A rocket sled accelerates from rest to top speed at 282 m/s in 5.0 seconds and was brought to rest in only 1.40 seconds. What is the acceleration rate while slowing down in m/s^2? What is the displacement on the accelerating part of the trio and what is the displacement during the decelerating part of the trip?

We can solve this problem in three parts:

1. **Acceleration while slowing down:**
   We are given:
   - Initial velocity during deceleration, \( v_i = 282 \, \text{m/s} \)
   - Final velocity, \( v_f = 0 \, \text{m/s} \)
   - Time taken for deceleration, \( t = 1.40 \, \text{seconds} \)

   The formula for acceleration is:

   \[
   a = \frac{{v_f - v_i}}{{t}}
   \]

   Substituting the values:

   \[
   a = \frac{{0 - 282}}{1.40} = \frac{{-282}}{1.40} = -201.43 \, \text{m/s}^2
   \]

   The acceleration rate while slowing down is \( -201.43 \, \text{m/s}^2 \) (the negative sign indicates deceleration).

---

2. **Displacement during the accelerating part of the trip:**
   We are given:
   - Initial velocity during acceleration, \( v_i = 0 \, \text{m/s} \)
   - Final velocity, \( v_f = 282 \, \text{m/s} \)
   - Time taken for acceleration, \( t = 5.0 \, \text{seconds} \)

   First, we calculate the acceleration during this part using the same acceleration formula:

   \[
   a = \frac{{v_f - v_i}}{{t}} = \frac{{282 - 0}}{5.0} = 56.4 \, \text{m/s}^2
   \]

   Now, to find the displacement, we use the equation:

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

   Since \( v_i = 0 \), the formula simplifies to:

   \[
   d = \frac{1}{2} a t^2 = \frac{1}{2} (56.4) (5.0)^2 = \frac{1}{2} (56.4) (25) = 705 \, \text{meters}
   \]

   The displacement during the accelerating part is \( 705 \, \text{meters} \).

---

3. **Displacement during the decelerating part of the trip:**
   We can use the following kinematic equation:

   \[
   d = \frac{v_i + v_f}{2} \times t
   \]

   Substituting the values:

   \[
   d = \frac{282 + 0}{2} \times 1.40 = \frac{282}{2} \times 1.40 = 141 \times 1.40 = 197.4 \, \text{meters}
   \]

   The displacement during the decelerating part is \( 197.4 \, \text{meters} \).

---

### Summary of Results:
- **Acceleration while slowing down:** \( -201.43 \, \text{m/s}^2 \)
- **Displacement during acceleration:** \( 705 \, \text{meters} \)
- **Displacement during deceleration:** \( 197.4 \, \text{meters} \)

---

Would you like further details or have any other questions?

### Related Questions:
1. What would be the total distance covered during the entire trip?
2. How does the time of deceleration compare with the time of acceleration in terms of displacement covered?
3. What is the average velocity during the decelerating phase?
4. If the rocket sled were to slow down over 2.0 seconds instead of 1.40, what would the deceleration rate be?
5. How does the negative acceleration (deceleration) affect the total energy of the system?

### Tip:
Always ensure to keep units consistent throughout calculations to avoid errors.

A rocket sled accelerates from rest to top speed at 282 m/s in 5.0 seconds and was brought to rest in only 1.40 seconds. What is the acceleration rate while slowing down in m/s^2? What is the displacement on the accelerating part of the trip and what is the displacement during the decelerating part of the trip?

Learn how to calculate the acceleration while slowing down a rocket sled from 282 m/s in 1.40 seconds and the displacement during acceleration and deceleration. This problem involves kinematic equations and is suitable for Grades 9-12.

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