Chapter 2b Answer Key - WordPress.com

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Chapter 2 Answer Key BC Science Physics 11 Page 39 Practice Problems 2.1.1 1a. 1100 m 1b. 500 m 36o S of W 2a. 503 m 2b. 209 m 3. Total Distance = 19 m. Displacement = 5.0 m 53o E of N Page 41 Quick Check 1a. Average speed is total distance over total time. Instantaneous speed is speed at a give point in time. 1b. When an object is moving at a constant speed. 2. 89 km/h 3. 0.76 hr or 46 min 4. 460 km Page 44 Practice Problems 2.1.3 1a. Object at rest. 1b. Object moving away from origin at constant velocity. 1c. Object moving towards origin at a constant velocity. 2a. 2b.

2c. Person moving with ball so graph line horizontal 3a. Time (s) 0.0 1.0 2.0 3.0 4.0 Distance 8.3 16.6 24.9 33.2 41.5 (m)

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Page 46 2.1 Review Questions 1. Yes, when the direction is different 2. Right and left. Compass directions – north, south, east, west 3. 0.64 km 4. 2.3 hr 5a. 1.3x102 s 5b. 2.2 min 6. 114 km/hr 7a. 95 km 7b. 86 km/hr 8a. 0.50 sec 8b. 28 m/s Page 49 Practice Problems 2.2.1 1a. 20km/h/s 1b. 3 m/s2 2. 2.00 km/h/s 3. 15 km/h/s Page 52 2.2 Review Questions 1a. 16.7 m/s 1b. 25.6 m/s; 92.1 km/h 2. 3.0 s 3a. v f = 20m/s + 14.0m/s 2t 3b. 14.0 m/s2t 3c. 14.0 m/s2 3d. 14.0 m/s2 3e. Observers were at different locations. The aircraft was already moving when observer (a) recorded data. 4. 5.0 m/s Chapter 2 Kinematics

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Version 13-10-04 5. 8.5 m/s2 6. Graph to come 7a. 10 m/s2 7b. 0 7c. -7.5 m/s2 Page 56 Practice Problems 2.3.1 1a. 15.0 m/s 1b. 4.00 m/s2 1c. acceleration 1d. v f = 15m/s + (4.00m/s 2 )t 2a. 5.0 m/s 2b. 9.8 m/s2 2c. 17 m/s Page 58 Quick Check 1. 10 m/s [E] 2. 64.0 m 3. 41.6 m Page 62 2.3 Review Questions 1a. 5.5 m/s 1b. -7.9 m/s2 1c. v f = 5.5m/s − (7.90m/s 2 )t 2a.6.6 m/s 2b. -2.2 m/s2 2c. 3.0 s 3. -1.23 m/s2 4a. 4.0 m/s 4b. 10 m 5a. 3.00 m/s2 5b. 45.0 m/s 5c. 3.38 x 102 m 6.167 m or 1.7 x 102 m 7. 1.0 x 101 s 8. 9.0 x 1015 m/s2 Page 65 Practice Problems 2.4.1 1a. 1.0 x 102 m/s 1b. air resistance slows the ball down 2. 62.6 m 3a. 18.6 m/s 3b. 17.6 m 4. 2.4 x 102 m

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Version 13-10-04 Page 67 Quick Check 1a. 9.8 m/s2 1b. -9.8 m/s2 1c. 9.8 m/s2 2. (1.2, 0); at time 1.2 s, the ball reaches the highest point of its motion and comes to rest before falling back down. 3. 4.9 m Page 68 2.4 Review Questions 1. 0.40 m/s2 2a. 196 m/s2 2b. -9.8 m/s2 2c. 196 m/s 2d. At B, because vy=0 at peak of flight 2e. As soon as the ball leaves the pitcher’s hand, the only force is gravity, which means a = g = -9.8 m/s2 2f. Direction is as important, as well as speed 2g. 82.3 m 2h. 82.3 2i. 0 – ball returns to original place 3a. 4.9 m 3b. 14.7 m Page 69 Chapter 2 Review Questions 1. Velocity is speed and direction 2. 79 km/h 3a. 20.8 3b. 55.13 3c. 0.2 m/s 4. 2.56 s 5. When acceleration constant 6. 70 m/s or 252 km/h or 2.5 x 102 km/h 7a. 8.00 m/s2 7b. 40.0 m/s2 7c. 2.00 x 102 m 7d. v f = 20.0m/s + (8.0m/s 2 )t 8. 4.0 m/s 9. 77.3 s 10. 66 m/s 11. 49 m 12a. 16 m/s 12b. 6.8 s 12c. 6.4 s 13. 0.10 s to pass window 14a. 4.00 m/s2 14b. 9.00 s

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Version 13-10-04 15. 7.67 m/s 16a. 3.1 m/s 16b. 0.64 s 17a. Graph of d vs t is a parabola 17b. Graph of d vs t2 is a straight line 17c. A slope of 2.5 cm/s2. So d = kt2. Since d =

1 2 1 at , the slope must equal a . 2 2

Therefore, a = 2k = 5.0cm/s 2 Page 73 Chapter 2 Extra Practice 1a. 60 s 1b. 3600 s 1c. 86400 s 1d. 3.15 x 107 s 2. 622 km/h or 6.2 x 102 km/h 3. 1.0 x 10-2 mm/s 4. 3.84 x 105 km 5. 5.3 x 102 km 6a. 15.0 m/s 6b. 4.00 m/s2 6c. v f = 15.0m/s + (4.00m/s 2 )t 7a. 37.5 m/s 7b. 47 m 7c. 120 m 8a. 46.6 m – moose is saved 8b. 59.2 m – moose needs to move! 9. 10.4 s 10a. 24.6 m/s 10b. 88.5 km/h 11. -20.0 m/s2 12. 8.2 s 13a. 3.50 m/s 13b. -0.25 m/s2 13c. 24.5 m 13d. v = 3.5m/s − (0.25m/s 2 )t 15. 6.3 m/s2

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