differential equation about a forensic test

Question

So in a forensic test a bullet is fired into a box containing a resistive medium ( for example.Defining $x(t)$ to be the horizontal distance covered by the bullet over time t,it is known that

$$\ddot{x}=-\mu \dot{x}^{(3/2)}$$

for some given constant $\mu >0$ if $x(0)=0$ $x'(0)=V$ (where $V >0)$ find an expression for $x(t)$ for $t \geq 0$ how far does the bullet travel before it stops?

I don't know how to solve this using any of the methods I've learned any help is much appreciated

Answer 1

Set

$y = \dot x; \tag 1$

then the equation

$\ddot x = -\mu \dot x^{3/2} \tag 2$

may be written

$\dot y = - \mu y^{3/2}, \tag 3$

or

$y^{-3/2} \dot y = - \mu; \tag 4$

we have

$\dfrac{d}{dt}(y^{-1/2}) = -\dfrac{1}{2} y^{-3/2} \dot y, \tag 5$

or

$-2 \dfrac{d}{dt} (y^{-1/2}) = y^{-3/2} \dot y; \tag 6$

substituting (6) into (3) we obtain

$-2 \dfrac{d}{dt} (y^{-1/2}) = -\mu, \tag 7$

whence

$\dfrac{d}{dt} (y^{-1/2}) = \dfrac{1}{2} \mu; \tag 8$

therefore, since

$y(0) = \dot x(0) = V, \tag 9$

$y^{-1/2}(t) - V^{-1/2} = y^{-1/2}(t) - y^{-1/2}(0)$ $= \displaystyle \int_0^t \dfrac{d}{ds} (y^{-1/2}(s)) \; ds = \int_0^t \dfrac{\mu}{2} \; ds = \dfrac{\mu}{2} t, \tag{10}$

whence

$y^{-1/2}(t) = \dfrac{\mu}{2}t + V^{-1/2} = \dfrac{\mu t + 2V^{-1/2}}{2}, \tag{11}$

$y^{1/2}(t) = \dfrac{2}{\mu t + 2V^{-1/2}}; \tag{12}$

thus,

$\dot x(t) = y(t) = \dfrac{4}{(\mu t + 2V^{-1/2})^2}; \tag{13}$

we see from this equation that $\dot x(t) > 0$ for all finite $t$ and that $\dot x(t) \to 0$ as $t \to \infty$.

The distance the bullet travels in time $t$ is thus, since $x(0) = 0$,

$x(t) = x(t) - x(0) = \displaystyle \int_0^t \dot x(s) \; ds = \int_0^t \dfrac{4}{(\mu s + 2V^{-1/2})^2} \; ds; \tag{14}$

this integral is relatively easy to evaluate:

$\displaystyle \int_0^t \dfrac{4}{(\mu s + 2V^{-1/2})^2} \; ds = \dfrac{1}{\mu^2} \int_0^t \dfrac{4}{(s + 2\mu^{-1} V^{-1/2})^2} \; ds = \dfrac{1}{\mu^2} \left [ \dfrac{-4}{(s + 2 \mu^{-1} V^{-1/2})} \right ]_0^t$ $= \dfrac{1}{\mu^2} \left [ \dfrac{-4}{(t + 2 \mu^{-1} V^{-1/2})} - \dfrac{-4}{2 \mu^{-1} V^{-1/2}} \right ] = \dfrac{1}{\mu^2} \left [ \dfrac{4}{2 \mu^{-1} V^{-1/2}} - \dfrac{4}{(t + 2 \mu^{-1} V^{-1/2})} \right ]; \tag{15}$

finally,

$x(t) = \dfrac{1}{\mu^2} \left [ \dfrac{4}{2 \mu^{-1} V^{-1/2}} - \dfrac{4}{(t + 2 \mu^{-1} V^{-1/2})} \right ]; \tag{16}$

in the limit as $t \to \infty$, the bullet stops, and

$x(\infty) = \dfrac{1}{\mu^2} \left [ \dfrac{4}{2 \mu^{-1} V^{-1/2}} - \right ] = \dfrac{2V^{1/2}}{\mu} \tag{17}$

is the total distance it travels.