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

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.

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

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

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

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.

2. Originally Posted by 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.

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

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

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

Originally Posted by prepforcalc
… transform 1/2at^2 into x …
That could be just a substitution. It might help, to know the context.

3. Originally Posted by mmm4444bot
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

4. Originally Posted by prepforcalc
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?

5. Originally Posted by 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.

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

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

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

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

integrate both sides with respect to 't'

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

Apply known conditions to evaluate C1.

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

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. Originally Posted by Subhotosh Khan
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. Originally Posted by Subhotosh Khan
In case of constant acceleration,

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

integrate both sides with respect to 't'

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

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.

9. Originally Posted by prepforcalc
… 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.

10. Originally Posted by prepforcalc
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?

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