# Mechanical acceleration and veloctiy relationship

1. Nov 30, 2011

### awesome4444

1. The problem statement, all variables and given/known data

The motion of a particle is expressed by a=42-12x2 where a is in m/s2 and X is in m, with the initial condition: v=0 when x=0 and t=0
Determine v when X=6
Determine X when v becomes 0 again

2. Relevant equations
I tried using a=dv/dt so i took the anti derivative of a to find the equation of v which turned out to be 42x-4x^3 + c and then pluged in 6 for x and found v=-612m/s

but then for the second part when i tried finding the roots of the equation b i have two answers +3.2 and -3.2

So i dont know if i should use +3.2 or -3.2
Can somone tell me if im doin anything wrong? :(

3. The attempt at a solution

2. Dec 1, 2011

### grzz

Note that a = $\frac{dv}{dt}$

and so v = $\int$adt but v $\neq$$\int$adx.

3. Dec 1, 2011

### awesome4444

so we would integrate (a*dt) but im not sure wat dt is :P

4. Dec 1, 2011

### grzz

I think that you know that the symbol

$\int$...dt

means

'integrate ... with respect to t'

5. Dec 1, 2011

### grzz

adt does not mean that a is multiplied by anything.

6. Dec 1, 2011

### TaxOnFear

You can't integrate your equation with respect to t as your equation doesn't rely on t (which it should in an ideal world). Maybe you copied down wrong?

7. Dec 1, 2011

### grzz

One can use the following:

a = $\frac{dv}{dt}$ = $\frac{dv}{dx}$$\frac{dx}{dt}$ = v$\frac{dv}{dx}$

hence ∫a dx = ∫v dv = $\frac{v^{2}}{2}$ + constant

8. Dec 2, 2011

Clever.

9. Dec 2, 2011

### grzz

I was reminded about that other method by a recent post on PHYSICS FORUMS!