# Kinematics moving particle problem

1. Sep 11, 2010

### crybllrd

1. The problem statement, all variables and given/known data
The position of a particle moving along the x axis depends on the time according to the equation

x=ct$$^{2}$$-bt$$^{3}$$

where x is in meters and t in seconds. What are the units of (a) constant c and (b) constant b? Let their numerical values be 3.0 and 2.0 respectively.(c) At what time does the particle reach its maximum positive x position? From t=0 to t=2.0, (d) what distance does the particle move and (e) what is its displacement? Find its velocity at times (f) 1.0 (g) 2.0 (h) 3.0 and (i)4.0. Find its acceleration at times (j) 1.0 (k) 2.0 (l) 3.0 and (m)4.0.

2. Relevant equations

x=ct$$^{2}$$-bt$$^{3}$$

3. The attempt at a solution

I can't even get started on part a. I am not sure how to find the constants. The only thing I could think of would be using the final unit (m), so in that case one would have to be meters and the other seconds?
Am I on the right track?

2. Sep 11, 2010

### rock.freak667

You have x=ct2-bt3

'x' is in meters(m), so that the quantities ct2 and bt3 must also have units meters(m).

You know that 't' is in seconds (s), so 't2' has units s2, so what should the units of 'c' be such that (units of c)*(units of t2)= meters?

3. Sep 12, 2010

### crybllrd

OK so for part A I have c=$$\frac{m}{s^{2}}$$ and B as b=$$\frac{m}{s^{3}}$$. I believe the unit s should cancel out.
On part c I need to find a relative maximum. I can look at the graph and tell that t should equal 1. I was thinking to use the formula x=$$\frac{-b}{2a}$$ then plug x in for t in the given equation, but I don't think I can use that because it is not a quadratic.