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A particle starting from rest so that its velocity varies as the nth power of the distance described from the commencement of the motion. Prove that $\mathbb{n <= 0.5}$.

I know we need to express $\mathbb{v = k(x-a)^n}$ and need to prove that a particle starting from rest cannot have acceleration directly proportional to distance covered. I am unable to see why.

subrat verma
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2 Answers2

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Hint You can define the starting point as zero, getting rid of the $a$. Now you have $\frac {dx}{dt}=kx^n$. What happens if $n=1$ and you integrate this? I don't see why there is a problem on $(0.5,1)$

Ross Millikan
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Place origin at the starting point. We are to assume that the velocity $dx/dt=v(t)=kx^n$ for all $t\in[0,\infty)$ for some constant $k$. It follows $$ \frac{dx}{x^n}=k\,dt. $$ Integrating gives $kt+C=\frac{1}{1-n}x^{1-n}$, so the initial condition $t=0, x=0$ implies $C=0$, $n<1$. Thus $$ x=((1-n)kt)^{1/(1-n)}. $$ Differentiating twice gives $a(t)=K t^{1/(1-n)-2}$ (I absorbed a number of things to another constant $K$). Infinite accelerations are physically impossible, but what happens if $t\to0$ from above and $n>1/2$.

Jyrki Lahtonen
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  • Can we think of it like this: let $\mathbb{v=kx^n}$ then a=vdv/dx=knx^(2n-1), so if 2n-1>0; both v and a will become zero at time t=0, hence the particle remains at rest. – subrat verma Oct 05 '13 at 06:37