Answer to Question #137469 in Classical Mechanics for bimal

Question

A particle leaves the 'origin with velocity u. If
it moves with an acceleration =m/v^2,where v be
the velocity at any instant, show that the
distance x travelled in time t is given by
4 cx= (3 ct+ c^2)4/3 - u^4.

Answer 1

As per the question,

Initial velocity of the particle is u,

Acceleration of the particle $(a)=\frac{m}{v^2}$

Let the distance traveled by the particle in the time t is x

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

$\Rightarrow dv= a dt$

Now, substituting the values of a in the above,

$dv=\frac{m}{v^2}dt$

Now, taking the integration,

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

$\Rightarrow \int_u^v v^2dv=\int_0^t mdt$

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

$\Rightarrow \frac{v^3-u^3}{3}=mt$

$v=(3mt+u^3)^{1/3}$

Now, we know that $v= \frac{dx}{dt}$

$dx=vdt$

Now, taking the integration of both side,

$\int_0^xdx=\int_0^tvdt$

Now , substituting the values,

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

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

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

$\Rightarrow 4mx=(3mt+u^3)^{4/3}-u^4$