Formula for time in terms of acceleration and distance

[mondo]

Reading Griffiths book on electromagnetics I stumbled upon his analogy to Newton and acceleration. Author claims that the formula for time when acceleration and distance are given is: $$t= \sqrt{\frac{2\lambda}{a}}$$, all is clear beside 2 in the nominator. My calculations goes as follows: $$a = \frac{v}{t}$$ next $$V = \frac{s}{t}$$ hence $$a = \frac{s}{t^2} -> t = \sqrt{\frac{s}{a}}$$.

What have I missed?

 

[Nugatory]

What have I missed?

The velocity is zero at time zero, so the average speed over time $t$ is $at/2$, and that’s where the two is coming from. Google for “SUVAT equations” for more.

 

[pasmith]

If an object is accelerating, its velocity is not constant. Its displacement is therefore not given by vt, but by \int v(t)\,dt = \int v_0 + at\,dt = s_0 + v_0t + \frac12 at^2.

 

[kuruman]

You have also missed that your expression for time works only when the object starts accelerating from rest.

 

[A.T.]

My calculations goes as follows: $$a = \frac{v}{t}$$ next $$V = \frac{s}{t}$$ hence $$a = \frac{s}{t^2} -> t = \sqrt{\frac{s}{a}}$$.

What have I missed?

Ironically you have correctly used different case of 'v' for velocity change vs. mean velocity, but then incorrectly assumed they are the same.

 

[mondo]

Thanks, I have correctly got the equation by correcting my acceleration formula, with a formula for an average acceleration. However, in the very next equation author writes this "and hence your average velocity is: $$v_{ave} = \frac{1}{2} at = \sqrt{\lambda a}{2}$$ How did he derive it? If I multiply the formula for t: $$\sqrt{\frac{2\lambda}{a}}$$ by $$\frac{1}{2}a$$ I get $$\frac{1}{2}a\sqrt{\frac{2 \lambda}{a}}$$

 

[kuruman]

Thanks, I have correctly got the equation by correcting my acceleration formula, with a formula for an average acceleration. However, in the very next equation author writes this "and hence your average velocity is: $$v_{ave} = \frac{1}{2} at = \sqrt{\lambda}{2}$$ How did he derive it? If I multiply the formula for t: $$\sqrt{\frac{2\lambda}{a}}$$ by $$\frac{1}{2}a$$ I get $$\frac{1}{2}a\sqrt{\frac{2\lambda}{a}}$$

You could be clearer by writing equations with an equals sign separating the two sides that are equal instead of just one of the sides. I think what you are trying to say is this. Start with
$v_{\text{avg.}}=\dfrac{1}{2}at.~~$ This is correct for the average velocity over interval $t$ under constant acceleration $a$.

Now it seems that $\lambda$ is the distance traveled in time $t$ starting from rest under constant acceleration $a$. Then $$\lambda =\frac{1}{2}at^2\implies t=\sqrt{\frac{2\lambda}{a}}$$which is the first equation that you posted in post #1. So far so good.

Going back to the expression for the average velocity $$\cancel{v_{\text{avg.}}t=\frac{1}{2}at=\frac{1}{2}\sqrt{\frac{2\lambda}{a}}}$$ $$v_{\text{avg.}}=\frac{1}{2}at=\frac{1}{2}a\sqrt{\frac{2\lambda}{a}}=\sqrt{\left(\frac{a}{2}\right)^2\times\frac{2\lambda}{a}}=\sqrt{\frac{\lambda a}{2} }.$$If you're wondering why the average velocity is what Griffiths says it is, you start with the definition for the average velocity $$v_{\text{avg.}}=\frac{\lambda}{t}$$ and note that $$\lambda=\frac{1}{2}at^2\implies \frac{\lambda}{t}=\frac{1}{2}at=v_{\text{avg.}}.$$BTW the second equation in parentheses that you wrote $$v_{ave} = \left(\frac{1}{2} at = \sqrt{\lambda}{2}\right)$$ makes no sense because it is dimensionally incorrect.

(Edited to fix typos. See Post #10.)

 

[mondo]

You could be clearer by writing equations with an equals sign separating the two sides that are equal instead of just one of the sides

Agree, how to put math equation inlined with a text, I tried with single $ but it doesn't work.

Going back to the expression for the average velocity vavg.t=12at=122λaIf you're wondering why the average velocity is what Griffiths says it is, you start with the definition for the average velocity

I don't agree with your formula for V_ave - why did you get t on the left hand side next to v_ave. Plus, what happened to the multiplication by 'a' in the middle term? Anyway, in the book the formula is yet different - $$V_{ave} = \frac{1}{2}at = \sqrt{\frac{\lambda a}{2}}$$

 

[PeroK]

Agree, how to put math equation inlined with a text, I tried with single $ but it doesn't work.


I don't agree with your formula for V_ave - why did you get t on the left hand side next to v_ave. Plus, what happened to the multiplication by 'a' in the middle term? Anyway, in the book the formula is yet different - $$V_{ave} = \frac{1}{2}at = \sqrt{\frac{\lambda a}{2}}$$

Where is this in Griffiths book?

 

[kuruman]

Agree, how to put math equation inlined with a text, I tried with single $ but it doesn't work.


I don't agree with your formula for V_ave - why did you get t on the left hand side next to v_ave. Plus, what happened to the multiplication by 'a' in the middle term? Anyway, in the book the formula is yet different - $$V_{ave} = \frac{1}{2}at = \sqrt{\frac{\lambda a}{2}}$$

Yes, I had an error in the formula which I thought I fixe but it was late at night (for me) and didn't fix it completely. Thank you for pointing out the error and sorry about the confusion. Here is what I meant to post. $$v_{\text{avg.}}=\frac{1}{2}at=\frac{1}{2}a\sqrt{\frac{2\lambda}{a}}=\sqrt{\left(\frac{a}{2}\right)^2\times\frac{2\lambda}{a}}=\sqrt{\frac{\lambda a}{2} }.$$ If you want to post an in-line expression in LaTeX enclose it with two # on each side. For example, \sqrt{2} becomes $\sqrt{2}$.

 

[jbriggs444]

If you want to post an in-line expression in LaTeX enclose it with two # on each side. For example, \sqrt{2} becomes $\sqrt{2}$.

When demonstrating $\LaTeX$ there is a trick that can be employed so that the "#" and "$" delimiters can be seen rather than interpreted by the rendering engine. What one does is to change the color on one of the "#" or "$" signs in a pair. Change it to black -- the default.

Start by keying in your $\LaTeX$ with the doubled delimiters. For instance:

Square root of two is $\sqrt{2}$
Square root of two is: $$\sqrt{2}$$

Select the second delimiter in one of the pairs. If you are on a computer, that's left-click and hold while sweeping over it so that the single delimiter is highlighted).

Click the palette icon ("text color") and select black (RGB #000000).

Repeat for all of the delimiter pairs. Since black is the default color, the change is invisible.

Internally, the effect is that BB code tags for text color are inserted before and after the selected character. If one turns off BB code interpretation in the editor (click the [ ] icon for "toggle BB code") the result will look like this:

Square root of two is $\sqrt{2}$
Square root of two is: $$\sqrt{2}$$

The presence of the color tags means that the $\LaTeX$ delimiters are no longer back to back. So the $\LaTeX$ rendering engine (MathJax) does not activate. Without the color tags the delimiters are back to back, the rendering engine triggers, the $\LaTeX$ is rendered and the reader sees the expected result:

Square root of two is $\sqrt{2}$
Square root of two is: $$\sqrt{2}$$

If one wants to disable the BB code engine to allow BB codes to be seen by the reader rather than rendered, a different trick can be employed. Surround the codes with [PLAIN] and [/PLAIN] tags. However, this trick is ineffectual in fooling the MathJax engine.

 

[mondo]

kuruman, thank you. All is clear now.

 

[A.T.]

When demonstrating $\LaTeX$ there is a trick that can be employed so that the "#" and "$" delimiters can be seen rather than interpreted by the rendering engine.

You can wrap the LaTeX code in CODE-tags:

$$\sqrt{2}$$