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The brake mechanism used to reduce recoil

  1. May 27, 2017 #1
    1. The problem statement, all variables and given/known data
    The brake mechanism used to reduce recoil in certain types of guns consists essentially of a piston which is attached to the barrel and may move in a fixed cylinder filled with oil as the barrel recoils with an initial velocity v0, the piston moves and oils is forced through orifices in the piston, causing the piston and the barrel to decelerate at a rate proportional to their velocity; that is, ##a=-kv##. Express v in terms of t, x in terms of t, v in terms of x. Draw the corresponding motion curves.

    2. Relevant equations
    ##\ \displaystyle a=\frac{dv}{dt} \ ##
    ##\ \displaystyle v=\frac{ds}{dt}##

    3. The attempt at a solution

    Substituting -kv for a in the fundamental formula defining acceleration, a=dv/dt, we write
    ##\ \displaystyle -kv=\frac{dv}{dt} \ ## ; ##\ \displaystyle \frac {dv}{dt}=-kdt##

    ##\ \displaystyle \ int_v_0^v \, dx = \left. -k \int_0^t \, dt \right \ ##

    ## \ \displaystyle ln\fracv_{v_0}=-kt ##

    I have made mistakes when doing latex. So I paste official solution. I understand most of it but the confusing part is that when integrating why the author did not use integration constant?

    v in terms of t.png

    Source: Vector Mechanics for Engineers by Beer/Johnston.

    Thank you.
     
  2. jcsd
  3. May 27, 2017 #2

    BvU

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    He did use one: ##v_0## and the other, ##t_0## is zero (the moment of firing)

    Another way to look at this: if F is a primitive of f, then ##\int_a^b f = F(b) - F(a)## (no integration constants: they cancel)

    \ \displaystyle \int_{v_0}^v \, dx = \left. -k \int_0^t \, dt \right \
    $$ \ \displaystyle \int_{v_0}^v \, dx = -k \int_0^t \, dt $$

    \fracv --> \frac v
     
  4. May 27, 2017 #3
    I have forgotten some rules of integration. Isn't there any integration constant in definite integral?

    Thank you.
     
  5. May 27, 2017 #4
    But there are left and right in this example. This is confusing for me.

    left and right.png

    Thank you.
     
  6. May 27, 2017 #5

    BvU

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    Ah, the \right is a vertical line ## \ \ ## | ##\ \ ## (I think it's called a strut) , not a backslash. The \left . is then a kind of dummy.

    And this ##\TeX## loses track when two underscores are parsed, so the { } are needed -- don't know if that's universal.

    $$\ \displaystyle \int_{v_0}^v \, dx = \left. -k \int_0^t \, dt \right |_1^3 $$

    $$F'(x) = f(x) \Rightarrow \int^x f(u) \, du = F(x) + C \ \ {\rm and }\ \left . \int^b_a f(x) \, dx = F \right |^b_a = F(b)-F(a) $$
     
  7. May 27, 2017 #6
    Because definite integral would be a number which is the diffference of two values of a function and these two values have the same constant, these constants will cancel out, right?

    Thank you.
     
  8. May 28, 2017 #7
    Yes, it seems that the last backslash in \ \ displaystyle is overwritten. So why do some people use it?

    Thank you.
     
  9. May 28, 2017 #8

    BvU

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    No: ##TeX## sees a right without an argument such as ], ) or |.
    The backslash is not an argument but a sign for ##TeX## to parse the next character(s) as keywords
     
  10. May 28, 2017 #9
  11. May 28, 2017 #10

    BvU

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    \left and \right are grouping characters for ##TeX##. It needs to know where a group begins and ends (e.g. for vertical sizing), so it can't have unpaired instances. But if you want a single delimiter you can pair with \left . or \right . : the dot gives you an invisible delimiter.
    (it doesn't do anything, it acts as a dummy, a placeholder).
     
  12. May 30, 2017 #11
    I am too amateurish in this topic that when you mentioned TeX I am quite puzzled. Would you explain what the relation between TeX and LaTeX?

    Thank you.
     
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