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Thread: deriving v^2=v^2 + 2ax with calculus

  1. #1

    deriving v^2=v^2 + 2ax with calculus

    In the lines below, where does the 2Vo come from? The 2Vo I speak of is to the right of the first equal sign.

    . . . . .[tex]v(t)^2\, =\, v_0^2\, +\, a^2(t\, -\, t_0)^2\, +\, 2v_0 a(t\, -\, t_0)[/tex]

    . . . . . . . . .[tex]=\, v_0^2\, +\, 2a\, \left(\frac{1}{2}a(t\, -\, t_0)^2\, +\, v_0 (t\, -\, t_0)\right)[/tex]

    . . . . . . . . .[tex]=\, v_0^2\, +\, 2a(x(t)\, -\, x_0)[/tex]

    Also, which calculus steps are required to transform 1/2at^2 into x? I am referring to the part of the equation to the right of the second equal sign for this second question.
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    Last edited by stapel; 02-02-2018 at 02:57 PM. Reason: Typing out the text in the graphic.

  2. #2
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by prepforcalc View Post
    In the lines below, where does the 2Vo come from? The 2Vo I speak of is to the right of the first equal sign.

    . . . . .[tex]v(t)^2\, =\, v_0^2\, +\, a^2(t\, -\, t_0)^2\, +\, 2v_0 a(t\, -\, t_0)[/tex]

    . . . . . . . . .[tex]=\, v_0^2\, +\, 2a\, \left(\frac{1}{2}a(t\, -\, t_0)^2\, +\, v_0 (t\, -\, t_0)\right)[/tex]

    . . . . . . . . .[tex]=\, v_0^2\, +\, 2a(x(t)\, -\, x_0)[/tex]
    It looks like they started with:

    v(t) = v0 + a(t - t0)

    and squared each side -- so v0 got squared.

    Symbol v0 probably represents an initial velocity in some exercise. Can you post the original?

    Quote Originally Posted by prepforcalc View Post
    … transform 1/2at^2 into x …
    That could be just a substitution. It might help, to know the context.
    Last edited by stapel; 02-02-2018 at 02:58 PM. Reason: Copying typed-out graphical content into reply.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  3. #3
    Quote Originally Posted by mmm4444bot View Post
    It looks like they started with:

    v(t) = v0 + a(t - t0)

    and squared each side -- so v0 got squared.

    In that case, where did the following come from? This is from the screenshot in my first (original) post and it is to the right of the first equal sign? I think what you say sounds plausible, but leaves this question. For anyone else reading this, I would still like to know the answers to the questions in the OP. The other possible starting equation might be, x=x+vt+1/2at^2, or something like that. Obviously, taking the derivative of that would lead into my OP quite nicely, but what I would most like to know are the answers to the questions in the original post.

    Screenshot 2018-02-01 at 6.15.35 AM.png

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    Quote Originally Posted by prepforcalc View Post
    In that case, where did the following come from? This is from the screenshot in my first (original) post and it is to the right of the first equal sign? I think what you say sounds plausible, but leaves this question. For anyone else reading this, I would still like to know the answers to the questions in the OP. The other possible starting equation might be, x=x+vt+1/2at^2, or something like that. Obviously, taking the derivative of that would lead into my OP quite nicely, but what I would most like to know are the answers to the questions in the original post.

    Screenshot 2018-02-01 at 6.15.35 AM.png
    v(t) = v0 + a(t - t0)

    square both sides

    [v(t)]2 = [v0 + a(t - t0)]2

    What do you get after simplifying the right-hand-side?
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

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    Quote Originally Posted by prepforcalc View Post
    In the lines below, where does the 2Vo come from? The 2Vo I speak of is to the right of the first equal sign.

    . . . . .[tex]v(t)^2\, =\, v_0^2\, +\, a^2(t\, -\, t_0)^2\, +\, 2v_0 a(t\, -\, t_0)[/tex]

    . . . . . . . . .[tex]=\, v_0^2\, +\, 2a\, \left(\frac{1}{2}a(t\, -\, t_0)^2\, +\, v_0 (t\, -\, t_0)\right)[/tex]

    . . . . . . . . .[tex]=\, v_0^2\, +\, 2a(x(t)\, -\, x_0)[/tex]

    Also, which calculus steps are required to transform 1/2at^2 into x? I am referring to the part of the equation to the right of the second equal sign for this second question.
    In case of constant acceleration,

    a = [tex]\dfrac{d^2x}{dt^2}[/tex]

    integrate both sides with respect to 't'

    a*t = [tex]\dfrac{dx}{dt} \ + \ C_1[/tex]

    Apply known conditions to evaluate C1.

    Integrate it again..... tell us what did you get ....
    Last edited by stapel; 02-02-2018 at 02:59 PM. Reason: Copying typed-out graphical content into reply.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  6. #6

    Fyi

    FYI - This is not a homework assignment. Instead, I am trying to understand something I am reading while self-studying calculus. It's been a while since I've taken algebra, so actually working out equations step by step is very helpful. Of course, I understand if you don't have time.

  7. #7
    Quote Originally Posted by Subhotosh Khan View Post
    v(t) = v0 + a(t - t0)

    square both sides

    [v(t)]2 = [v0 + a(t - t0)]2

    What do you get after simplifying the right-hand-side?

    Here's what I got: Vo^2 + 2Voa(t-to) + 2a(t-t0) Note, it's been a while since I've taken algebra and I'm working off 3 hours of sleep right now, so if this isn't right, I'm not surprised.

  8. #8
    Quote Originally Posted by Subhotosh Khan View Post
    In case of constant acceleration,

    a = [tex]\dfrac{d^2x}{dt^2}[/tex]

    integrate both sides with respect to 't'

    a*t = [tex]\dfrac{dx}{dt} \ + \ C_1[/tex]

    Apply known conditions to evaluate C1.

    Integrate it again..... tell us what did you get ....
    For this, I got, 1/2at^2=dx + c. I'm pretty sure dx + c is incorrect. I've had no trouble getting 1/2at^2 from at, the trouble is getting x out of that. For anyone reading this, my questions are in the original post, I am only responding in hopes of finding the answer to those questions through the Socratic method.

  9. #9
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by prepforcalc View Post
    … I would still like to know the answers to the questions in the OP …
    I made a start. I asked whether you could provide the original context. I thought you were a calculus student, so I responded accordingly.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  10. #10
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by prepforcalc View Post
    FYI - This is not a homework assignment. Instead, I am trying to understand something I am reading while self-studying calculus. It's been a while since I've taken algebra, so actually working out equations step by step is very helpful. Of course, I understand if you don't have time.
    Next time, please provide this type of information in your OPs.


    What is that, exactly? Can you provide a link? Attach a (legible) screen shot? Type it out?


    One reason why this forum has guidelines is to reduce the amount of time volunteers need to spend guessing.


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    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

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