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.


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.


1 Answer 1


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.

In this sense, what you say about treating some vectors as positive and others as negative is not entirely wrong. Really, what you're doing is since your problem is essentially one-dimensional (because the particle moves in a circle, which is a one-dimensional figure), you can just forget about vectors and use scalars, as long as you are careful in picking a "positive" and a "negative" direction.

  • 1
    $\begingroup$ 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 . . . $\endgroup$
    – JDG
    Feb 7, 2014 at 2:41

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