An object's initial velocity is 1.25  m/s
  in the +x direction. It slows down with a constant acceleration whose magnitude is 1.32  m/s2
  . After it reaches a momentary stop it reverses its direction of motion, to the -x, and speeds up with the same magnitude of the acceleration. What is its displacement (in meters) from the initial moment to  t
  = 7.00  s
  ? Keep 3 digits after the decimal point.

We are given the following information:

- Initial velocity, \( v_0 = 1.25 \, \text{m/s} \) (in the +x direction).
- Magnitude of acceleration, \( a = 1.32 \, \text{m/s}^2 \).
- The motion involves decelerating to a stop, reversing direction, and accelerating back in the opposite direction.

### Step 1: Divide the motion into two parts:
- **First part (0 to stop)**: The object slows down, then stops.
- **Second part (reverse motion)**: After stopping, the object reverses direction and accelerates.

### First Part: Deceleration to a stop
We can use the equation of motion:

\[
v = v_0 + at
\]

At the momentary stop, the velocity becomes zero. So, \( v = 0 \), and we have:

\[
0 = 1.25 + (-1.32) \cdot t_1
\]

Solve for \( t_1 \), the time it takes to stop:

\[
t_1 = \frac{1.25}{1.32} \approx 0.946 \, \text{seconds}
\]

Now, to find the displacement during this time, we use:

\[
x_1 = v_0 t_1 + \frac{1}{2} a t_1^2
\]

Substitute the values:

\[
x_1 = 1.25 \cdot 0.946 + \frac{1}{2} (-1.32) \cdot (0.946)^2
\]

\[
x_1 \approx 1.180 - 0.593 \approx 0.587 \, \text{meters}
\]

### Second Part: Acceleration in the -x direction
In the second part, the object reverses direction and accelerates at \( a = 1.32 \, \text{m/s}^2 \) in the negative x direction. The time for this part of the motion is:

\[
t_2 = 7.00 - t_1 = 7.00 - 0.946 = 6.054 \, \text{seconds}
\]

The initial velocity for this part is 0, so we can find the displacement using:

\[
x_2 = \frac{1}{2} a t_2^2
\]

Substitute the values:

\[
x_2 = \frac{1}{2} \cdot 1.32 \cdot (6.054)^2
\]

\[
x_2 \approx \frac{1}{2} \cdot 1.32 \cdot 36.67 \approx 24.22 \, \text{meters}
\]

### Total Displacement:
The total displacement is the sum of the displacements in both parts:

\[
x_{\text{total}} = x_1 + x_2
\]

\[
x_{\text{total}} = 0.587 + 24.22 \approx 24.807 \, \text{meters}
\]

Thus, the displacement after 7 seconds is approximately **24.807 meters**.

---

Would you like any further details or explanations on the steps? Here are some related questions you might find useful:

1. What is the formula for displacement when acceleration is constant?
2. How do we calculate the time it takes for an object to stop when decelerating?
3. How do you find the total displacement when the motion involves both deceleration and acceleration?
4. What are the key kinematic equations used in these types of problems?
5. How would the solution change if the acceleration was negative throughout the entire motion?

**Tip**: When solving problems involving deceleration followed by acceleration, it's helpful to break the motion into separate intervals to simplify calculations.

An object's initial velocity is 1.25 m/s in the +x direction. It slows down with a constant acceleration whose magnitude is 1.32 m/s². After it reaches a momentary stop it reverses its direction of motion, to the -x, and speeds up with the same magnitude of the acceleration. What is its displacement (in meters) from the initial moment to t = 7.00 s? Keep 3 digits after the decimal point.

Learn how to calculate displacement for an object moving under uniform acceleration, including deceleration, stopping, and reversing direction. This problem involves kinematic equations and is suitable for students in Grades 10-12.

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