# Acceleration question

1. May 4, 2004

### LD_90

The acceleration of a particle moving at constant speed is perpendicular to its instantaneous velocity. How can I show this by using the fact that
d(v^2)/dt=d(v. v )/dt=0?

This doesn't seem like the most direct way to look at the concept. Help me out.

2. May 4, 2004

### arildno

I think that it is, indeed, the most direct way, since you utilize the sole information you have been given, namely that the speed is constant.
Since the speed is constant, you have the equation d/dt(v^(2))=0 immediately at your disposal.

3. May 4, 2004

### arildno

I've thought a bit more of your complaint; possibly you would prefer a more
"physical" argument, rather than what perhaps seems like mathematical "trickery" and "waving with the magic wand".
So, I'll give a second argument, based on Newton's second law:
(Forgive me for not "LATEX"-ing the code, I'll use the symbol "." for the dot product in the remainder)

Newton's second law states:
F=ma

Form the dot product:

F.v=ma.v

Integrate this from 0 to an arbitrary T:

int(0 to T)F.vdt=1/2mv^(2)(t=T)-1/2mv^(2)(t=0)

Since the speed is given as constant in time, we necessarily have:

int(0 to T)F.vdt=0, for ARBITRARY CHOICE OF T!!!!

This means, that we necessarily have F.v=0 for all times!
Or, going back to Newton's 2 law again:
0=ma.v-->a.v=0, a is perpendicular on v

4. May 6, 2004

### LD_90

Thank you arildno. It all clicked after I read your first post.