Answer to Question #137468 in Classical Mechanics for bimal

Question

A particle leaves the 'origin with velocity u. If
it moves with an acceleration -Liv being
v2
the velocity at any instant, show that the
distance x travelled in time t is given by
4 px= (3 lit+ 112)4/3 - u4.

Answer 1

As per the given question,

It is given that acceleration of the particle $(a)=\frac{Li}{v^2}$

We know that,

$a=\frac{dv}{dt}$

$\Rightarrow dv= adt$

Now, substituting the value of a,

$\int_u^v dv=\int_0^t\frac{Li}{v^2}dt$

$\Rightarrow \int_u^vv^2dv=\int_0^tLidt$

$\Rightarrow [\frac{v^3}{3}]_u^v=Lit$

$\Rightarrow v^3=3Lit+u^3$

$\Rightarrow v=(Lit+u^3)^{1/3}$

We know that,

$v=\frac{dx}{dt}$

$\Rightarrow dx=vdt$

Integrating both side with respect to t,

$x=\int_0^t (Lit+u^3)^{1/3}dt$

$\Rightarrow x =[\frac{(Lit+u^3)^{4/3}}{4Li/3}]_0^t$

$\Rightarrow x=\frac{3}{4Li}[(Lit+u^3)^{4/3}-u^4]$

$\Rightarrow4Lix=3[(Lit+u^3)^{4/3}-u^4]$