1. **Problem:** A car goes 6 km east and then 4 km north. Find its displacement. 2. **Formula:** Displacement in two dimensions is found using the Pythagorean theorem: $$d = \sqrt{x^2 + y^2}$$ where $x$ and $y$ are the components of the displacement. 3. **Calculation:** Here, $x=6$ km and $y=4$ km. $$d = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 7.21 \text{ km}$$ 4. **Answer:** The closest option is A. 7 km. 1. **Problem:** A student walks 2 m east, 2 m west, 2 m east, and 2 m west. Find the displacement. 2. **Explanation:** Displacement is the net change in position. East and west movements cancel each other. 3. **Calculation:** Total east = 2 + 2 = 4 m, total west = 2 + 2 = 4 m. Net displacement = 4 m east - 4 m west = 0 m. 4. **Answer:** C. 0 m. 1. **Problem:** Which statement is TRUE about displacement? 2. **Explanation:** Displacement can be zero even if distance is not zero, for example, when an object returns to its starting point. 3. **Answer:** C. It completes one lap, her displacement is 0 m. 1. **Problem:** If a runner moves around a circular track of radius 50 m, what is the displacement after one lap? 2. **Explanation:** Displacement is the straight-line distance from start to end. After one lap, the runner returns to the start. 3. **Answer:** Displacement = 0 m (not listed, so question incomplete or options incorrect). 1. **Problem:** A moving object changes direction but maintains the same speed. What changes? 2. **Explanation:** Velocity is speed with direction. Changing direction changes velocity. 3. **Answer:** A. Velocity only. 1. **Problem:** A car moves 60 km east in 2 hours, then 40 km west in 1 hour. Find average velocity. 2. **Formula:** Average velocity = total displacement / total time. 3. **Calculation:** Displacement = 60 km east - 40 km west = 20 km east. Total time = 2 + 1 = 3 hours. Average velocity = 20 km / 3 hr = 6.67 km/hr east. 4. **Answer:** Closest is C. 6.7 km/hr west (direction should be east, so answer likely a typo, correct is 6.7 km/hr east). 1. **Problem:** Define acceleration. 2. **Answer:** B. Change in velocity per unit time. 1. **Problem:** A body moving in a straight line with constant speed has? 2. **Answer:** B. Constant velocity. 1. **Problem:** Car speeds up from 10 m/s to 30 m/s in 5 s. Find acceleration. 2. **Formula:** $$a = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t}$$ 3. **Calculation:** $$a = \frac{30 - 10}{5} = \frac{20}{5} = 4 \text{ m/s}^2$$ 4. **Answer:** B. 4 m/s2. 1. **Problem:** If acceleration is negative, the object is? 2. **Answer:** B. Slowing down. 1. **Problem:** Horizontal velocity of a projectile (neglecting air resistance) is? 2. **Answer:** B. Remains constant. 1. **Problem:** Vertical motion of a projectile is affected by? 2. **Answer:** B. Gravity only. 1. **Problem:** Path followed by a projectile is called? 2. **Answer:** A. Parabola. 1. **Problem:** Projectile launched at 45 degrees will have? 2. **Answer:** A. Max horizontal and vertical velocity components. 1. **Problem:** Time of flight of a projectile depends on? 2. **Answer:** B. Only vertical velocity. 1. **Problem:** Vertical velocity of a projectile at max height is? 2. **Answer:** A. 0 m/s. 1. **Problem:** Drop feather and stone in vacuum, they will? 2. **Answer:** B. Fall together. 1. **Problem:** Displacement can be zero even when? 2. **Answer:** C. Object returns to starting point. 1. **Problem:** Instantaneous velocity is found by? 2. **Answer:** B. The slope of the tangent to the position-time graph. 1. **Problem:** When does a freely falling object reach terminal velocity? 2. **Answer:** B. When air resistance balances weight. **Test II. Problem Solving:** 1. Alvin free falls for 3.6 s. (a) Final velocity: $$v = gt = 9.8 \times 3.6 = 35.28 \text{ m/s}$$ (b) Distance fallen: $$d = \frac{1}{2}gt^2 = 0.5 \times 9.8 \times 3.6^2 = 63.5 \text{ m}$$ 2. Brick dropped for 4.0 s. (a) Velocity: $$v = gt = 9.8 \times 4 = 39.2 \text{ m/s}$$ (b) Distance: $$d = 0.5 \times 9.8 \times 4^2 = 78.4 \text{ m}$$ 3. Coin flipped straight up. (a) Velocity at top = 0 m/s. (b) Initial speed: Using $$v^2 = u^2 - 2gd$$, at top $v=0$, so $$u = \sqrt{2gd} = \sqrt{2 \times 9.8 \times 0.25} = 2.21 \text{ m/s}$$ (c) Time in air: $$t = \frac{2u}{g} = \frac{2 \times 2.21}{9.8} = 0.45 \text{ s}$$ 4. Ball thrown horizontally from 6.5 m height at 45 m/s. (a) Time to ground: $$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 6.5}{9.8}} = 1.15 \text{ s}$$ (b) Horizontal distance: $$x = vt = 45 \times 1.15 = 51.75 \text{ m}$$ (c) Vertical velocity at impact: $$v_y = gt = 9.8 \times 1.15 = 11.27 \text{ m/s}$$ (d) Magnitude of velocity: $$v = \sqrt{v_x^2 + v_y^2} = \sqrt{45^2 + 11.27^2} = 46.4 \text{ m/s}$$ Direction angle: $$\theta = \tan^{-1}(\frac{v_y}{v_x}) = \tan^{-1}(\frac{11.27}{45}) = 14.1^\circ \text{ below horizontal}$$ **Test III. Essay:** 1. The book falls freely because gravity is the only force acting on it, while the paper is affected by air resistance causing it to fall slower. 2. The ball's velocity decreases going up, is zero at the top, then increases downward. Acceleration remains constant downward at 9.8 m/s2 throughout.