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In problem 3, you calculated the average acceleration of the train. To calculate the distance the train traveled before stopping (problem 4), we can use the kinematic equation:

\[
v_f^2 = v_i^2 + 2a \cdot d
\]

Where:
- \( v_f = 0 \, \text{m/s} \) (final velocity, since the train comes to a stop)
- \( v_i = 16.67 \, \text{m/s} \) (initial velocity from the problem)
- \( a = -16670 \, \text{m/s}^2 \) (acceleration from problem 3)
- \( d \) is the distance we want to find

Rearranging the equation to solve for \( d \):

\[
d = \frac{v_f^2 - v_i^2}{2a}
\]

Substituting the values:

\[
d = \frac{0 - (16.67)^2}{2 \times (-16670)}
\]

Now, let's calculate it:

\[
d = \frac{-278.0889}{-33340} = 0.00834 \, \text{m}
\]

So, the correct distance is approximately **0.00834 meters**, not 0.01667 m as selected.

Would you like more details or have any questions?

Here are 5 relative questions to consider:
1. How does the sign of acceleration affect the distance calculation?
2. What does negative acceleration indicate in this context?
3. How would the distance change if the initial velocity were doubled?
4. Why is the final velocity zero in this problem?
5. What other kinematic equations could be used to solve similar problems?

**Tip:** When solving kinematic equations, always double-check unit consistency to avoid calculation errors.

Using the information from problem number 3, calculate the distance that the train traveled before it was brought to a stop by the Man of Steel.

Learn how to solve a kinematics problem involving acceleration and distance. In this scenario, a train moving at 16.67 m/s is brought to rest by Superman in 0.001 seconds. We use kinematic equations to calculate the distance traveled before stopping, with an emphasis on unit consistency and negative acceleration.

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