A bullet going with speed 350 m/s enters concrete wall and penetrates a distance of 5.0 cm before coming to rest. Find the deceleration.

Explanation

To find the deceleration of the bullet as it penetrates the concrete wall, we can use the third equation of motion:

\[ V^2 = U^2 + 2aS \]

Where:
- \( V \) = final velocity (0 m/s, since the bullet comes to rest)
- \( U \) = initial velocity (350 m/s)
- \( a \) = acceleration (which will be negative since it's deceleration)
- \( S \) = distance penetrated (5 cm = 0.05 m)

### Step-by-Step Solution:

1. **Convert the distance from centimeters to meters:**
\[ S = 5 \, \text{cm} = 5 \times 10^{-2} \, \text{m} = 0.05 \, \text{m} \]

2. **Substitute the known values into the equation:**
\[ V^2 = U^2 + 2aS \]
\[ 0^2 = (350)^2 + 2a(0.05) \]

3. **Calculate \( U^2 \):**
\[ 0 = 122500 + 0.1a \]

4. **Rearranging the equation to solve for \( a \):**
\[ 0.1a = -122500 \]
\[ a = \frac{-122500}{0.1} \]
\[ a = -1225000 \, \text{m/s}^2 \]

5. **Express the deceleration in a more manageable form:**
\[ a = -1.225 \times 10^6 \, \text{m/s}^2 \]

### Final Answer:
The deceleration of the bullet is \( 1.225 \times 10^6 \, \text{m/s}^2 \).