# Can an object have constant speed and tangential acceleration without undergoing circular motion?

Is it possible for an object to be moving with constant speed and a constant magnitude of acceleration tangent to this velocity but NOT move in a circular path? If there are other situations could I please have an example?

• Acceleration 'tangent to velocity' is just straight-line forward (or backward) acceleration. – Whit3rd Feb 6 at 1:02

There’s a really simple but useful theorem in these cases: if a particle has a constant speed, it’s acceleration is orthogonal to its velocity.

The proof is also quite simple (if you know basic vector calculus). If an object has velocity vector $$\vec{v}$$, then its squared speed is $$s^2=\vec{v}\cdot\vec{v}$$. Since the speed is constant, its time derivative is zero. Thus,

$$\frac{\mathrm{d}}{\mathrm{d}t}s^2=\left(\frac{\mathrm{d}}{\mathrm{d}t}\vec{v}\right)\cdot\vec{v}+\vec{v}\cdot\frac{\mathrm{d}}{\mathrm{d}t}\vec{v}=\vec{a}\cdot\vec{v}+\vec{v}\cdot\vec{a}=0.$$

Thus, $$\vec{v}\cdot\vec{a}=0$$, and there can be no tangential component of acceleration. This makes a lot of sense, too. A tangential component would speed the particle up, thus contradicting the assumption of having a tangential component.

Now, if your definition of “constant tangential component” includes a vanishing one, than literally any trajectory with a constant speed satisfies your criteria, not just circular motion.

I hope this helps!

• Thanks, turns out the teacher wrote the question wrong so i was right! – Zach Feb 6 at 16:58

Just to clear out some misconceptions here. Unless the acceleration is 0 it is impossible to have constant velocity. What you're talking about is constant speed, which is the absolute value of the velocity.

Also, no it is not possible. If the acceleration is tangent to the velocity, it would just accelerating or breaking linearly, so a circular movement would not even be possible.

What I assume that you mean is an acceleration in a direction perpendicular to the velocity. Even here, it is not necessarily a circle. Just consider the 2D version of this problem. Then you will always have two possible directions for the acceleration.

A function that has constant magnitude of acceleration perpendicular to the velocity, given that the linear speed along the curve is constant is $$y=\sqrt{1-(\arccos(\cos(\pi x))/\pi)^2}$$ and it looks like repeating half circles. I found that example here. You could argue that the constant magnitude of acceleration does not hold for $$x=2n+1$$ where $$n$$ is an integer, but in 3D it would be possible to get around that. Moving along a helix is one example. The movement can be described with the parametric equation $$x=r\cos t$$, $$y=r\sin t$$, $$z=ct$$ . This movement has constant speed and constant magnitude of the acceleration.

I don't know if it is possible in 2D.

Also, I might add that if the acceleration has constant magnitude and is perpendicular to the velocity, then the speed will automatically be constant, so the condition about constant speed is not necessary.

• Thanks, I meant constant speed. And that was a question on our test, it said tangential acceleration which didn't make sense to me because I didn't get how it could be tangent to another vector. – Zach Feb 6 at 1:02
• The curvature is constant, except for $x=kn+2$ – klutt Feb 6 at 11:07
• @AaronStevens Well, that was incorrect. I meant $x=2n+1$. I have corrected that now. Maybe I should stop answering while I'm having a fever. – klutt Feb 6 at 12:37