How Do You Calculate Time Ratios in Motion with Drag Force?

[tanaygupta2000]

Homework Statement

At t=0, a particle of mass m having velocity u starts moving through a liquid kept in a horizontal tube and experiences a drag force F = -kdx/dt. It covers a distance L before coming to rest. If the times taken to cover the distances L/2 and L/4 are t2 and t4 respectively, then the ratio t2/t4 (ignoring gravity) is ?

Relevant Equations

F = -kdx/dt

Since given F = -kdx/dt
so I equated mx'' = -kx'
which gave x(t) = A + B exp(-kt/m)
hence v(t) = (-kB/m) exp(-kt/m)
and using v(0) = u, v(t) = u exp(-kt/m)

then I integrated dx = v(t)dt, dx from 0 to L and v(t)dt from 0 to t to find the distance covered L in terms of time taken t.
From this I got L = mu/k (1 - exp(-kt/m) )

Then I substituted L/2 = mu/k (1 - exp(-kt2/m) )
and L/4 = mu/k (1 - exp(-kt4/m) )

Now after this, I'm not getting how to find the ratio t2/t4 from the last two expressions of L/2 and L/4.

 

[Chestermiller]

If it ultimately travels a distance L, then at infinite time, x = L

 

[haruspex]

dx from 0 to L and v(t)dt from 0 to t to find the distance covered L in terms of time taken t.

You are in danger of confusing yourself by using the same symbol, L, for both the given distance covered at t=∞ and the distance covered at some arbitrary time t.