A particle moves in a medium under the influence of a retard

A particle moves in a medium under the influence of a retarding force equal to mk(vA3+aA2v) where k and a are constants. Show that for any value of the initial speed the particle will never move a distance greater than Pi/2ka and that the particle comes to rest only for t rightarrow infinity. Apply Newton\'s second law. Replace acceleration (a) with (v dv/dx), separate variables and integrate to get x=x(v). What is the largest x can be? Go back to the newtons law equation, replace a with (dv/dt). Separate and integrate. At what time(s) is v=0?

Solution

a)

F = -mk(v^3 + a^2*v)

So,

a = F/m = -k(v^3 + a^2*v)

Now,

a = v*dv/dx

So, v*dv/dx = -k*(v^3+a^2*v)

So, dv/dx = -k*(v^2 + a^2)

So, dv/(v^2 +a^2) = -k*dx

So, by integrating both sides, we get

atan(v/a)/a = -k*x + P <------ P = some constant

Let the initial velocity be u = 0

So, at x = 0 , atan(u/a)/a = P

So, u = tan(Pa)*a = 0

So, P = 0

maximum value of atan(v/a) = Pi/2

So, for maximum distance covered, Xmax ,

So, k*Xmax = Pi/2a <------- assuming P to be 0 <---- for initial velocity =0

So, Xmax = Pi/2ka <--------answer(Proved)

b)

a = -k*(v^3 + a^2*v)

Now, a = dv/dt

So, -dv/(v^3 + a^2*v) = k*dt

Integrating both sides, we get :

(1/2a^2)*ln((v^2+a^2)/v^2) = k*t

So, (1/2a^2)*ln(1 + a^2/v^2) = k*t

So, for v = 0 <------- it comes to rest,

So, (1/2a^2)*ln (infinite) = k*t

So, t = inifinite <------- PROVED