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A point is moving from the start of a curve to its end. This curve is defined by moving independently its two axis: x and y. "y" moves in a linear way: 1 + (-0.5) * t;

Where 1 is the start coordinate, -0.5 its movement and t is a time number that goes from 0 to 1.

On the other hand, x isn´t linear and is defined as this: 1 / (1 + (-0.5) * t);

---However, this equation is causing a lot of trouble since "t" is not known yet. Simplifying the equation and getting rid of that division is my goal. Am asking if there´s an acceleration formula that may replace this: 1 / (1 + (-0.5) * t);

That same value is this over time:

  • 1.052632___(when t = 0.1);

  • 1.111111____ (when t = 0.2);

  • 1.176471____(when t = 0.3);

  • 1.25_______(when t = 0.4);

  • 1.333333___ (when t = 0.5);

  • 1.428571___ (when t = 0.6);

  • 1.538462___ (when t = 0.7);

  • 1.666666___ (when t = 0.8);

  • 1.818181____(when t = 0.9);

  • 2_________(when t = 1);

The acceleration isn´t constant, but I can´t figure its equation.

Bernard
  • 175,478
Fernando
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  • Wzlcome to TeX SX! The acceleration at time $t$ is the second derivative. – Bernard Sep 29 '20 at 21:14
  • Am sorry. I don´t know enough of derivatives and am not understanding the tutorials. What would be the second derivative of this? Can you at least tell me where to find good info on this subject? I don´t want to sound like the type of guy that doesn´t even bother to investigate, but am really not that good at math. – Fernando Sep 29 '20 at 22:16
  • Do you know the formula for the derivative of a composition of functions? – Bernard Sep 29 '20 at 22:20
  • No... sorry. The only thing that I know that may be useful to solve this problem is how does constant acceleration work. And all I found on derivatives is confusing. – Fernando Sep 29 '20 at 23:33
  • Ok, so I´ve done some research and I think I understand most of how the derivatives work, but it´s still confusing. When you say that acceleration at time t is the second derivative, do you mean that I need to get the derivative of this: 1 + (-0.5) * t; or to get the derivatuve of 1 over that? At any case, is the second derivative the derivative of that result? – Fernando Sep 30 '20 at 03:16
  • And just to be sure, could you give the derivative formula in this case? Because am doubting and I need to consider that I don´t know the value of t (because I think that the formula needs that as a constant, but again, am not entirely sure). – Fernando Sep 30 '20 at 03:24
  • The functio has the form$\frac1u$, with $u=1+t/2$. The fist derivative is $-1/u^2$ (w.r.t. $u$) and $-1/u^2\cdot u'$ (w.r.t. $t$). The second derivative w.r.t. $u$ is $-2/u^3$. W.r.t. $t$, you have to differentiate the product $-1/u^2\cdot u'$. Is that clear? – Bernard Sep 30 '20 at 07:47
  • Am sorry. I know it must be annoying to deal with someone like me, so thanks a lot. But is not clear. To begin with, where does u come from? I had 1 / (1 -0.5 * t), not 1 / (1 + t / 2). (Sorry if it´s a stupid question). Then, I actually tried getting the derivative of (1 + t/2), but it gave me 0.5 instead of −1/u2; (I replaced the t by (t + dt) in u, then subtracted u and then divided all by dt). I think that´s the right formula, but it didn´t gave your result. And I tried to do the rest w.r.t t, but because of this I didn´t understood. – Fernando Sep 30 '20 at 18:36
  • It's not annoying at all, we're here to help. $u$ is just the name of another variable, because one has to compute the derivatives by substitution. As to the function I wrote, it's an error: it should have been $1/(1-t/2)$, of course, and, yes, the derivative of $1+t/2$ is $1/2$. But it's the derivative of the inverse that you want. – Bernard Sep 30 '20 at 18:44
  • Ok, so I´ve got the derivative of 1 / (1 - t/2), which is 2 / (4 - 4t + sqr t). At first I thought it would give me -1/u sqr (which was your result 3 comments above), and that caused me a lot of confusion, but I know it´s right since I compared it to a calculator´s right result, but then what? Am I supposed to get the derivative of that result and that´s the final result, the acceleration at time t? (By the way, how do you write square? Am trying alt 0178 but it doesn´t work). – Fernando Sep 30 '20 at 20:14
  • Your derivative is wrong; it is $\frac 1{2\bigl(1-\tfrac t2\bigr)^2}$, which can be simplified to $\frac{2}{4-4t+t^2}$. – Bernard Sep 30 '20 at 21:04
  • That´s... just what I wrote. sqr t means squared t. I originally had 0.5 / (1 - t + sqrT/ 4), which is the same as your simplification. So, which would be the following step? – Fernando Sep 30 '20 at 21:09
  • How can you obtain a square root from a rational function? – Bernard Sep 30 '20 at 21:15
  • Is not a square root. Is t * t... sorry if my way of writing is weird. – Fernando Sep 30 '20 at 21:16
  • Why don't you code it t^2? Anyway, to compute the second derivative, it will be simpler to have the first derivative factored, so do not expand the denominator. Also, more generally, you should know by heart the formula $\bigl(\frac 1{x^n}\bigr)'=-\frac{n}{x^{n+1}}$. – Bernard Sep 30 '20 at 21:22
  • So then it is getting the derivative of the derivative. However, is your formula right? Based on the results of a calculator, it should be -4/(x - 2)^3; But according to your method, it would be -2/(2 - t)^3; – Fernando Oct 01 '20 at 01:13
  • I've taught it for more than 20 years! You're forgetting that the function is rewritten as $\frac{\color{red}2}{2-x}$, and the derivative is linear, so the 1st derivative is $\color{red}2\cdot\frac1{(2-x)^2}$, &c. – Bernard Oct 01 '20 at 07:42
  • Ok, so I think that I get it now. The second derivative, or acceleration, of 1 / (1 -0.5 * t) would be -4 / (2 - t)^3; However... I still have some doubts. First, the results are negative, even though the displacement goes from 1 to 2. For example, when t = 0.5, displacement is 1.333333 and the acceleration (2nd derivative) is -1.185185. Then, what I really want is the formula for that displacement. And I need to apologize, since I originally asked for the acceleration instead. Sometimes i don´t know how to express myself. (Now is the perfect time to remember that am not annoying). – Fernando Oct 01 '20 at 19:03
  • The second derivative is poditive: $4/(2-t)^3$. You seem to have forgotten to apply the formula for a composition of functions. With more details, this derivative is $:2\cdot -2/(2-t)^3\cdot -1$. – Bernard Oct 01 '20 at 19:16
  • I have the same derivative, but I didn´t multiply it by -1 at the end. Can you explain where does it come from? You are saying that it´s some formula for a composition of functions, but that I know, that´s just getting the derivative. Anyway, regardless of sign, that´s just the acceleration. How can I get the displacement from that? – Fernando Oct 01 '20 at 19:51
  • It is simply the derivative of $2-t$. The formula for the composition $\frac1{u(t)^2}$ is $\frac{-2}{u(t)^3}\cdot\color{red}{u'(t)}$. – Bernard Oct 01 '20 at 19:54
  • The displacement is the integral of the speed (first derivative), and the speed is the integral of the acceleration (second derivative), but I don't see where it will lead you to. – Bernard Oct 01 '20 at 19:58

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