A train approaches a small town with constant velocity of `+28.6` m/s. The operator applies the brake, reducing the train's velocity to `+11.4` m/s . If the average acceleration of the the train during braking is `-1.35 m//s^(2)`, for what elapsed time does the operator apply the brake ?

Options

A. 8.44 s
B. 3.38 s
C. 12.7 s
D. 5.92 s

Answer

To solve the problem of the train's braking, we will use the formula for average acceleration. The average acceleration \( a \) is defined as the change in velocity \( \Delta v \) divided by the time taken \( \Delta t \):

\[ a = \frac{\Delta v}{\Delta t} \]

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial velocity \( v_1 = +28.6 \, \text{m/s} \)
- Final velocity \( v_2 = +11.4 \, \text{m/s} \)
- Average acceleration \( a = -1.35 \, \text{m/s}^2 \)

2. **Calculate the Change in Velocity:**
- The change in velocity \( \Delta v \) can be calculated as:
\[ \Delta v = v_2 - v_1 = 11.4 \, \text{m/s} - 28.6 \, \text{m/s} = -17.2 \, \text{m/s} \]

3. **Use the Average Acceleration Formula:**
- Rearranging the formula for average acceleration gives us:
\[ \Delta t = \frac{\Delta v}{a} \]

4. **Substitute the Known Values:**
- Now, substituting the values we have:
\[ \Delta t = \frac{-17.2 \, \text{m/s}}{-1.35 \, \text{m/s}^2} \]

5. **Calculate the Time:**
- Performing the calculation:
\[ \Delta t = \frac{17.2}{1.35} \approx 12.74 \, \text{s} \]

6. **Conclusion:**
- The elapsed time during which the operator applies the brake is approximately \( 12.74 \, \text{s} \).

### Final Answer:
The elapsed time during which the operator applies the brake is approximately **12.74 seconds**.

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