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 kinematic equation:

\[ v^2 = u^2 + 2as \]

Where:
- \( v \) = final velocity (0 m/s, since the bullet comes to rest)
- \( u \) = initial velocity (350 m/s)
- \( a \) = acceleration (which we are trying to find)
- \( s \) = distance penetrated (5 cm, which we need to convert to meters)

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial velocity, \( u = 350 \, \text{m/s} \)
- Final velocity, \( v = 0 \, \text{m/s} \)
- Distance, \( s = 5 \, \text{cm} = 0.05 \, \text{m} \) (conversion from cm to m)

2. **Substitute the values into the kinematic equation:**
\[ v^2 = u^2 + 2as \]
Plugging in the known values:
\[ 0^2 = (350)^2 + 2a(0.05) \]

3. **Simplify the equation:**
\[ 0 = 122500 + 0.1a \]

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

5. **Express the answer in scientific notation:**
\[ a = -1.225 \times 10^6 \, \text{m/s}^2 \]

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