A train is traveling down a straight track at 28 m/s when the engineer applies the brakes, resulting in an acceleration of −1.0 m/s2 as long as the train is in motion. How far does the train move during a 56-s time interval starting at the instant the brakes are applied?

To solve this problem, we can use the following kinematic equation for uniformly accelerated motion:

\[
d = v_0 t + \frac{1}{2} a t^2
\]

where:
- \(d\) is the distance traveled,
- \(v_0\) is the initial velocity,
- \(t\) is the time,
- \(a\) is the acceleration.

### Given values:
- Initial velocity: \(v_0 = 28 \, \text{m/s}\),
- Acceleration: \(a = -1.0 \, \text{m/s}^2\) (negative due to braking),
- Time: \(t = 56 \, \text{seconds}\).

Let's calculate the distance traveled in this 56-second time interval:

\[
d = 28 \times 56 + \frac{1}{2} \times (-1.0) \times 56^2
\]

Now, calculating the terms:

\[
d = 28 \times 56 + \frac{1}{2} \times (-1.0) \times 56^2
\]

\[
d = 1568 + \frac{1}{2} \times (-1) \times 3136
\]

\[
d = 1568 - 1568
\]

\[
d = 0 \, \text{meters}
\]

Since the calculation yields zero distance, this indicates the train comes to a stop before 56 seconds. To determine the actual stopping time, we need to calculate how long it takes for the train to stop.

The stopping time \(t_s\) can be found using the equation:

\[
v = v_0 + at
\]

where \(v = 0\) (the final velocity when the train stops).

\[
0 = 28 + (-1.0) \times t_s
\]

Solving for \(t_s\):

\[
t_s = \frac{28}{1.0} = 28 \, \text{seconds}
\]

So, the train stops after 28 seconds. Now, we calculate the distance traveled during these 28 seconds using the same kinematic equation:

\[
d = 28 \times 28 + \frac{1}{2} \times (-1.0) \times 28^2
\]

\[
d = 784 + \frac{1}{2} \times (-1) \times 784
\]

\[
d = 784 - 392
\]

\[
d = 392 \, \text{meters}
\]

Therefore, the train travels 392 meters before it comes to a stop.

### Final Answer:
The train moves **392 meters** during the 56-second time interval.

---

Would you like any further details or have any questions? Here are 5 related questions you might find interesting:

1. How would the result change if the train's acceleration were different?
2. What is the significance of negative acceleration in this context?
3. How do you calculate the stopping distance for different initial velocities?
4. Can you determine the speed of the train after any given time of braking?
5. What if the brakes were applied for a shorter time interval?

**Tip:** Always double-check units when using kinematic equations to avoid calculation errors!

A train is traveling down a straight track at 28 m/s when the engineer applies the brakes, resulting in an acceleration of −1.0 m/s² as long as the train is in motion. How far does the train move during a 56-s time interval starting at the instant the brakes are applied?

Solve a kinematics problem where a train decelerates from 28 m/s with an acceleration of -1.0 m/s². Learn how to calculate the stopping distance and time using kinematic formulas for uniformly accelerated motion. Suitable for students in Grades 9-12.

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