Forces and calculus problem

• Mar 7th 2010, 05:17 AM
NeilT
Forces and calculus problem
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.

Q: A particle of mass m kg moves in a horizontal straight line from the origin O with initial velocity U i $ms^{-1}$, where i is the unit vector in the direction of motion. A resistive force $-mkv^3$i acts on the particle, where $k$ is a constant and $v$i is the velocity of the particle at time $t$ seconds measured from the start of the motion.

(i) Show that the velocity of the particle satisfies the differential equation
$\frac{dv}{dx} = -kv^2$,
where $x$ is the distance of the particle from $O$.
Hence show that $v = \frac{U}{1 + kUx}$.

(ii) Using (i) or otherwise, show that
$kUx^2 + 2x = 2Ut$.

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

Solution:

(i) $F = -mkv^3$

$ma = -mkv^3$

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

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

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

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

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

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

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

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

$kvxU + v - U = 0$

$v(kUx + 1) = U$

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

(ii)
$\frac{dx}{dt} = \frac{U}{1+kUx}$

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

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

so $c=0$

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

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

From $v= \frac{U}{1+ kUx}$, v will be U/2 when $\frac{U}{2}= \frac{U}{1+ kUx}$. Solve that for x.
Put that value of x into $kUx^2+ 2x= 2Ut$ and solve for t.
Thankyou Halls of Ivy, I got $t=\frac{4}{kU^2}$