Here is a problem i've been working on. I think i've got the first two parts but I'm stuck on the last section. I'd appreciate anyone checking my solution so far and even better helping me with the third section.

Thanks in advance!

Q: A particle of massmkg moves in a horizontal straight line from the originOwith initial velocityUi$\displaystyle ms^{-1}$, whereiis the unit vector in the direction of motion. A resistive force $\displaystyle -mkv^3$iacts on the particle, where $\displaystyle k$ is a constant and $\displaystyle v$iis the velocity of the particle at time $\displaystyle t$ seconds measured from the start of the motion.

(i) Show that the velocity of the particle satisfies the differential equation

$\displaystyle \frac{dv}{dx} = -kv^2$,

where $\displaystyle x$ is the distance of the particle from $\displaystyle O$.

Hence show that $\displaystyle v = \frac{U}{1 + kUx}$.

(ii) Using (i) or otherwise, show that

$\displaystyle kUx^2 + 2x = 2Ut$.

(iii) Find an expression, in terms ofkandU, for the time taken for the speed of the particle to reduce to half its initial value.

Solution:

(i) $\displaystyle F = -mkv^3$

$\displaystyle ma = -mkv^3$

$\displaystyle v\frac{dv}{dx} = -kv^3$

$\displaystyle \frac{dv}{dx} = -kv^2$ as required

then

$\displaystyle \int \frac{1}{kv^2} dv = \int - dx$

$\displaystyle - \frac{1}{kv} = -x + c$ but at $\displaystyle x=0 , v=U$

so $\displaystyle c = -\frac{1}{kU}$

and $\displaystyle \frac{1}{kv} = -x - \frac{1}{kU}$

$\displaystyle -1 = -kvx - \frac{kv}{kU}$

$\displaystyle kvx - 1 = -\frac{v}{U}$

$\displaystyle kvxU + v - U = 0$

$\displaystyle v(kUx + 1) = U$

$\displaystyle v = \frac{U}{1+kUx}$as required

(ii)

$\displaystyle \frac{dx}{dt} = \frac{U}{1+kUx} $

$\displaystyle \int (1 + kUx)dx = \int U dt$

$\displaystyle x + \frac{1}{2} kUx^2 = Ut + c$ but at $\displaystyle x=0, t=0$

so $\displaystyle c=0$

and $\displaystyle \frac{1}{2} kUx^2 + x = Ut$

so $\displaystyle kUx^2 + 2x = 2Ut$ as required.

(iii) Help please!