# Totally confused acceleration and velocity of a particle in vectors(cross product)

1. Nov 12, 2008

### CmbkG

Totally confused!!! acceleration and velocity of a particle in vectors(cross product)

1. The problem statement, all variables and given/known data
The acceleration of a particle is given by a=vx(cxr) where r is the position, v is the velocity and c is a constant.

Show that the following are constants
(a)|v|
(b)c.(rxv)
(c)c.v-1/2|cxr|^2

3. The attempt at a solution
a=c(v.r)-r(v.c)

d/dta=d/dt[c(v.r)-r(v.c)]

2. Nov 12, 2008

### Dick

Re: Totally confused!!! acceleration and velocity of a particle in vectors(cross prod

Something is constant if its time derivative is zero. |v| is constant if v.v is constant. Take d/dt(v.v). Remember d/dt(v)=a.

3. Nov 12, 2008

### CmbkG

Re: Totally confused!!! acceleration and velocity of a particle in vectors(cross prod

Thanks for the help.

Ive got part (a) and it equals zero. Its just im not too sure on what way to even start part (b) or (c). Should i use the cartesian components or how do i find out what r and c actually equal?

4. Nov 12, 2008

### weejee

Re: Totally confused!!! acceleration and velocity of a particle in vectors(cross prod

(b) d(c.(rxv))/dt = c.(vxv) + c.(rxa) = c.(rxa)
Try to show that this equals zero.

(c) d(c.v)/dt = c.a
d(1/2|cxr|^2)/dt = (cxr).(cxv)
Try to show that these two equals each other.

5. Nov 12, 2008

### Dick

Re: Totally confused!!! acceleration and velocity of a particle in vectors(cross prod

How did you get a)? A very similar technique will get you b) and c). Start by taking d/dt of the expressions. For c) you might find the identity (axb).c=a.(bxc) useful.

6. Nov 12, 2008

### CmbkG

Re: Totally confused!!! acceleration and velocity of a particle in vectors(cross prod

Thanks for both your help there, appreciate it alot.

Just wondering though weejee, how did you get d(c.(rxv))/dt to be equal to c.(vxv)+c.(rxa)? where did the vxv and the rxa come from?

7. Nov 12, 2008

### weejee

Re: Totally confused!!! acceleration and velocity of a particle in vectors(cross prod

The differentiation can act on either 'r' or 'v'. If it acts on 'r' it gives c.(vxv) and if it acts on 'v' it leads to c.(rxa).

8. Nov 12, 2008

### CmbkG

Re: Totally confused!!! acceleration and velocity of a particle in vectors(cross prod

ahh, rite, i see what you mean now. Thanks alot.

9. Nov 12, 2008

### Dick

Re: Totally confused!!! acceleration and velocity of a particle in vectors(cross prod

It's called the 'product rule'.