Difference: mean velocity & modulus of mean velocity vector

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

[/B]
A)

Mean velocity is defined as <v> = total distance traveled/ total time taken = πR/Γ = 0.5 m/s

B) How is part a) different from part b)?

I think what Irodov means by mean velocity is mean speed in part a.
It is mean speed which is defined as <v> = total distance traveled/ total time taken .

In part b,
<v> is defined as <v> = total displacement traveled/ total time taken in the direction of total displacement
|<v>| = R/Γ = 0.32 m/s

Have I understood the difference between mean velocity and modulus of mean velocity vector properly?

How to calculate part c?

 

[scottdave]

I am not sure about this terminology, either. But if you go halfway around the circle, your displacement will be the diameter of the circle, not the radius.

 

[jbriggs444]

I think what Irodov means by mean velocity is mean speed in part a.

I think he means what he says -- mean velocity. Which is a vector.
Part B is then trivial -- take the magnitude of the vector.
Part C...

The question implies that it does not matter what tangential acceleration is as long as it is constant. So cheat and assume that tangential acceleration is zero.

 

[Pushoam]

But if you go halfway around the circle, your displacement will be the diameter of the circle, not the radius.

Yes, I calculated that way , but didn't write 2 before R. Thanks for pointing it out.
So, |<v>| = 2R/Γ = 0.32 m/s

 

[haruspex]

Mean velocity is defined as <v> = total distance traveled/ total time taken

I'm unsure whether you are saying that is how you believe it is defined or how Irodov defines it.
Distance traveled is generally taken to be the integral of a scalar, $\int |\vec {ds}|$. Dividing that by $\Delta t$ gives average speed.
The integral $\int\vec{ds}$ gives the displacement, a vector. Dividing that by $\Delta t$ gives the average velocity.

 

[Pushoam]

I'm unsure whether you are saying that is how you believe it is defined or how Irodov defines it.

This is how I believe it is defined.

But, even if I take
<v> = $\frac {\int |\vec {ds}|} {Δt}$,

How will I calculate $\int |\vec {ds}|$ ?

I have to take $\int |\vec {ds}|$ as total distance traveled in the time Γ.

So, how does this formulation make a difference?

 

[haruspex]

This is how I believe it is defined.

Then you are wrong.
The question asks for average velocity, which is $<\vec v>=\frac {\int \vec {ds}} {Δt} = \frac {\vec r(t_0+\Delta t)-\vec r(t_0)}{\Delta t}$, where $vec r = \vec r(t)$ is the vector position. $\vec r(t_0+\Delta t)-\vec r(t_0)$ is the displacement.
What is the displacement in this case?

But, even if I take
<v> = $\frac {\int |\vec {ds}|} {Δt}$

As I wrote, that is average speed. In the present case, $\int |\vec {ds}|=\pi R$.

B) How is part a) different from part b)?

It is a bit odd, but part b seems to be asking for the magnitude of the vector found in part a.

 

[Pushoam]

Then you are wrong.
The question asks for average velocity,

I am using Irodov's book.
And I feel that by the term" mean velocity", Irodov means "mean speed "and by the term "mean velocity vector", he means "mean velocity".
You can read the question for reference.
Assuming that the mean velocity is mean speed, Is what I wrote correct?

 

[haruspex]

I feel that by the term" mean velocity", Irodov means "mean speed "

Ah, I see. He writes v for $|\vec v|$ and v for $\vec v$. Then yes, qn a is asking for mean speed and your answer is correct.

 

[Pushoam]

For part c,

Taking the constant tangent acceleration to be 0,⇒ speed is constant ⇒ mean speed = instantaneous speed

<w> =( v_f - v_i )/ Δt
= (v$\hat y$ - v$(-\hat y$))/10 = 2*0.5/10 = 0.1 m/s² $\hat y$
| <w>| = 0.1 m/s²

But, how to show that this is true for any constant tangent acceleration?

 

[haruspex]

Taking the constant tangent acceleration to be 0,⇒ speed is constant ⇒ mean speed = instantaneous speed

I suspect the question intends that it starts from rest, so the tangential acceleration is not zero.

 

[Pushoam]

I suspect the question intends that it starts from rest, so the tangential acceleration is not zero.

It is not said so in the question.
Nonetheless, I will try to solve it for non-zero constant tangent acceleration.

 

[Pushoam]

Taking the constant tangent acceleration to be c.
speed v = ct + v_i
<v> = (1/τ)∫₀^τ(ct + v_i)dt
= (cτ/2) + v_i) = 0.5 m/s
<w> =( v_f - v_i )/ Δt
= [ (ct + v_i)$\hat y$ - v_i(- $\hat y$)]/t

Taking v_i =0,
cτ/2 = 0.5 m/s
So,| <w> |= c = 0.1 m/s²

How to solve it with non - zero v_i?

<v> = (cτ/2) + v_i = 0.5 m/s
2 v_i = 0.1 - cτ
|<w> | = [ ct + 2 v_i]/τ = 0.1 m/s²Is this correct?

 

[haruspex]

I suspect the question intends that it starts from rest, so the tangential acceleration is not zero.

Thinking about it some more, that assumption is unnecessary.

Taking the constant tangent acceleration to be c.
speed v = ct + v_i
<v> = (1/τ)∫₀^τ(ct + v_i)dt
= (cτ/2) + v_i) = 0.5 m/s
<w> =( v_f - v_i )/ Δt
= [ (ct + v_i)$\hat y$ - v_i(- $\hat y$)]/t

Taking v_i =0,
cτ/2 = 0.5 m/s
So,| <w> |= c = 0.1 m/s²

How to solve it with non - zero v_i?

<v> = (cτ/2) + v_i = 0.5 m/s
2 v_i = 0.1 - cτ
|<w> | = [ ct + 2 v_i]/τ = 0.1 m/s²Is this correct?

Your method has found the constant tangential acceleration, but that is a scalar. The question (this time) clearly defines the acceleration sought as a vector.
Suppose the initial speed is v_i in the positive x direction. What is the final speed, and in what direction? What is the change in velocity?

 

[Pushoam]

Your method has found the constant tangential acceleration, but that is a scalar. The question (this time) clearly defines the acceleration sought as a vector.

The question asks for constant tangent acceleration.
I am using Irodov's book.
And I feel that by the term"constant tangent acceleration ", Irodov means "magnitude of constant tangent acceleration "and by the term "constant tangent acceleration vector", he means "constant tangent acceleration".
You can read the question for reference.
Assuming that the constant tangent acceleration is magnitude of constant tangent acceleration, Is what I wrote correct?

Please, see post #8 and 9.
Even if you take constant tangent acceleration to be a vector, then this vector can't be constant as tangential direction itself goes on changing.
So, by constant tangent acceleration, what Irodov means is the magnitude of the tangent acceleration is constant.

Because of this acceleration, the speed will go on changing.
So, I have to calculate final speed.
Now, the direction of final velocity is opposite to the direction of the initial velocity. This is what I have donein #13.

Suppose the initial speed is vi in the positive x direction. What is the final speed, and in what direction? What is the change in velocity?

I have calculated it.

<w> =( v_f - v_i )/ Δt
= [ (ct + v_i)$\hat y$ - v_i(- $\hat y$)]/t

 

[haruspex]

The question asks for constant tangent acceleration.

No it doesn't.
It tells you the tangent acceleration, $\frac{d|\vec v|}{dt}$, is constant, but asks for modulus of the average total acceleration, $|\frac{\vec{\Delta v}}{\Delta t}|$

<w> =( vf - vi )/ Δt
$= [ (ct + v_i)\hat y - v_i(- \hat y)]/t$

Ok, but the distance and time information you are given creates a relationship between v_i and c. Try eliminating one from that expression.

 

[Pushoam]

No it doesn't.

Sorry, what I meant by

The question asks for constant tangent acceleration.

is that the question tells to take constant tangent acceleration.

but asks for modulus of the average total acceleration, |$\frac{\vec{\Delta v}}{\Delta t}$|

This is what I have calculated.
Isn't |$\frac{\vec{\Delta v}}{\Delta t}$| = <w> =( v_f - v_i )/ Δt
= [ (ct + v_i)$\hat y$ - v_i(- $\hat y$) ]/t, Here t corresponds to Δt?

 

[haruspex]

Sorry, what I meant by

is that the question tells to take constant tangent acceleration.

This is what I have calculated.
Isn't |$\frac{\vec{\Delta v}}{\Delta t}$| = <w> =( v_f - v_i )/ Δt
= [ (ct + v_i)$\hat y$ - v_i(- $\hat y$) ]/t, Here t corresponds to Δt?

Sorry, I edited my previous post in parallel with your last post. Please see my edit.

 

[Pushoam]

Ok, but the distance and time information you are given creates a relationship between vi and c. Try eliminating one from that expression.

I have done it in #13.

<v> = (cτ/2) + v_i = 0.5 m/s
2 v_i = 0.1 - cτ
|<w> | = [ ct + 2 v_i]/τ = 0.1 m/s²

 

[haruspex]

I have done it in #13.

I'm very sorry, I didn't read it carefully enough.
I saw

How to solve it with non - zero vi?

and mistakenly assumed you had not found a way.
Yes, your answer in post #13 is fine.