# Uniform Circular Motion: v = v_0 + (a_T)*t as a vector VS as a scalar equation

First, I'll explain my confusion, which may be enough for someone to clarify my misunderstanding.

Second, to give context, I will give the homework problem (which I've solved successfully) which caused my confusion.

Note: I'll use boldface to denote vector quantities. Also, I am in my third semester of Calculus.

A particle accelerates around a track. Velocity is given by the below equation.

$$v = v_0 + a_Tt$$

When I substitute in values for this equation to solve homework problems, I use scalar values—in other words, I use magnitudes. I find $|\mathbf{a_T}|$ to use in the above equation.

However, if the equation is written as a sum of vectors:

$$\mathbf{v} = \mathbf{v_0} + \mathbf{a_T}t$$

Then I take the magnitude of both sides . . .

($\theta$ is the angle between the positive $x$-axis and $\mathbf{a_C}$.)

\begin{align} |\mathbf{v}| &= |\mathbf{v_0} + \mathbf{a_T}t|\\ &\text{. . . algebra . . .}\\ &= \sqrt{|\mathbf{v_0|}^2 + 2|\mathbf{v_0}||\mathbf{a_T}|t + |\mathbf{a_t}|t^2}\\ &\ne |\mathbf{v_0}| + |\mathbf{a_T}|t\ \ \text{*} \end{align}

But the starred equation is what I am using for homework problems and I get the answers with no problems.

https://i.stack.imgur.com/mVjCV.png

To summarize my solution:

1. Use $\mathbf{a}$ to find $|\mathbf{a_C}|$ and $|\mathbf{a_T}|$.
2. Substitute $|\mathbf{a_C}|$ into $a_C = \frac{{v_0}^2}{r}$ and solve for $v_0$.
3. Substitute $v_0$ and $\mathbf{a_T}$ into $v_1 = v_0 + a_T t$ and solve for $t_1$ when $v_1 = 0$.

I also run into sign issues because I'm treating the value I get for tangential acceleration as the magnitude of a vector. This means that the value for tangential acceleration is positive, but it needs to subtract from velocity. There is a similar issue with the centripetal acceleration.

The only way to deal with it properly seems to be to treat vectors counter clock-wise around the circle or pointing away from the circle as positive, and other vectors as negative . . . which is akin to treating the circle as a straight line almost . . .

It feels like a hell of a lot of hand-waving, though I am sure that's because I don't understand what is happening.

It is certainly not true that $\mathbf{v} = \mathbf{v_0} + \mathbf{a_T}t$, and it is not true that $a_C = \frac{v_0^2}{r}$. First, the acceleration is not all tangential: $\mathbf{a} \neq \mathbf{a_T}$. Second, it is changing with time, since $a_C$ points from the particle towards the center, and as such it keeps on rotating. Therefore, we should write the velocity as $\mathbf{v} = \mathbf{v_0} + \int_0^t \mathbf{a}(t')\ dt'$. (I don't know if you've been introduced to integrals yet; if you haven't, you just need to know that when the acceleration changes over time we can't use the formula $\mathbf{v} = \mathbf{v_0} + \mathbf{a}t$.) And since the magnitude of the velocity is changing with time, we should write $a_C = \frac{|\mathbf{v}|^2}{r}$. $a_C = \frac{v_0^2}{r}$ is only true at $t=0$.
The formula $v = v_0 + a_T t$ indeed works (as long as $a_T$ is constant, which in your problem it is), because $a_T$ is defined as the time rate of change of the magnitude $v$ of the velocity. It's not always true that $a_T = |\mathbf{a_T}|$, because as you say there is a sign issue. The issue arises from the fact that $v$, $v_0$ and $a_T$ aren't the magnitude of their respective vectors; rather, they're their tangential components. The sign will depend on which way around the circle you consider to be forwards and which you consider to be backwards.
• Yes, I should have mentioned that I am in my final semester of Calculus. Your third paragraph was the piece I was missing: $v = v_0 + a_T t$ works because this equation concerns only the tangential components of these vectors, which means the equation itself is one-dimensional . . .