We can break the problem down into three distinct stages of motion and calculate the displacement during each stage. We'll then add these displacements to the initial position of the object to find its final position.

Initial conditions:

• Initial position: $x_0 = +4.5 \, \text{m}$
• Time intervals:
  • Stage 1: $t_1 = 1.8 \, \text{s}$
  • Stage 2: $t_2 = 2.0 \, \text{s}$
  • Stage 3: $t_3 = 3.5 \, \text{s}$

Stage 1: Negative velocity

• Velocity: $v_1 = -2.1 \, \text{m/s}$
• Displacement during Stage 1: $\Delta x_1 = v_1 \cdot t_1 = (-2.1) \cdot (1.8) = -3.78 \, \text{m}$

Stage 2: Positive velocity

• Velocity: $v_2 = +6.3 \, \text{m/s}$
• Displacement during Stage 2: $\Delta x_2 = v_2 \cdot t_2 = (6.3) \cdot (2.0) = 12.6 \, \text{m}$

Stage 3: Velocity equals zero

• Velocity: $v_3 = 0 \, \text{m/s}$
• Displacement during Stage 3: $\Delta x_3 = v_3 \cdot t_3 = 0 \cdot (3.5) = 0 \, \text{m}$

Final position calculation:

The total displacement is the sum of the displacements from each stage: $\Delta x_{\text{total}} = \Delta x_1 + \Delta x_2 + \Delta x_3 = -3.78 + 12.6 + 0 = 8.82 \, \text{m}$

The final position is the initial position plus the total displacement: $x_f = x_0 + \Delta x_{\text{total}} = 4.5 + 8.82 = 13.3 \, \text{m}$

Final answer:

The final position of the object is 13.3 meters.

Would you like more details or have any questions?

Here are 5 questions related to this problem:

1. How would the final position change if the initial velocity was positive instead of negative in Stage 1?
2. What would be the total displacement if the object did not stop in Stage 3?
3. How would air resistance affect the final position in this scenario?
4. Could we calculate the average velocity of the object over the entire motion?
5. How would the problem change if the object had a constant acceleration instead of constant velocity?

Tip: When dealing with motion problems, always break the motion into segments and calculate displacement for each separately.