1. **Problem statement:** Determine the signs and changes of velocity, displacement, and acceleration vectors for Marie-Philip Poulin skating toward the Team USA net, with positive direction defined toward the Team USA net. 2. **Key concepts:** - Velocity vector sign indicates direction of motion. - Displacement vector sign indicates position relative to origin. - Acceleration vector sign indicates direction of change in velocity. - If velocity magnitude decreases but direction is positive, acceleration is negative. 3. **Step-by-step solution:** **(a) Velocity vector sign:** Marie-Philip is moving toward the Team USA net, which is positive direction, so velocity vector sign is **+**. **(b) Displacement vector sign:** Since displacement is measured from Team Canada net toward Team USA net and she is moving toward Team USA net, displacement vector sign is **+**. **(c) Acceleration vector sign:** She is slowing down while moving in positive direction, so acceleration is opposite to velocity, thus acceleration vector sign is **-**. **(d) Velocity magnitude change:** She is slowing down, so velocity magnitude is **decreasing**. **(e) Displacement magnitude change:** She is still moving toward Team USA net, so displacement magnitude is **increasing**. --- 2. **Problem statement:** Tyler McGregor moves 5m right, 8m forward, then 2m left. Find net displacement. 3. **Key concepts:** - Displacement is vector sum of individual movements. - Right and left are along x-axis; forward is along y-axis. 4. **Step-by-step solution:** - Total x-displacement: $5 - 2 = 3$ m - Total y-displacement: $8$ m - Net displacement magnitude: $$\sqrt{3^2 + 8^2} = \sqrt{9 + 64} = \sqrt{73} \approx 8.54\text{ m}$$ - Direction angle $\theta$ from x-axis: $$\theta = \tan^{-1}\left(\frac{8}{3}\right) \approx 69.44^\circ$$ forward of right. --- 3. **Problem statement:** Andres Borregales kicks a football at 25 m/s at 44° above horizontal. Find (a) maximal height, (b) horizontal range. 4. **Key concepts:** - Vertical velocity component: $v_{y0} = 25 \sin 44^\circ$ - Horizontal velocity component: $v_{x} = 25 \cos 44^\circ$ - Max height formula: $$h_{max} = \frac{v_{y0}^2}{2g}$$ where $g=9.81$ m/s² - Range formula: $$R = \frac{v_x \times 2 v_{y0}}{g}$$ (time of flight $t=\frac{2 v_{y0}}{g}$) 5. **Step-by-step solution:** **(a) Maximal height:** Calculate $v_{y0}$: $$v_{y0} = 25 \times \sin 44^\circ \approx 25 \times 0.6947 = 17.37\text{ m/s}$$ Calculate max height: $$h_{max} = \frac{(17.37)^2}{2 \times 9.81} = \frac{301.7}{19.62} \approx 15.38\text{ m}$$ **(b) Horizontal range:** Calculate $v_x$: $$v_x = 25 \times \cos 44^\circ \approx 25 \times 0.7193 = 17.98\text{ m/s}$$ Calculate time of flight: $$t = \frac{2 \times 17.37}{9.81} = \frac{34.74}{9.81} \approx 3.54\text{ s}$$ Calculate range: $$R = v_x \times t = 17.98 \times 3.54 \approx 63.65\text{ m}$$ --- **Final answers:** 1. (a) + (b) + (c) - (d) decreasing (e) increasing 2. Net displacement magnitude $\approx 8.54$ m at $69.44^\circ$ forward of right. 3. (a) Max height $\approx 15.38$ m (b) Horizontal range $\approx 63.65$ m