I am given the following question:

The trajectory of a charged particle moving in a magnetic field is given by $\textbf{r}=b\cos(\beta t)\textbf{i}+b\sin(\beta t)\textbf{j}+ct\textbf{k}$ where $b,\beta$ and $c$ are positive constants. Show that the particle moves with constant speed and find the magnitude of its acceleration.

I have calculated $v(t)$ and $a(t)$ using first and second order derivatives. As far as I understand, if speed is constant, then acceleration should be zero, so the second derivative of $r$ is zero. Also, I think that the slope of the first derivative should be constant if we are given constant speed. However, I don't know how to combine all of these facts to show that speed is constant and then find the magnitude of acceleration (which is, in my opinion, zero). How does the acceleration relate to the rate of change of the speed, and how is the acceleration not $0$ for a constant speed?


1 Answer 1


The speed of a particle is defined to be the magnitude of its velocity vector. But you already found the velocity vector; so...

Note, however, that the magnitude of a particle's acceleration is not equal to the rate of change of its speed. This problem seems to be designed to illustrate why this is the case.

  • $\begingroup$ I understood the velocity and speed part. And for the acceleration, it is not zero, because our speed is constant and not the velocity, so we have an expression of acceleration depending on time, and from that, we find the magnitude of acceleration? $\endgroup$
    – draregon
    Aug 30, 2021 at 20:30
  • $\begingroup$ @AregKarapetyan: That's right. $\endgroup$ Aug 30, 2021 at 20:31
  • $\begingroup$ Thank you very much!! $\endgroup$
    – draregon
    Aug 30, 2021 at 20:33

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