of a circle, word distance is used, not displacement.
Displacement is rather small change or difference of molecule coordinates in vibration or elasticity in many cases a body would come back to its original position.
 
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  • #11
ggandy
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Yes. In your sketch you calculate the instantaneous velocity and instantaneous speed after 1 second. If you calculate the average velocity and the average speed for the duration of this 1second you ll find that the average speed is greater than the magnitude of the average velocity.
Thanks a lot. Now I see that I just calculated only the instantaneous speed and velocity.
 
  • #12
ggandy
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As a clear and extreme case, say a body starts from the origin at time 0 and go around a path and come back to the origin at time T.
There is no displacement so Displacement / T = 0 < Length of path / T

Then let us do Galilean (or Lorentz for SR) transformation so that
starting point x=0, time=0
goal point x=X, time =T
Displacement is a straight line meaning inertial, constant velocity motion. Path of the body are curves. The line is the shortest one among the curves. So in this new frame also
Displacement < Length of path (=Distance though I myself am not accustomed to saying so)
Displacement /T < Length of path/T (=average speed)


About the terminology, I confess my coarse usage was :
Path length is the length that a body moved in some way = Distance here
Distance is the length of line connecting the two points. The shortest distance.= Displacement here
ref.
A circle is a shape consisting of all points in a plane that are a given distance from a given point, the centre;
Here distance is used, not displacement.

Displacement is rather small change or difference of coordinates in vibration or elasticity in many cases a body would come back.
Now I think that the straight line is not the real motion in gravity, the curve line is the real motion in gravity.
In my sketch the values of the speed and velocity are just instantaneous values.
So If I calculate the lengths of the straight line and curve line, I would got the results of the average speed and velocity. Then I would know that the average speed is greater than the average velocity.
 
  • #13
anuttarasammyak
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Now I think that the straight line is not the real motion in gravity, the curve line is the real motion in gravity.
In my sketch the values of the speed and velocity are just instantaneous values.

For peace of your mind you may think that in infinitesimal time all the motions are line motions where speed and (magnitude of ) velocity coincide.

In post #10 motions of real, natural or artificial, e.g. grab and move, does not matter. But I should not have referred to Galilean or Lorentz transformation under gravity that would cause interesting but missing the point of the thread discussions.
 
  • #14
ggandy
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For peace of your mind you may think that in infinitesimal time all the motions are line motions where speed and (magnitude of ) velocity coincide.

In post #10 motions of real, natural or artificial, e.g. grab and move, does not matter. But I should not have referred to Galilean or Lorentz transformation under gravity that would cause interesting but missing the point of the thread discussions.
I appreciate your kind explanation.
 
  • #15
Doc Al
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Then if I calculate the average speed from the total distance traveled, would it be bigger than the average velocity?
Sure. I was going to give the same example that @anuttarasammyak gave as an extreme case: Imagine a car driving around a racetrack at 100 mph and returning to its starting point: The average speed was 100 mph; the average velocity was zero.

Another example: Toss a ball straight up and catch when it comes down.
 
  • #16
Delta2
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Come to think of it, from a purely mathematical viewpoint this thread is about the inequality
$$\frac{1}{T}\left \|\int_0^T \mathbf{v(t)}dt\right\|\leq\frac{1}{T}\int_0^T\left\|\mathbf{v(t)}\right\|dt$$ which is just a property of integration and the modulo (norm) ##\|\mathbf{v}\|## of a vector , which is a generalization of the property
$$|\int f(x)dx|\leq\int|f(x)|dx$$ for real valued functions of a real variable.
 
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  • #17
Leo Liu
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Speed is just the magnitude of the velocity vector. In this case the displacement is less than the displacement because some components of the velocity vectors throughout the motion could cancel out one another.

Bonus: In a circular motion, the displacement is essentially 0, as the distance traveled is not 0. In this case both the velocity and the speed is not 0 or ##\left\langle 0, 0 \right\rangle##bcause the velocity changes direction and the speed is constant.
 
  • #18
ggandy
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Sure. I was going to give the same example that @anuttarasammyak gave as an extreme case: Imagine a car driving around a racetrack at 100 mph and returning to its starting point: The average speed was 100 mph; the average velocity was zero.

Another example: Toss a ball straight up and catch when it comes down.
Thank you for the good examples.
 
  • #19
ggandy
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Come to think of it, from a purely mathematical viewpoint this thread is about the inequality
$$\frac{1}{T}\left \|\int_0^T \mathbf{v(t)}dt\right\|\leq\frac{1}{T}\int_0^T\left\|\mathbf{v(t)}\right\|dt$$ which is just a property of integration and the modulo (norm) ##\|\mathbf{v}\|## of a vector , which is a generalization of the property
$$|\int f(x)dx|\leq\int|f(x)|dx$$ for real valued functions of a real variable.
Thanks a lot.
 
  • #20
ggandy
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Speed is just the magnitude of the velocity vector. In this case the displacement is less than the displacement because some components of the velocity vectors throughout the motion could cancel out one another.

Bonus: In a circular motion, the displacement is essentially 0, as the distance traveled is not 0. In this case both the velocity and the speed is not 0 or ##\left\langle 0, 0 \right\rangle##bcause the velocity changes direction and the speed is constant.
I'm sorry I can't see what you mean.
In your second line, " cancel out one another"; What are the components that are canceled out?
In your last line, "speed is constant"; Why does the speed constant in gravity?
 
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  • #21
Leo Liu
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In your second line, " cancel out one another"; What are the components that are canceled out?
Sorry if that wan't clear. Every infinitesimal displacement vector can be added together, but they have directions which make the total displacement less than or equal to the distance.

If you want some intuition, you can imagine a triangle with three sides a, b, c, in which side a and b are infinitesimal displacement (##\lim_{x,y\to 0} b(x,y)## ##\lim_{x,y\to 0} a(x,y)##, where x and y are the components of the vectors). The side c is a resultant vector of a and b, denoting the total displacement. So we can write ##\lim_{x,y\to 0} c(x,y) = \lim_{x,y\to 0} b(x,y) + \lim_{x,y\to 0} a(x,y)## (again, vectors can be added together). In comparison, the tiny distance traveled is sum of the length of a and the length of b. An axiom of geometry tells us that the length of two sides in a triangle added together must be greater than the length of the other side; therefore, the length of the displacement vector c must be smaller than the distance which is the length of a and the length of b added together.

Now back to your question--why does the length of the velocity vector equal value the speed? It is because both of them are instantaneous (not average) in this case. So it is not legitimate to divide (not rigorously saying) c by a unit of time to get the instantaneous rate of change, otherwise you are calculating the average velocity in this tiny time interval. If we think a bit further, the speed at a point is equal to the tangent of the distance function (the change in the length of vector, which is b-a if we assume the object travels from a to b), which is also equal to the length of the velocity vector. The displacement vectors merely give the velocity vector a direction.

I hope this helps.
In your last line, "speed is constant"; Why is the speed constant in gravity?
"In Circular Motion (Bonus)"
 
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  • #22
Delta2
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therefore, the length of the displacement vector c must be greater than the distance which is the length of a and the length of b added together.
You must mean smaller there not greater. You are right that the triangular inequality is behind all this. The proof of inequality in my post #16 is based on triangular inequality of the modulus (norm) of a vector $$ \|\mathbf{v_1}+\mathbf{v_2}\|\leq\|\mathbf{v_1}\|+\|\mathbf{v_2}\|$$
 
  • #23
ggandy
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Sorry if that wan't clear. Every infinitesimal displacement vector can be added together, but they have directions which make the total displacement less than or equal to the distance.

If you want some intuition, you can imagine a triangle with three sides a, b, c, in which side a and b are infinitesimal displacement (##\lim_{x,y\to 0} b(x,y)## ##\lim_{x,y\to 0} a(x,y)##, where x and y are the components of the vectors). The side c is a resultant vector of a and b, denoting the total displacement. So we can write ##\lim_{x,y\to 0} c(x,y) = \lim_{x,y\to 0} b(x,y) + \lim_{x,y\to 0} a(x,y)## (again, vectors can be added together). In comparison, the tiny distance traveled is sum of the length of a and the length of b. An axiom of geometry tells us that the length of two sides in a triangle added together must be greater than the length of the other side; therefore, the length of the displacement vector c must be smaller than the distance which is the length of a and the length of b added together.

Now back to your question--why does the length of the velocity vector equal value the speed? It is because both of them are instantaneous (not average) in this case. So it is not legitimate to divide (not rigorously saying) c by a unit of time to get the instantaneous rate of change, otherwise you are calculating the average velocity in this tiny time interval. If we think a bit further, the speed at a point is equal to the tangent of the distance function (the change in the length of vector, which is b-a if we assume the object travels from a to b), which is also equal to the length of the velocity vector. The displacement vectors merely give the velocity vector a direction.

I hope this helps.

"In Circular Motion (Bonus)"
Thank you for your kind explanation, but I'm sorry I can't still understand your added explanation. Maybe the reason why I can't understand is it just consists of texts without images. I'm going to try again to understand that, thank you.
 
  • #24
Leo Liu
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Thank you for your kind explanation, but I'm sorry I can't still understand your added explanation. Maybe the reason why I can't understand is it just consists of texts without images. I'm going to try again to understand that, thank you.
1593478297693.png

If you zoom in on the trajectory of the object, it will become many some line segments. The diagram is simply a representation of a piece of the trajectory. Since the motion has direction, we just consider them as vectors.
 

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