celticcat:

A particle, P, moves on the x-axis for time t ≥ 0, in seconds, with velocity: v = 2 /(1 + 3x) , where x, in centimetres, is the displacement from the origin x = 0. Find an expression for acceleration a, and hence, show that a varies directly with v^3. Its the last part which has got me on this one.

satellite73:

second derivative is $\frac{6}{(1+3x)^3}$ with is a multiple of the cube of $\frac{2}{(1+3x)}$

celticcat:

think you

sillybilly123:

nah, the 2nd derivative of v(x) wrt x has no physical meaning there is a very common physics shortcut/ "trick": $$a = v \frac{dv}{dx}$$. that is what you should be using here. and you should see the cubic relationship from the product of $$v(x)$$ and $$v'(x)$$ simple explanation is of "trick" that, here $$v = v(x)$$, so suppose that $$x=x(t)$$, which is of course true but we don't know, nor do we need to know, what that function actually is. so $$v = v(x(t))$$ and we chain it wrt t as: $$\frac{dv}{dt} = a = \frac{dv}{dx}\frac{dx}{dt} = v \frac{dv}{dx}$$ tbc, this give acceleration in terms of displacement x