The velocity of a bullet is reduced from `400 ms^(-1)` to `100 ms^(-1)` while travelling thought material of thickness of 1 m. The retardation will be

Options

A. `7.5 * 10^4 m/s^2`
B. `8.5 * 10^3 m/s^2`
C. `7.5 * 10^2 m/s^2`
D. `8.5 * 10^2 m/s^2`

Answer

To solve the problem of finding the retardation of a bullet that slows down from a velocity of 400 m/s to 100 m/s while traveling through a material of thickness 1 m, we can use the equations of motion.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial velocity (U) = 400 m/s
- Final velocity (V) = 100 m/s
- Distance traveled (s) = 1 m

2. **Use the Third Equation of Motion:**
The third equation of motion states:
\[ V^2 = U^2 + 2as \]
Here, \(a\) is the acceleration (or retardation in this case).

3. **Rearranging the Equation:**
Since we are dealing with retardation, we can rewrite the equation as:
\[ V^2 = U^2 - 2as \]
This is because retardation will have a negative effect on the velocity.

4. **Substituting the Known Values:**
Substitute \(V\), \(U\), and \(s\) into the equation:
\[ (100)^2 = (400)^2 - 2a(1) \]

5. **Calculating the Squares:**
\[ 10000 = 160000 - 2a \]

6. **Rearranging to Solve for \(a\):**
\[ 2a = 160000 - 10000 \]
\[ 2a = 150000 \]
\[ a = \frac{150000}{2} = 75000 \, \text{m/s}^2 \]

7. **Finding the Retardation:**
Since retardation is the negative acceleration, we have:
\[ \text{Retardation} = -a = -75000 \, \text{m/s}^2 \]

### Final Answer:
The retardation of the bullet is \(75000 \, \text{m/s}^2\).