Why does the author derive $v=v(0)+At$ in this way? In the book "understanding physics" by Mansfield, the author derives the constant acceleration equations and at the start he derives the equation $v=v_0+A(t-t_0)$. I immediately noticed that setting $t=0$ in this gives $v(0)=v_0-At_0$, ie the equation is equivalent to $v=v(0)+At$.
However, the author seems to get this by a more complicated method. He first says "It is often convenient to choose the origin and starting instant so that x=0 when t=0. In this case, $x_0=0$ and $t_0=0$, and the equation $v=v_0+A(t-t_0)$ becomes $v=v_0+At$."
The reason this is confusing me, is because the equation he gets is simply $v=v(0)+At$ (since he defines $v_0$ as the value of $v$ at $t_0$, but he just said that $t_0=0$, therefore his $v_0$ has now become $v(0)$ in the last equation).
Therefore, he essentially is saying: "It is often convenient to choose the origin and starting instant so that x=0 when t=0. In this case, $x_0=0$ and $t_0=0$, and the equation $v=v_0+A(t-t_0)$ becomes $v=v(0)+At$."
I am not sure if you see why this is confusing, but he seems to be saying that "in this case", ie in the case where $x_0=0$ when $t_0=0$, then $v=v(0)+At$. But this is not true, since it is not only true in this case, because I clearly showed above that his initial equation $v=v_0+A(t-t_0)$ already implies $v=v(0)+At$. So we do not need the special case of $x_0=0$ when $t_0=0$ for $v=v(0)+At$ to result from $v=v_0+A(t-t_0)$.
Therefore, I am not sure what the author meant by saying "in this case" and am wondering if I have gone wrong somewhere.
 A: You are correct, but you miss one subtlety. His equation after the assumption of $t_0=0$ was
$$ v = v_0 + At $$
While yours was
$$ v = v(0) + At $$
You can probably see it now, but the distinction is that $v_0$ is not necessarily equal to $v(0)$. In fact $v_0$ is defined to be $v(t_0)$, so for a nonzero acceleration you can show that $v_0=v(0)$ only if $t_0=0$.
(On the other hand the assumption that $x_0=0$ is completely unnecessary)
A: I too find the quoted paragraph puzzling.
If we consider $x$ and $v$ to be functions of $t,$
and say that $t_0$, $x_0,$ and $v_0$ are the "starting" time, position, and velocity,
then $x_0 = x(t_0)$ and $v_0 = v(t_0).$
If we now set our coordinate system so that $x=0$ when $t=0$
(as the author writes), it is true that $x(0) = 0,$
but let's consider an example in which some particle moves according to
$x_0 = -1,$ $t_0 = -1,$ $v_0 = 1,$ and $A = 0.$
In that example we will also have $x=0$ when $t=0.$
So I do not see how "$x=0$ when $t=0$" implies that $x_0 = 0$
or that $t_0 = 0.$
Here is a different version of the quote from the book,
which I have modified so that I think it is correct:

It is often convenient to choose the starting instant so that $t_0=0$. Then the equation $v=v_0+A(t-t_0)$ becomes $v=v_0+At$.

