# 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.

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)

• I didn't say $v_0=v(0)$ in general, I said it is true based on the author's assumption to let $x_0=0$ and $t_0=0$. This is why I was confused, because he already had the equation from the start. – Raghib Jan 1 '18 at 17:10
• Here is how I derived $v(0)$ equation: The author has $v=v_0+A(t-t_0)$, so letting $t=0$ we get $v(0)=v_0+A(-t_0)$, and substituting this back into the equation gives $v=v(0)+At$. I am not saying that $v_0=v(0)$, I am using the substitution $v(0)=v_0+A(-t_0)$. However, at the end it actually does become $v_0=v(0)$ since the author lets $t_0=0$. Please let me know if I am wrong. – Raghib Jan 1 '18 at 17:20

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 $v=v_0+At$.