Position, velocity and acceleration of a particle

drew.walker · #1 $Old$ June 1st, 2009, 05:58 AM

I have a problem for uni that I don't really know how to approach. The problem is:

"The velocity of a particle at time t = v(t) = sinti + costj - 10tk. What is the particle's acceleration, a(t) = dv/dt as a function of time? Given that at t = 0 the particle is at (1,0,0), find its position at time t."

I understand the relationship between position, velocity and acceleration. i.e. - If x(t) is the position at time t, then the velocity v(t) = dx/dt and the acceleration a(t) = dv/dt. What I don't understand is how to treat the i, j and k in the expression? Just for my own sanity I'm thinking of t like I'd usually think about an x variable, but are i, j and k supposed to be treated as constants? If they didn't exist, I'd have no problem integrating/differentiatiating, but their occurence has completely thrown me off. Any help would be greatly appreciated.

mr fantastic · #2 $Old$ June 1st, 2009, 06:07 AM

Quote:

Originally Posted by drew.walker $View Post$

I have a problem for uni that I don't really know how to approach. The problem is:

"The velocity of a particle at time t = v(t) = sinti + costj - 10tk. What is the particle's acceleration, a(t) = dv/dt as a function of time? Given that at t = 0 the particle is at (1,0,0), find its position at time t."

I understand the relationship between position, velocity and acceleration. i.e. - If x(t) is the position at time t, then the velocity v(t) = dx/dt and the acceleration a(t) = dv/dt. What I don't understand is how to treat the i, j and k in the expression? Just for my own sanity I'm thinking of t like I'd usually think about an x variable, but are i, j and k supposed to be treated as constants? If they didn't exist, I'd have no problem integrating/differentiatiating, but their occurence has completely thrown me off. Any help would be greatly appreciated.

If $r = x_1(t) i + x_2(t) j + x_3(t) k$ then $\frac{dr}{dt} = \frac{d x_1}{dt} i + \frac{d x_2}{dt} j + \frac{dx_3}{dt} k$ .

drew.walker · #3 $Old$ June 1st, 2009, 06:15 AM

Your response is a little over my head, but would I be right in saying $\frac {dv}{dt} = costi - sintj - 10k$

mr fantastic · #4 $Old$ June 1st, 2009, 06:16 AM

Quote:

Originally Posted by drew.walker $View Post$

Your response is a little over my head, but would I be right in saying $\frac {dv}{dt} = costi - sintj - 10k$

Yes.

drew.walker · #5 $Old$ June 1st, 2009, 06:29 AM

Thanks for the help. My only remaining confusion lies with the second part of the question in relation to the position at time t.

I think I'm right in saying that $x(t) = \int v(t) dt$ , which I thought would mean that $x(t) = -costi + sintj - 5t^2k$

However, -cos(0) = -1 (which contradicts the position at t = 0 of (1,0,0)). Is there something wrong with the logic I've applied here? Unfortunately I'm mainly doing this based on intuition as our course material is pretty poor.

mr fantastic · #6 $Old$ June 1st, 2009, 06:50 AM

Quote:

Originally Posted by drew.walker $View Post$

Thanks for the help. My only remaining confusion lies with the second part of the question in relation to the position at time t.

I think I'm right in saying that $x(t) = \int v(t) dt$ , which I thought would mean that $x(t) = -costi + sintj - 5t^2k$

However, -cos(0) = -1 (which contradicts the position at t = 0 of (1,0,0)). Is there something wrong with the logic I've applied here? Unfortunately I'm mainly doing this based on intuition as our course material is pretty poor.

$\vec{r} = -\cos (t) \vec{i} + \sin (t) \vec{j} - 5 t^2 \vec{k} + \vec{C}$ where $\vec{C}$ is a constant vector.

Given: At t = 0, $\vec{r} = \vec{i}$ .

But at t = 0, $\vec{r} = -\vec{i} + \vec{C}$ .

Therefore $\vec{i} = -\vec{i} + \vec{C} \Rightarrow \vec{C} = 2 \vec{i}$ .

Therefore $\vec{r} = -\cos (t) \vec{i} + \sin (t) \vec{j} - 5 t^2 \vec{k} + 2 \vec{i} = [2 -\cos (t)] \vec{i} + \sin (t) \vec{j} - 5 t^2 \vec{k}$ .