Having gone through a plank of thickness h, a bullet changed its velocity from `v_(0)` to `v_(1)`. Find the time of motion of the bullet on the plank, assuming the resistance force to be proportional to the square of the velocity.

Answer

To solve the problem of finding the time of motion of a bullet as it passes through a plank of thickness \( h \), we will follow these steps:

### Step 1: Understand the relationship between force and velocity
The resistance force \( F \) acting on the bullet is proportional to the square of its velocity \( v \). Therefore, we can express this as:
\[ F = -k v^2 \]
where \( k \) is a constant of proportionality, and the negative sign indicates that the force opposes the motion.

### Step 2: Relate force to acceleration
Using Newton's second law, we know that:
\[ F = m a \]
where \( m \) is the mass of the bullet and \( a \) is its acceleration. Since \( F = -k v^2 \), we can write:
\[ m a = -k v^2 \]
This implies:
\[ a = \frac{-k}{m} v^2 \]

### Step 3: Express acceleration in terms of velocity
We can express acceleration \( a \) as:
\[ a = \frac{dv}{dt} \]
Thus, we have:
\[ \frac{dv}{dt} = -\frac{k}{m} v^2 \]

### Step 4: Rearrange and integrate
Rearranging the equation gives:
\[ \frac{dv}{v^2} = -\frac{k}{m} dt \]
Now, we will integrate both sides. The left side will be integrated from \( v_0 \) to \( v_1 \) (initial to final velocity), and the right side will be integrated from \( 0 \) to \( T \) (time of motion):
\[ \int_{v_0}^{v_1} \frac{dv}{v^2} = -\frac{k}{m} \int_{0}^{T} dt \]

### Step 5: Perform the integration
The integral on the left side yields:
\[ \left[-\frac{1}{v}\right]_{v_0}^{v_1} = -\frac{k}{m} T \]
This simplifies to:
\[ -\frac{1}{v_1} + \frac{1}{v_0} = -\frac{k}{m} T \]
or:
\[ \frac{1}{v_0} - \frac{1}{v_1} = \frac{k}{m} T \]

### Step 6: Relate displacement to velocity
Next, we relate the displacement \( ds \) to the velocity:
\[ a = \frac{dv}{dt} = v \frac{dv}{ds} \]
Thus, we can write:
\[ -k v^2 = v \frac{dv}{ds} \]
This simplifies to:
\[ -k v = \frac{dv}{ds} \]

### Step 7: Integrate with respect to displacement
Rearranging gives:
\[ \frac{dv}{v} = -k ds \]
Integrating from \( v_0 \) to \( v_1 \) and \( 0 \) to \( h \):
\[ \int_{v_0}^{v_1} \frac{dv}{v} = -k \int_{0}^{h} ds \]
This yields:
\[ \ln\left(\frac{v_1}{v_0}\right) = -kh \]
or:
\[ k = -\frac{\ln\left(\frac{v_1}{v_0}\right)}{h} \]

### Step 8: Substitute \( k \) back into the time equation
Substituting \( k \) into the time equation gives:
\[ \frac{1}{v_0} - \frac{1}{v_1} = \frac{-\ln\left(\frac{v_1}{v_0}\right)}{h} \cdot T \]
Rearranging for \( T \):
\[ T = \frac{h \left(\frac{1}{v_0} - \frac{1}{v_1}\right)}{-\ln\left(\frac{v_1}{v_0}\right)} \]

### Final Result
Thus, the time of motion of the bullet through the plank is given by:
\[ T = \frac{h (v_1 - v_0)}{v_1 v_0 \ln\left(\frac{v_0}{v_1}\right)} \]