# Math Help - Differential Equation Problem

1. ## Differential Equation Problem

I have been struggling with this problem for a while:

A bullet passes through a piece of wood 0.1 m thick. It enters the wood at 400 m/s and leaves at 180 m/s. The velocity v of the bullet is assumed to obey the differential equation dv/dt = -kv^2 where k is a positive constant. How long does it take the bullet to pass through the wood?

I got this far:

dv/dt = -kv^2
-1/v = -kt + C
v = 1/(kt + C)

400 = v(0) = 1/(k(0) + C)
400 = 1/C
.0025 = C

v = 1/(kt + .0025)

From here, I am not sure what to do. I tried to find k by making v = dx/dt and solving the displacement equation, which turned out to be another dead end.

Any help would be appreciated. Thanks.

2. Originally Posted by machi4velli
I have been struggling with this problem for a while:

A bullet passes through a piece of wood 0.1 m thick. It enters the wood at 400 m/s and leaves at 180 m/s. The velocity v of the bullet is assumed to obey the differential equation dv/dt = -kv^2 where k is a positive constant. How long does it take the bullet to pass through the wood?

I got this far:

dv/dt = -kv^2
-1/v = -kt + C
v = 1/(kt + C)

400 = v(0) = 1/(k(0) + C)
400 = 1/C
.0025 = C

v = 1/(kt + .0025)

From here, I am not sure what to do. I tried to find k by making v = dx/dt and solving the displacement equation, which turned out to be another dead end.

Any help would be appreciated. Thanks.
I don't know about the DE, but
(average velocity)*time = distance
So,
(180 +400)/2 *t = 0.1
t = 0.1 /290 = 0.000345 sec -------------answer.

------------------------
Edit.

Ooops, sorry, that is wrong because the acceleration, dv/dt, is not constant.

Now I have to solve that through the DE. Heck.

---------------------
dv/dt = -kv^2 ---------------------(i)
dv /v^2 = -k dt
-1/v = -kt +C
v = 1/(kt +C) ----------------------(ii)

When v=400m/sec, t=0,
So, in (i), 400 = 1/(k*0 +C)
C = 1/400 = 0.0025

Hence, v = 1/(kt +0.0025) -----------(iii)
Develop that,
kt +0.0025 = 1/v
kt = 1/V -0.0025
kt = (1 -0.0025v)/v
k = (1 -0.0025v) /(vt) ---------------(iv)
Substitute that into (i)
dv/dt = -[(1 -0.0025v) /(vt)](v^2)
dv/dt = [(0.0025v -1)(v)] /t
dv/dt = (0.0025v^2 -v) /t
Cross multiply,
(dv)t = (0.0025v^2 -v)dt
dv /(0.0025v^2 -v) = dt /t -------------(v)

According to the Wolfram Integrator,
INT[1 /(0.0025x^2 -x)]dx = ln(400 -0.1x) -ln(x), so,

Integrate both sides of (v),
ln(400 -0.1v) -ln(v) = ln(t) +C
ln[(400 -0.1v)/v] = ln(t) +C -------------(vi)

That pesky C!
Now I'm stalled.

Heck, suppose C = 0,
[I don't know the consequence of letting C=0 here. Would the answers be wrong? I don't know. I just want to get out of this trap.]
ln[(400 -0.1v)/v] = ln(t)
(400 -0.1v)/v = t --------------------------(vii)
When the bullet comes out, the v=180 m/sec, so,
(400 -0.1*180)/180 = t
t = 382/180 = 2.12 sec.

Therefore, the bullet will pass through the wood in 2.12 seconds after it hit the wood. ----answer.

3. Originally Posted by machi4velli
I have been struggling with this problem for a while:

A bullet passes through a piece of wood 0.1 m thick. It enters the wood at 400 m/s and leaves at 180 m/s. The velocity v of the bullet is assumed to obey the differential equation dv/dt = -kv^2 where k is a positive constant. How long does it take the bullet to pass through the wood?

I got this far:

dv/dt = -kv^2
-1/v = -kt + C
v = 1/(kt + C)

400 = v(0) = 1/(k(0) + C)
400 = 1/C
.0025 = C

v = 1/(kt + .0025)

From here, I am not sure what to do. I tried to find k by making v = dx/dt and solving the displacement equation, which turned out to be another dead end.

Any help would be appreciated. Thanks.
$\frac{dv}{dt} = -kv^2$

$\frac{dv}{v^2} = -kdt$

$\frac{-1}{v} = -kt + C$

So
$v = -\frac{1}{C - kt}$

$v = \frac{1}{kt - C}$ <-- My C is the negative of yours, is the only difference.

So v = 400 at t = 0, thus
$400 = \frac{1}{-C}$

$C = -\frac{1}{400}$

Thus
$v = \frac{1}{kt - \frac{-1}{400}} = \frac{400}{400kt + 1}$

We still have that pesky k.

$v = \frac{dx}{dt}$

$dx = v dt = \frac{400}{400kt + 1} dt$

$\int_{x_0}^x dx = \int_0^y \frac{400dt}{400kt + 1}$ <-- y is the time at which the bullet leaves the wood.

$x - x_0 = 0.1 = 400 \int_0^y \frac{dt}{400kt + 1}$

Let $T = 400kt + 1$ ==> $dT = 400k dt$

$0.1 = 400 \int_{1}^{400ky + 1} \frac{\frac{dT}{400k}}{T}$

$0.1 = \frac{1}{k} \int_{1}^{400ky + 1} \frac{dT}{T}$

$0.1 = \frac{1}{k} \cdot ln(T)|_1^{400ky+1}$

$0.1k = ln(400ky+1) - ln(1) = ln(400ky+1)$

So
$0.1k = ln(400ky + 1)$

We can get another equation in k and y by using the v(t) equation. We know that v = 180 when t = y, so
$180 = \frac{400}{400ky + 1}$

We have two equations in two unknowns.

Solve the v equation for $400ky + 1$:
$400ky + 1 = \frac{400}{180}$

Thus the ln equation says:
$0.1k = ln \left ( \frac{400}{180} \right )$

This gives a value for k, and you can use that in the v equation to get a value for y. I get that k = 7.98508 and y = 0.000383 s.

-Dan