Certainly! Let's revisit the solution with a focus on using the trigonometric identities to break down the vector components and calculate the displacement and average velocity.

---

Step 1: Breaking the motion into vector components

We already have the information for the three segments of motion:

• Segment 1: 1.29 km due east → This gives the displacement vector:

  • $x_1 = 1.29 \, \text{km}$
  • $y_1 = 0 \, \text{km}$

• Segment 2: 0.275 km due south → This gives the displacement vector:

  • $x_2 = 0 \, \text{km}$
  • $y_2 = -0.275 \, \text{km}$

• Segment 3: 3.32 km at 64.3° north of west

  We need to break this displacement into $x$ and $y$ components using trigonometric identities:

• The angle is 64.3° north of west. Since this is north of west, the x-component will be negative (westward), and the y-component will be positive (northward).

Using the trig identities:

• $\cos(64.3^\circ) = \cos(90^\circ - 64.3^\circ) = \sin(64.3^\circ)$
• $\sin(64.3^\circ) = \sin(90^\circ - 64.3^\circ) = \cos(64.3^\circ)$

The displacement vector for segment 3 can then be calculated as:

$x_3 = 3.32 \times \cos(64.3^\circ)$ $y_3 = 3.32 \times \sin(64.3^\circ)$

Let’s calculate the values using the known values of $\cos(64.3^\circ) \approx 0.438$ and $\sin(64.3^\circ) \approx 0.898$:

$x_3 = 3.32 \times (-0.438) \approx -1.45 \, \text{km}$ $y_3 = 3.32 \times 0.898 \approx 2.98 \, \text{km}$

---

Step 2