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

[prepforcalc]

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

. . . . .$\displaystyle v(t)^2\, =\, v_0^2\, +\, a^2(t\, -\, t_0)^2\, +\, 2v_0 a(t\, -\, t_0)$

. . . . . . . . .$\displaystyle =\, v_0^2\, +\, 2a\, \left(\frac{1}{2}a(t\, -\, t_0)^2\, +\, v_0 (t\, -\, t_0)\right)$

. . . . . . . . .$\displaystyle =\, v_0^2\, +\, 2a(x(t)\, -\, x_0)$

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.

 

[mmm4444bot]

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

. . . . .$\displaystyle v(t)^2\, =\, v_0^2\, +\, a^2(t\, -\, t_0)^2\, +\, 2v_0 a(t\, -\, t_0)$

. . . . . . . . .$\displaystyle =\, v_0^2\, +\, 2a\, \left(\frac{1}{2}a(t\, -\, t_0)^2\, +\, v_0 (t\, -\, t_0)\right)$

. . . . . . . . .$\displaystyle =\, v_0^2\, +\, 2a(x(t)\, -\, x_0)$

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?

… transform 1/2at^2 into x …

That could be just a substitution. It might help, to know the context. :cool:

 

[prepforcalc]

It looks like they started with:

v(t) = v₀ + a(t - t₀)

and squared each side -- so v₀ got squared.

:cool:

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.

 

[Deleted member 4993]

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.

View attachment 9089

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?

 

[Deleted member 4993]

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

. . . . .$\displaystyle v(t)^2\, =\, v_0^2\, +\, a^2(t\, -\, t_0)^2\, +\, 2v_0 a(t\, -\, t_0)$

. . . . . . . . .$\displaystyle =\, v_0^2\, +\, 2a\, \left(\frac{1}{2}a(t\, -\, t_0)^2\, +\, v_0 (t\, -\, t_0)\right)$

. . . . . . . . .$\displaystyle =\, v_0^2\, +\, 2a(x(t)\, -\, x_0)$

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 = $\displaystyle \dfrac{d^2x}{dt^2}$

integrate both sides with respect to 't'

a*t = $\displaystyle \dfrac{dx}{dt} \ + \ C_1$

Apply known conditions to evaluate C₁.

Integrate it again..... tell us what did you get ....

 

[prepforcalc]

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.

 

[prepforcalc]

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.

 

[prepforcalc]

In case of constant acceleration,

a = $\displaystyle \dfrac{d^2x}{dt^2}$

integrate both sides with respect to 't'

a*t = $\displaystyle \dfrac{dx}{dt} \ + \ C_1$

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.

 

[mmm4444bot]

… 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.

 

[mmm4444bot]

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.

:idea: 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.


Here's a link to the complete forum guidelines (and here's a link to a summary). Thanks!

 

[prepforcalc]

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.

 

[mmm4444bot]

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. :cool:

 

[prepforcalc]

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. :cool:

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.

 

[Dr.Peterson]

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:

$\displaystyle v(t)^2 = v_0^2 + a^2(t - t_0)^2 + 2v_0a(t - t_0) = v_0^2 + 2a(\frac{1}{2} a(t - t_0)^2) + v_0(t - t_0)) = v_0^2 + 2a(x(t) - x_0)$

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

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

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

This last bit, if they didn't show it to you before, is the definite integral of $\displaystyle at - v_0$, from $\displaystyle t_0$ 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, $\displaystyle \int dx = x$ 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.