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.

Question

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 · Answer

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

celticcat · Answer

think you

sillybilly123 · Answer

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