Mechanical acceleration and veloctiy relationship

[awesome4444]

Homework Statement

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

Homework 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 don't know if i should use +3.2 or -3.2
Can somone tell me if I am doing anything wrong? :(

The Attempt at a Solution

 

[grzz]

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

and so v = \intadt but v \neq\intadx.

 

[awesome4444]

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

 

[grzz]

I think that you know that the symbol

\int...dt

means

'integrate ... with respect to t'

 

[grzz]

adt does not mean that a is multiplied by anything.

 

[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?

 

[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

 

[TaxOnFear]

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

Clever.

 

[grzz]

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