# Is centripetal acceleration based on speed or velocity?

I know that the formula for centripetal acceleration is $\ a_c = v^2/r$. Does the $v$ represent the average velocity of the car which is the arc length traveled divided by the time? If the answer is yes then, I have a second question. Why do we use distance divided by time to calculate the velocity instead of displacement. For example if there was a car that traveled a unbanked curve of radius 200m and completed a 2pi radian rotation in 10 sec. Find the centripetal acceleration. Why do we do $v= (2pi*200)/10$ instead of $v= (0)/10$ to find the $v$. Since $v$ should be displacement divided by time not distance.

• Have you taken a calculus course yet? Feb 17 '17 at 1:13
• Nope, I have not taken calculus. We just started learning circular motion. Feb 17 '17 at 1:18

The conventions of differential calculus are subtle and hard to explain, and that seems to be the core of this conundrum. Note, 'velocity', a vector, is defined at every point of the trajectory of a moving object, and is defined as a limit of very-short-path-traveled divided by very-short-time-elapsed. So, the magnitude of the velocity, the speed, and the direction of the velocity (which, in circular motion, just takes on every value possible in two dimensions), are both important in general, but in the specific case of a complete circle, only magnitude has a non-trivial meaning. Expressing a formula with "v^2" reinforces that only the magnitude is intended, because there is no general way to 'square' a vector.

Acceleration, like velocity, is a vector defined at every point, by a similar limit of very-small-velocity-change divided by very-short-time-elapsed, and is inconsistent with using an AVERAGE of velocity (since it depends on the change over very short times). Acceleration has a direction just as velocity does, but (again, because this is a circle) it is described as 'centripetal' meaning it takes the direction toward the center of the circle.

It makes little sense to talk of average directed acceleration, because the full circular path makes that average vanish (by symmetry) in uniform circular motion: only the magnitude of the acceleration is constant in this kind of motion. So, the "centripetal acceleration" means only the magnitude of the instantaneous acceleration, without any averaging.

So, one cannot make use of the given formula while averaging vectors over a whole cycle of the circular motion. Differential calculus makes the hints, and therefore the formula for acceleration, more understandable.

Since I started writing this, you've gotten some good answers, but I'll post this anyways.

These things become much clearer once you take a course in calculus. The most important thing to know is that when you talk about acceleration or velocity, you're usually always talking about the instantaneous acceleration or velocity, which means the value at a single point in time, not averaged over a period of time. A quick explanation first: When you find the average velocity $\vec{v_{avg}}=\frac{\vec{\Delta{d}}}{\Delta{t}}=\frac{\vec{d_2}-\vec{d_1}}{t_2-t_1}$, the object does not necessarily have that velocity at any point in time. It's simply an average. As you point out, a complete circle causes $\vec{d_1}=\vec{d_2}$ and therefore $\vec{v_{avg}}=\vec{0}$, but the car never has an instantaneous velocity of $\vec{0}$ (it's always moving with some speed and in some direction).

Finding the instantaneous velocity often requires calculus, but the idea is simple enough. Use the equation for the average velocity $\vec{v_{avg}}=\frac{\vec{\Delta{d}}}{\Delta{t}}$ and make $\Delta{t}$ very small. So small that it's nearly zero (called taking the limit as $\Delta{t}$ approaches $0$). This is kind of like an average over a very small period of time, so small that it gives the instantaneous value at that specific time). Don't worry about the math, but try to understand the idea if you can.

Now on to your question. For the centripetal acceleration formula $a_c = \frac{v^2}{r}$, it assumes that the speed (the magnitude of the instantaneous velocity vector at any point in time) remains constant, even though the velocity is changing direction (instantaneous velocity direction is always tangent to the circle). The radius $r$ also remains constant. It can be written clearer as $\lvert \vec{a_c} \rvert = \frac{\lvert \vec{v} \rvert^2}{r}$, where the bars represent the magnitude of the vectors. So since $\vec{v}$ represents the instantaneous velocity, $\lvert \vec{v} \rvert$ represents the instantaneous speed. Since the speed is constant, it is simply the average speed, so the path distance divided by the total time $\lvert \vec{v} \rvert = v=\frac{2\pi r}{\Delta{t}}$, which is the speed at any given time. Now we have the equation $\lvert \vec{a_c} \rvert = (\frac{2\pi r}{\Delta{t}})^2 \cdot \frac{1}{r}$.

• Just to clarify, this that the $v_c$ is the instantaneous velocity and that I can't use the centripetal acceleration formula if the tangential acceleration is not constant. Feb 18 '17 at 3:44
• @coderhk Actually the acceleration $\vec{a_c}$ is always perpendicular to the circle tangent when an object is moving in a circle, while the velocity $\vec{v}$ is tangent to the circle (velocity and acceleration are perpendicular). So to use this equation, the magnitude of (instantaneous) acceleration $a_c$ must be constant, the magnitude of (instantaneous) velocity $v$ (the speed) must be constant, and the radius must be constant, so basically all the magnitudes must be constant. Feb 18 '17 at 3:56

In the car example you gave, the problem would most likely assume uniform angular velocity and would be asking for the magnitude of linear acceleration. This being the case, the velocity should indeed $= \frac{2\pi \cdot 200}{10}$. Because there is constant angular velocity, there is also constant linear velocity, and hence the instantaneous velocity magnitude also equals the average magnitude of velocity in this case.

If the problem asked you to give the average velocity with v represented as v or $\vec{v}$ (vector form) for a full rotation, the vector could not be drawn as it would have zero magnitude.

So $\frac{distance}{time}$ is only used in cases where the angular velocity (and hence linear velocity and acceleration) is constant and the magnitude, not the vector, is being asked for.