1. **Stating the problem:** Given the displacement vector $\Delta \vec{r} = \langle 1.10, 1.05, -1.70 \rangle$ m over a time interval $\Delta t = 780$ s, and velocity vector $\vec{v} = \langle 8.46, 8.07, -13.07 \rangle \times 10^{-3}$ m/s, we want to analyze the motion and find the angle $\theta_x$ between the velocity vector and the x-axis. 2. **Relevant formulas:** - Speed $v = |\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$ - Unit vector $\hat{v} = \frac{\vec{v}}{|\vec{v}|}$ - Angle with x-axis $\theta_x = \cos^{-1}\left(\frac{v_x}{|\vec{v}|}\right)$ 3. **Calculate the magnitude of velocity $|\vec{v}|$:** $$|\vec{v}| = \sqrt{(8.46 \times 10^{-3})^2 + (8.07 \times 10^{-3})^2 + (-13.07 \times 10^{-3})^2}$$ $$= \sqrt{(7.156 \times 10^{-5}) + (6.512 \times 10^{-5}) + (1.708 \times 10^{-4})}$$ $$= \sqrt{3.075 \times 10^{-4}} = 0.01753 \text{ m/s}$$ 4. **Check given speed:** Given speed is $2.92 \times 10^{-3}$ m/s, which differs from calculated magnitude $0.01753$ m/s, so we use the calculated magnitude for angle calculation. 5. **Calculate angle $\theta_x$:** $$\theta_x = \cos^{-1}\left(\frac{8.46 \times 10^{-3}}{0.01753}\right) = \cos^{-1}(0.4827)$$ $$\theta_x \approx 61.2^\circ$$ 6. **Interpretation:** The velocity vector makes an angle of approximately $61.2^\circ$ with the positive x-axis. **Final answer:** $$\theta_x \approx 61.2^\circ$$