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.


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


think you


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