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I am posting my derivation since I have been very confused by how the books treat this topic. I hope someone can let me know if my derivation is correct.

I will derive it in the most general manner without assuming things like $x(0)=0$ etc.

Derivation: From the definition of acceleration we have $a=\frac{dv}{dt}$ hence $v=\int a dt =at+c$. Letting $t=0$ gives $c=v(0)$, hence $\boxed{v=v(0)+at}$.

Then using $v=\frac{dx}{dt}$ and integrating we get $x=v(0)t+\frac{1}{2} at^2+c$, and letting $t=0$ we find that $c=x(0)$. Therefore, $\boxed{x=x(0)+v(0) t+\frac{1}{2} at^2}$.

At this point, I am confused about something. Basically, the book goes from here to say that "therefore, $\Delta x = v(0) t+\frac{1}{2} at^2$".

The reason I don't see why this is generally true is because $\Delta x$ isn't automatically the same as $x-x(0)$, because $\Delta x$ simply means change in $x$. It could be $x-x(1)$ for all I know. So why do the authors claim that $x=x(0)+v(0) t+\frac{1}{2} at^2$ is the same as $\Delta x = v(0) t+\frac{1}{2} at^2$.

The most I am seeing is that $\Delta x = x-x(0)$ would indeed be true but only if the initial time, $t_0$ was $0$, ie $t_0=0$. Then "change in position" would generally be $x-x(0)$ (am I right in thinking this?). Otherwise, for all we know, $t_0$ could be nonzero so $\Delta x$ would be $x-x(t_0)$, not $x-x(0)$. Again, I am not sure if I am right about this.

If someone can explain this point it would clear up a lot of the confusion I've been having about the constant acceleration equations.

Raghib
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  • Generally we take $\Delta x= x-x(0)$ only. This is generally called displacement. – jonsno Jan 01 '18 at 18:08
  • @samjoe: Please read my edit (the second last paragraph). I don't see why $\Delta=x-x(0)$ if we haven't chosen $t_0=0$ yet. I understand that $\Delta = x-x(t_0)$ where $t_0$ is initial time. What I don't understand is how the authors leapt from there to $t_0=0$. – Raghib Jan 01 '18 at 18:09

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$\Delta x$ has the definite meaning "final position minus initial position", but which position you call the "final position" and which you call the "initial position" are up to you. If you have already derived the equation $$x=x(0)+v(0) t+\frac{1}{2} at^2$$ then it is very natural to think of the final position to be $x$ and the initial position to be $x(0)$. In that case you certainly do get $$\Delta x = x-x(0) = v(0)t+\frac 12at^2$$

So basically, the answer is that $\Delta x$ is context dependent. In the book's derivations it means $x-x(0)$, but in other situations it may mean $x(t_2) - x(t_1)$ for any other times $t_2$ and $t_1$.

user517641
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  • Here is a very compact form of my question: Is the book wrong in saying $\Delta x = x-x(0)$ without saying that $t_0=0$ is the initial time? This is what I don't understand. – Raghib Jan 01 '18 at 18:15
  • I wouldn't say it's wrong, per se. But it would certainly be more clear if they specified that they're taking $\Delta x$ to be $x-x(0)$ (or equivalently the initial time to be $t_0 = 0$). – user517641 Jan 01 '18 at 18:16
  • Ok that clears it up then. Thank you very much – Raghib Jan 01 '18 at 18:17