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

  1. #11

    Lets start over with the original post

    I am more confused than when I started this thread. If someone could just show me step by step, the answers to my questions in the OP, I would really appreciate that. I provided all the relevant information and cannot provide the whole document due to copyright. Again, I am not taking any classes so nothing I post is homework. Efficient help would be most appreciated.

  2. #12
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by prepforcalc View Post
    Efficient help would be most appreciated.
    It's not easy to be efficient, when people requesting help keep secrets or leave us guessing. We're not asking for you to publish an entire document. We would simply like to know the context of what you're reading. I'm not sure what the real issue is; after all, you have already published some excerpts.

    In general, we do not post step-by-step solutions right away. We prefer to have a conversation with you about your difficulties, first. Most of the volunteers in this forum do not have time for typing up lessons/classroom material or to provide lectures on broad concepts, as this type of information is already available at thousands of locations on-line.

    Were I to know better what you need, in the way of lessons, I might be able to provide some links.

    Otherwise, let's give Subhotosh a chance to respond to your rejoinders.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  3. #13
    Quote Originally Posted by mmm4444bot View Post
    It's not easy to be efficient, when people requesting help keep secrets or leave us guessing. We're not asking for you to publish an entire document. We would simply like to know the context of what you're reading. I'm not sure what the real issue is; after all, you have already published some excerpts.

    In general, we do not post step-by-step solutions right away. We prefer to have a conversation with you about your difficulties, first. Most of the volunteers in this forum do not have time for typing up lessons/classroom material or to provide lectures on broad concepts, as this type of information is already available at thousands of locations on-line.

    Were I to know better what you need, in the way of lessons, I might be able to provide some links.

    Otherwise, let's give Subhotosh a chance to respond to your rejoinders.
    This is my last post about anything other than the solution to the math question. The screenshot in the original post is self explanatory, other than missing the steps that would explain why what's on the left and right of each equal sign is equal. There's nothing more to it. This is simple first year college calculus based physics. Granted, I'm learning, but this is very basic. Hopefully someone on this forum will recognize this and help out. That being said, I am exhausted today and apologize if my tone or explanations have been anything less than what they should be. I'm not sure either way at time moment. I am beat. Help would be great, but if not, I will figure this out another way.

  4. #14
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    Quote Originally Posted by prepforcalc View Post
    In the attached screenshot, where does the 2Vo come from? The 2Vo I speak of is to the right of the first equal sign.

    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.
    Since things have been tangled up by too many people jumping on you, let me summarize what you need to know about this work:

    First, here is your equation:

    [tex]v(t)^2 = v_0^2 + a^2(t - t_0)^2 + 2v_0a(t - t_0) = v_0^2 + 2a(\frac{1}{2} a(t - t_0)^2) + v_0(t - t_0)) = v_0^2 + 2a(x(t) - x_0)[/tex]

    The first step, [tex]v(t)^2 = v_0^2 + a^2(t - t_0)^2 + 2v_0a(t - t_0)[/tex], clearly arises because they started with [tex]v(t) = v_0 + a(t - t_0)[/tex] and squared both sides, using the fact that [tex](a+b)^2 = a^2 + 2ab + b^2[/tex]. Are you okay there?

    The second step, [tex]v_0^2 + a^2(t - t_0)^2 + 2v_0a(t - t_0) = v_0^2 + 2a(\frac{1}{2} a(t - t_0)^2) + v_0(t - t_0))[/tex], comes from factoring 2a out of the last two terms; an intermediate step they could have shown is [tex]v_0^2 + 2a\cdot \frac{1}{2} a(t - t_0)^2 + 2a\cdot v_0(t - t_0)[/tex]. Are you okay there?

    The last step, [tex]v_0^2 + 2a(\frac{1}{2} a(t - t_0)^2) + v_0(t - t_0)) = v_0^2 + 2a(x(t) - x_0)[/tex] comes from the fact that they presumably gave earlier, that [tex]x(t) - x_0 = \frac{1}{2} a(t - t_0)^2 + v_0(t - t_0)[/tex].

    This last bit, if they didn't show it to you before, is the definite integral of [tex]at - v_0[/tex], from [tex]t_0[/tex] to t. It would have helped if you'd just shown us where they said that, to confirm.

    When you wrote "1/2at^2=dx + c", you forgot that the integral of dx/dt with respect to t is just x, not dx; the result of integration can never be a differential. In effect, [tex]\int dx = x[/tex] because integration undoes a differential (or, more properly, because the integral of 1 is x).

    I hope I got everything right here; it's a lot of work trying to type all this up for you. If there's anything still lacking, let us know specifically.

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