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Difference: mean velocity & modulus of mean velocity vector

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  1. Jul 18, 2017 #1
    1. The problem statement, all variables and given/known data

    upload_2017-7-18_21-33-34.png

    2. Relevant equations


    3. The attempt at a solution

    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?
     

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  3. Jul 18, 2017 #2

    scottdave

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    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.
     
  4. Jul 18, 2017 #3

    jbriggs444

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    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.
     
  5. Jul 18, 2017 #4
    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
     
  6. Jul 19, 2017 #5

    haruspex

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    I'm unsure whether you are saying that is how you believe it is defined or how Irodov defines it.
    Distance travelled 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.
     
  7. Jul 19, 2017 #6
    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?
     
  8. Jul 19, 2017 #7

    haruspex

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    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?
    As I wrote, that is average speed. In the present case, ##\int |\vec {ds}|=\pi R##.
    It is a bit odd, but part b seems to be asking for the magnitude of the vector found in part a.
     
  9. Jul 19, 2017 #8
    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?
     
  10. Jul 19, 2017 #9

    haruspex

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    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.
     
  11. Jul 19, 2017 #10
    For part c,

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

    <w> =( vf - vi )/ 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?
     
    Last edited: Jul 19, 2017
  12. Jul 19, 2017 #11

    haruspex

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    I suspect the question intends that it starts from rest, so the tangential acceleration is not zero.
     
  13. Jul 19, 2017 #12
    It is not said so in the question.
    Nonetheless, I will try to solve it for non-zero constant tangent acceleration.
     
  14. Jul 19, 2017 #13
    Taking the constant tangent acceleration to be c.
    speed v = ct + vi
    <v> = (1/τ)∫0τ(ct + vi)dt
    = (cτ/2) + vi) = 0.5 m/s
    <w> =( vf - vi )/ Δt
    = [ (ct + vi)##\hat y## - vi(- ##\hat y##)]/t

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

    How to solve it with non - zero vi?

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


    Is this correct?
     
  15. Jul 19, 2017 #14

    haruspex

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    Thinking about it some more, that assumption is unnecessary.
    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 vi in the positive x direction. What is the final speed, and in what direction? What is the change in velocity?
     
  16. Jul 19, 2017 #15
    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.
    I have calculated it.
     
  17. Jul 19, 2017 #16

    haruspex

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    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}|##

    Ok, but the distance and time information you are given creates a relationship between vi and c. Try eliminating one from that expression.
     
  18. Jul 19, 2017 #17
    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> =( vf - vi )/ Δt
    = [ (ct + vi)##\hat y ## - vi(- ##\hat y##) ]/t, Here t corresponds to Δt?
     
  19. Jul 19, 2017 #18

    haruspex

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    Sorry, I edited my previous post in parallel with your last post. Please see my edit.
     
  20. Jul 20, 2017 #19
    I have done it in #13.
     
  21. Jul 20, 2017 #20

    haruspex

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    I'm very sorry, I didn't read it carefully enough.
    I saw
    and mistakenly assumed you had not found a way.
    Yes, your answer in post #13 is fine.
     
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