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### Given information:
- Initial velocity \( v_0 = 83 \) feet per second.
- Constant deceleration \( a = 0.8 \) feet per second squared.

#### 1) **How long will it take the train to stop?**
We can use the formula for time under constant acceleration (or deceleration):

\[
v = v_0 + a \cdot t
\]

Since the final velocity \( v \) is 0 (when the train stops), the equation becomes:

\[
0 = 83 - 0.8 \cdot t
\]

Solve for \( t \):

\[
0.8 \cdot t = 83
\]

\[
t = \frac{83}{0.8}
\]

Now, let's calculate this value.

#### 2) **How many feet will the train travel before it stops?**
We can use the kinematic equation for distance traveled under constant acceleration:

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

Since \( a \) is negative (deceleration), the equation becomes:

\[
d = 83 \cdot t - \frac{1}{2} \cdot 0.8 \cdot t^2
\]

Now, let’s calculate these values.Here are the results:

1) The train will take **103.75 seconds** to stop.

2) The train will travel **4305.625 feet** before it stops.

Would you like any further details or clarifications? 

Here are 5 related questions to deepen your understanding:
1. What would happen if the deceleration rate increased?
2. How would you calculate the stopping time if the initial velocity was halved?
3. What if the deceleration rate changed halfway through the stop—how would the equations change?
4. How is the stopping distance affected by an increase in velocity?
5. How can we calculate the average velocity during the stopping period?

**Tip:** When working with deceleration problems, always remember to treat deceleration as negative acceleration in the formulas.

A train is traveling at a velocity of 83 feet per second when it hits its brakes. It slows down at a constant rate of 0.8 feet per second each second until it stops. How long will it take the train to stop? After it hits its brakes, how many feet will it travel before it stops?

Learn how to calculate the stopping time and distance of a train traveling at 83 feet per second with a constant deceleration of 0.8 feet per second squared. This problem uses kinematic equations to find both the stopping time of 103.75 seconds and the stopping distance of 4305.625 feet. Suitable for high school students in Grades 9-12.

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