**Physics Problem: Analyzing One-Dimensional Motion with Constant Acceleration**An object moves in one-dimensional motion with constant acceleration \( a = 7.4 \, \text{m/s}^2 \). At time \( t = 0 \, \text{s} \), the object is at \( x_0 = 3.2 \, \text{m} \) and has an initial velocity of \( v_0 = 4 \, \text{m/s} \).**Problem:**How far will the object move before it achieves a velocity of \( v = 6.6 \, \text{m/s} \)?Your answer should be accurate to the nearest 0.1 m.**Solution:**To solve this problem, we will use the kinematic equation for velocity and displacement:\[ v^2 = v_0^2 + 2a(x - x_0) \]**Step-by-Step Solution:**1. **Given Data:** - Initial position, \( x_0 = 3.2 \, \text{m} \) - Initial velocity, \( v_0 = 4 \, \text{m/s} \) - Final velocity, \( v = 6.6 \, \text{m/s} \) - Acceleration, \( a = 7.4 \, \text{m/s}^2 \) 2. **Plugging the values into the kinematic equation:**\[ (6.6)^2 = (4)^2 + 2 \times 7.4 \times (x - 3.2) \]3. **Simplifying the equation:**\[ 43.56 = 16 + 14.8(x - 3.2) \]4. **Solving for \( x \):**\[ 43.56 - 16 = 14.8(x - 3.2) \]\[ 27.56 = 14.8(x - 3.2) \]\[ x - 3.2 = \frac{27.56}{14.8} \]\[ x - 3.2 \approx 1.86 \]\[ x \approx 5.06 \, \text{m} \]Considering the initial position \( x_0 = 3.2 \, \

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Description: **Physics Problem: Analyzing One-Dimensional Motion with Constant Acceleration**An object moves in one-dimensional motion with constant acceleration \( a = 7.4 \, \text{m/s}^2 \). At time \( t = 0 \, \text{s} \), the object is at \( x_0 = 3.2 \, \te

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An object moves in one dimensional motion with constant acceleration a = 7.4 m/s². At time t = 0 s, the object is at xo = 3.2 m and has an initial velocity of vo = 4 m/s. How far will the object move before it achieves a velocity of v = 6.6 m/s? Your answer should be accurate to the nearest 0.1 m.

An object moves in one dimensional motion with constant acceleration a = 7.4 m/s². At time t = 0 s, the object is at xo = 3.2 m and has an initial velocity of vo = 4 m/s. How far will the object move before it achieves a velocity of v = 6.6 m/s? Your answer should be accurate to the nearest 0.1 m.

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**Physics Problem: Analyzing One-Dimensional Motion with Constant Acceleration**

An object moves in one-dimensional motion with constant acceleration \( a = 7.4 \, \text{m/s}^2 \). At time \( t = 0 \, \text{s} \), the object is at \( x_0 = 3.2 \, \text{m} \) and has an initial velocity of \( v_0 = 4 \, \text{m/s} \).

**Problem:**
How far will the object move before it achieves a velocity of \( v = 6.6 \, \text{m/s} \)?

Your answer should be accurate to the nearest 0.1 m.

**Solution:**

To solve this problem, we will use the kinematic equation for velocity and displacement:

\[ v^2 = v_0^2 + 2a(x - x_0) \]

**Step-by-Step Solution:**

1. **Given Data:**
- Initial position, \( x_0 = 3.2 \, \text{m} \)
- Initial velocity, \( v_0 = 4 \, \text{m/s} \)
- Final velocity, \( v = 6.6 \, \text{m/s} \)
- Acceleration, \( a = 7.4 \, \text{m/s}^2 \)

2. **Plugging the values into the kinematic equation:**

\[ (6.6)^2 = (4)^2 + 2 \times 7.4 \times (x - 3.2) \]

3. **Simplifying the equation:**

\[ 43.56 = 16 + 14.8(x - 3.2) \]

4. **Solving for \( x \):**

\[ 43.56 - 16 = 14.8(x - 3.2) \]

\[ 27.56 = 14.8(x - 3.2) \]

\[ x - 3.2 = \frac{27.56}{14.8} \]

\[ x - 3.2 \approx 1.86 \]

\[ x \approx 5.06 \, \text{m} \]

Considering the initial position \( x_0 = 3.2 \, \

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