(a)!! d = 17 m [W 63 S]!! d opposite. (b)!! d = 79 cm [E 56 N] = 79 cm [W 56 S] (c)!! d = 44 km [S 27 E] = 44 km [N 27 W] metres. 3.

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1 Chapter Reiew, pages Knowledge 1. (b). (d) 3. (b) 4. (a) 5. (b) 6. (c) 7. (c) 8. (a) 9. (a) 10. False. A diagram with a scale of 1 cm : 10 cm means that 1 cm on the diagram represents 10 cm in real life. 11. True 1. False. To add two ectors on a diagram, join them tip to tail. 13. False. The resultant ector is the ector that results from adding the gien ectors. 14. False. When gien the x- and y-component ectors, the Pythagorean theorem should be used to determine the magnitude of the displacement ector. 15. True 16. False. The x-component of the ector 8.0 m [S 45 W] is 5.7 m [W]. 17. True 18. True 19. False. When two objects are dropped from the same height at the same time, both objects will land at the same time when there is no air resistance. 0. (a)(ii) (b)(i) (c)(i) (d)() (e)(iii) Understanding 1. Answers may ary. Sample answers: (a) A map or a building blueprints would hae a scale that is smaller than the real-world measurement because you would want to see the whole area at a reasonable size. (a) A diagram of a cell, atoms, or a microchip would hae a scale that is larger than the realworld measurement because the original objects are too small to see in detail.. For each ector, the magnitude stays the same, but the cardinal directions are replaced by their opposites. (a) d 17 m [W 63 S] d opposite 17 m [E 63 N] (b) d 79 cm [E 56 N] d opposite 79 cm [W 56 S] (c) d 44 km [S 7 E] d opposite 44 km [N 7 W] 3. Diagram size Real-world size 3.4 cm 170 m 0.75 cm 37.5 m 85.0 mm 45 m 5.0 cm 150 m Row 1: Multiply by 50 and change the units to metres. 3.4 cm m Row : Diide by 50 and change the units to cenitmetres m cm Row 3: Multiply by 50 and change the units to metres cm m Row 4: Diide by 50 and change the units to cenitmetres. 150 m cm 4. (a) (b) Copyright 011 Nelson Education Ltd. Chapter : Motion in Two Dimensions -

2 (c) This figure shows the gien ectors, with the tip of d 1 joined to the tail of d. The resultant ector d T is drawn in black from the tail of d 1 to the tip of d. Using a compass, the direction of d T is [W 54 N]. d T measures 5. cm in length, so using the scale of 1 cm : 50 m, the actual magnitude of d T is 1300 m. Statement: The net displacement of the drier is 1300 m [W 54 N]. 8. Gien: d 1.0 m [N]; d 0.80 m [N 45 W] Required: d T 5. (a) Since 10 is 50 times.4, an appropriate scale is 1 cm : 50 m. (b) Since 360 is 150 times.4, an appropriate scale is 1 cm : 150 km. (c) Since 100 is 500 times.4, an appropriate scale is 1 cm : 500 m. 6. For each ector, determine the complementary angle, then reerse the order of the directions. (a) d 566 m [W 18 N] d 566 m [N 7 W] (b) d 37 cm [E 68 S] d 37 cm [S E] (c) d 7150 km [S 38 W] d 7150 km [W 5 S] 7. Gien: d m [W]; d 1050 m [N] Required: d T Analysis: d T d 1 + d Analysis: d T d 1 + d This figure shows the gien ectors, with the tip of d 1 joined to the tail of d. The resultant ector d T is drawn in black from the tail of d 1 to the tip of d. Using a compass, the direction of d T is [N 1 W]. d T measures 5. cm in length, so using the scale of 1 cm : 0.5 m, the actual magnitude of d T is.6 m. Statement: The net displacement of the cue ball is.6 m [N 1 W]. 9. d y d T Use the Pythagorean theorem to determine each missing magnitude: d x + d y d T. Copyright 011 Nelson Education Ltd. Chapter : Motion in Two Dimensions -3

3 Row 1: d T + d y d T + d y ( ) + ( 4.0) 3.0 d T 5.0 Row : d T + d y d y d T ( ) ( 8.0) 10.0 d y 6.0 Row 3: d T + d y d T d y ( ) ( 5.00) Row 4: d T + d y d y d T ( ) ( 4.00) 8.06 d y d y φ 3.0 [E] 4.0 [N] E 53 N 5.00 [W] 7.00 [N] W 54.5 N 8.0 [E] 1.1 [S] E 14.4 S 351 [W] 456 [N] W 5.4 N Use the tangent function: tan d y. Row 1: Find the missing angle. tan d y tan tan (two extra digits carried) tan 1 (1.333) 53 Row : Find the missing angle. tan d y tan tan 1.40 tan 1 (1.40) 54.5 Row 3: Find the missing component ector. tan d y tan 14.4 d y 8.0 (0.57)(8.0) d y 1.1 d y Row 4: Find the missing component ector. tan d y tan (a) Gien: d T 5 m [W 7 S] Required: ; d y Analysis: d T + d y Since the direction of d T is between west and south, the direction of is [W] and the direction of d y is [S]. sin d y d T d y d T sin ( )( sin7 ) 5 m d y 49 m cos d T d T cos ( )( cos7 ) 5 m 16 m Statement: The ector has a horizontal or x-component of 16 m [W] and a ertical or y-component of 49 m [S]. Copyright 011 Nelson Education Ltd. Chapter : Motion in Two Dimensions -4

4 (b) Gien: d T 38 km [E 14 N] Required: ; d y Analysis: d T + d y Since the direction of d T is between east and north, the direction of is [E] and the direction of d y is [N]. sin d y d T d y d T sin ( )( sin14 ) 38 km d y 9. km cos d T d T cos ( )( cos14 ) 38 km 37 km Statement: The ector has a horizontal or x-component of 37 km [E] and a ertical or y-component of 9. km [N]. (c) Gien: d T 9 m [S 8 W] Required: ; d y Analysis: d T + d y Since the direction of d T is between south and west, the direction of is [W] and the direction of d y is [S]. cos d y d T d y d T cos ( )( cos8 ) 9 m d y 13 m sin d T d T sin ( )( sin8 ) 9 m 91 m Statement: The ector has a horizontal or x-component of 91 m [W] and a ertical or y-component of 13 m [S]. 3. (a) Gien: 5.0 m [W]; d y.9 m [S] Required: d T Analysis: d T + d y Let φ represent the angle d T with the x-axis. d T + d y d T + d y d T + d y ( ) + (.9 m) 5.0 m d T 5.8 m tan d y tan.9 m 5.0 m 30 Statement: The sum of the two ectors is 5.8 m [W 30 S]. (b) Gien: 18 m [E]; d y 5. m [N] Required: d T Analysis: d T + d y Let φ represent the angle d T with the x-axis. d T + d y d T + d y d T + d y ( ) + ( 5. m) 18 m d T 19 m tan d y tan 5. m 18 m 16 Statement: The sum of the two ectors is 19 m [E 16 N]. (c) Gien: 64 km [W]; d y 31 km [N] Required: d T Analysis: d T + d y Let φ represent the angle d T with the x-axis. d T + d y d T + d y d T + d y ( ) + ( 31 km) 64 km d T 71 km Copyright 011 Nelson Education Ltd. Chapter : Motion in Two Dimensions -5

5 tan d y 31 km tan 64 km 6 Statement: The sum of the two ectors is 71 km [W 6 N]. 33. Gien: d 1.0 m [left]; d 5.0 m [up] Required: d T Analysis: d T d 1 + d d T d 1 + d d T d 1 + d d T d 1 + d ( ) + ( 5.0 m).0 m d T 5.4 m Statement: The magnitude of the resultant ector is 5.4 m. 34. Gien: d 1.0 m [left]; d 6.0 m [up] Required: d T Analysis: d T d 1 + d d T d 1 + d d T d 1 + d d T d 1 + d ( ) + ( 6.0 m).0 m d T 6.3 m Statement: The magnitude of the resultant ector is 6.3 m. 35. Gien: d m [down]; d 4.0 m [right] Required: d T Analysis: d T d 1 + d d T d 1 + d d T d 1 + d d T d 1 + d ( ) + ( 4.0 m) 7.0 m d T 8.1 m Statement: The magnitude of the resultant ector is 8.1 m. 36. Gien: d 1 4 m [W 1 S]; d 33 m [E 5 S] Required: d T Analysis: d T d 1 + d Determine the total x-component and y-component of d T : d Tx d 1x + d x 4 m ( )( cos1 ) [W] + ( 33 m) ( cos5 ) [E] 3.48 m [W] m [E] 3.48 m [W] 0.3 m [W] 3.16 m [W] d Tx 3. m [W] d Ty d 1y + d y 4 m ( )( sin1 ) [S] + ( 33 m) ( sin5 ) [S] 4.99 m [S] m [S] m [S] d Ty 31 m [S] Determine the magnitude of d T : d T d Tx + d Ty d T d Tx + d Ty ( ) + ( m) (two extra digits carried) 3.16 m d T 31 m Let φ represent the angle d T with the x-axis. tan d Ty d Tx tan m (two extra digits carried) 3.16 m 84 Statement: The dog s displacement is 31 m [W 84 S]. 37. (a) Gien: d 36 m;.0 m/s Required: t Analysis: d t t d t d 36 m.0 m s t 18 s Statement: It will take the student 18 s to cross the rier. (b) Gien: 1 6. m/s [W];.0 m/s [N] Required: T Analysis: T 1 + Copyright 011 Nelson Education Ltd. Chapter : Motion in Two Dimensions -6

6 Let φ represent the angle T with the x-axis. T 1 + T 1 + T 1 + ( ) + (.0 m/s) 6. m/s T 6.5 m/s tan 1.0 m/s tan 6. m/s 18 Statement: The resulting elocity of the boat is 6.5 m/s [W 18 N]. (c) Gien: x 6. m/s [W] Required: Analysis: x t x t d x x t 6. m # s [W] $ % 18 s m [W] d x 1.1' 10 m [W] ( ) Statement: The student lands m or 110 m downstream from her destination. 38. The beanbag thrown at an angle will hit the ground first. Both beanbags hae the same initial elocity but different ertical components of the elocity. Since both accelerate down at the same rate (graity), the beanbag thrown at angle will hae a shorter time of flight because its initial ertical elocity was less. 39. Gien: i 15 m/s; θ 50 Required: ix ; iy Analysis: i ix + iy iy sin i iy i sin ( )( sin50 ) 15 m/s iy 11 m/s cos ix i ix i cos ( )( cos50 ) 15 m/s ix 9.6 m/s Statement: The initial elocity has a horizontal or x-component of 9.6 m/s and a ertical or y-component of 11 m/s. 40. (a) Gien: d y 1.3 m; a y 9.8 m/s ; y 0 m/s Required: t Analysis: d y y t + 1 a y t t d y a y t t d y a y t d y a y d y a y ( ) 1.3 m 9.8 m % # $ s ' s t 0.5 s Statement: The time of flight of the beanbag should be 0.5 s. (b) Gien: t s; x 4. m/s Required: Analysis: x t x t 4. m $ # s % s. m ( ) (two extra digits carried) Statement: The range of the beanbag should be. m. 41. (a) Gien: a y 9.8 m/s ; t 3.78 s; θ 45.0 Required: i Analysis: i iy sin Copyright 011 Nelson Education Ltd. Chapter : Motion in Two Dimensions -7

7 Determine the y-component: fy iy + a y t iy fy a y t ( ) 3.78 s m/s 18.5 m/s iy 19 m/s # $ % ' Use the sine function: sin iy i i iy sin 18.5 m/s (two extra digits carried) sin45.0 i 6. m/s Statement: The cannon is fired with an initial elocity of 6. m/s. (b) Gien: iy 18.5 m/s; t 3.78 s Required: d y Analysis: ix t ix t Since the initial angle is 45.0, the horizontal and ertical components of the elocity hae the same magnitude. ix t iy t 18.5 m $ # s % 3.78 s 70.0 m ( ) (two extra digits carried) Statement: The range of the cannon is 70.0 m. This figure shows the gien ectors, with the tip of d 1 joined to the tail of d, then the tip of d joined to the tail of d 3. The resultant ector d T is drawn in black from the tail of d 1 to the tip of d 3. Using a compass, the direction of d T is [N 30 E]. d T measures 5.6 cm in length, so using the scale of 1 cm : 0 m, the actual magnitude of d T is 11 m. Statement: The total displacement for his trip is 11 m [N 30 E]. 43. Gien: d 1 0 m [E 40 N]; d 360 m [N 30 W]; t s Determine the displacement of the boat: Required: d T Analysis: d T d 1 + d Analysis and Application 4. (a) The trip would be represented by three ectors: a 50 m [N] ector, a 50 m [E] ector, and another 50 m [N] ector. A possible scale would be 1 cm : 0 m, so each ector is.5 cm long. (b) Gien: d 1 50 m [N]; d 50 m [E]; d 3 50 m [N] Required: d T Analysis: d T d 1 + d + d 3 This figure shows the gien ectors, with the tip of d 1 joined to the tail of d. The resultant ector d T is drawn in black from the tail of d 1 to the tip of d. Using a compass, the direction of d T Copyright 011 Nelson Education Ltd. Chapter : Motion in Two Dimensions -8

8 is [N 1.4 W]. d T measures 4.5 cm in length, so using the scale of 1 cm : 100 m, the actual magnitude of d T is 450 m. Statement: The boat s displacement is 450 m [N 1.4 W]. Determine the aerage elocity of the boat: Required: a Analysis: a d t a d t 450 m [N 1.4 W] s 0.45 m/s [N 1.4 W] a.0 10 m/s [N 1.4 W] Statement: The boat s aerage elocity is.0 10 m/s [N 1.4 W]. 44. Answers may ary. Sample answer: (a) Since I would draw c from the tail of a to the tip of b (when they were tip to tail), I would draw a and c touching tails, and the displacement from the tip of a to the tip of c would represent b. (b) As in part (a), I would draw what I knew of the ector addition to determine the magnitude and direction of the missing ector. This time, I would put the tips of b and c together, and a would be the ector from the tail of c to the tail of b. (c) It does not matter which component ector because ectors can be added in either order and you would always get the same resultant ector. (d) Gien: a.1 cm [W]; c 4.3 cm [W 45 N] Required: d T Analysis: d T d 1 + d 45. Gien: d 1 5 blocks [N]; d 5 blocks [E] Required: d T Analysis: d T d 1 + d d T d 1 + d 5 blocks + 5 blocks 50 m $ ( 10 blocks )# 1 block % d T 500 m Statement: The total distance she traels is 500 m or.5 km. 46. (a) The most direct route is 1.5 blocks [E], 9 blocks [S], and another 1.5 blocks [E]. (b) Gien: 40.0 km/h; d blocks [E]; d 9 blocks [S]; d blocks [E] Required: t Analysis: d t t d t d 1.5 blocks + 9 blocks blocks 40.0 km/h 1 blocks 50 m $ 40.0 km/h # 1 block % 3000 m 1 km $ 40.0 km # 1000 m % h 60 min $ h # 1 h % t 4.5 min Statement: It will take the student 4.5 min to get to the market. 47. Gien: d 1 3 blocks [N]; d 5 blocks [W] Required: d T This figure shows the gien ectors, with the tail of a joined to the tail of c. The missing ector b is drawn in black from the tip of a to the tip of c. Using a compass, the direction of b is [W 74 N]. b measures 3.4 cm in length. Statement: Vector b is 3.4 m [W 74 N]. Analysis: d T d 1 + d Let φ represent the angle d T with the x-axis. 3 blocks [N] 750 m [N] 5 blocks [W] 150 m [W] Copyright 011 Nelson Education Ltd. Chapter : Motion in Two Dimensions -9

9 d T d 1 + d d T d 1 + d d T d 1 + d ( ) + ( 150 m) 750 m d T 1460 m tan d d 1 tan 150 m 750 m 59 Statement: The student s net displacement is 1460 m [N 59 W]. 48. Gien: d 1 5 blocks [W]; d 11 blocks [S]; d 3 blocks [W] Required: d T Analysis: d T + d y Let φ represent the angle d T with the x-axis. 5 blocks [W] 150 m [W] 11 blocks [S] 750 m [S] blocks [W] 500 m [W] 150 m [W] m [W] 1750 m [W] d y 750 m [S] d T + d y d T + d y d T + d y ( ) + ( 750 m) 1750 m d T 360 m tan d y tan 750 m 1750 m 57.5 Statement: The student s net displacement is 360 m [W 57.5 S]. 49. (a) Gien: 5 km/h; d 1 5 blocks [W]; d 5 blocks [S] Required: t Analysis: d t t d t d 5 blocks + 5 blocks 5 km/h 10 blocks 50 m $ 5 km/h # 1 block % 500 m 1 km $ 5 km # 1000 m % h 60 min $ 0.1 h # 1 h % t 6.0 min Statement: It takes her 6.0 min to drie home from the school. (b) Gien: d 1 5 blocks [W]; d 5 blocks [S]; t 6.0 min Determine the total displacement: Required: d T Analysis: d T d 1 + d Let φ represent the angle d T with the x-axis. 5 blocks 150 m d T d 1 + d d T d 1 + d d T d 1 + d ( ) + ( 150 m) 150 m d T 1770 m tan d d 1 tan 150 m 150 m 45 Statement: The net displacement from school to home is 1770 m [W 45 S]. Determine the aerage elocity: Required: a Analysis: a d t a d t 1770 m [W 45 S] 6.0 min a 18 km/h # 60 min 1 h $ 1 km $ % # 1000 m % Copyright 011 Nelson Education Ltd. Chapter : Motion in Two Dimensions -10

10 Statement: Her aerage elocity from school to home is 18 km/h [W 45 S]. 50. Gien: 5. km/h; d 1 5 blocks [W]; d blocks [S]; d 3 5 blocks [E] Required: t Analysis: d t t d t d 5 blocks + blocks + 5 blocks 5. km/h 1 blocks 50 m $ 5. km/h # 1 block % 3000 m 1 km $ 60 min $ 5. km # 1000 m % # 1 h % h t 7.14 min Statement: It took the student 7.14 min to get from school to the library. 51. Gien: 750 m; d T 1100 m Determine how far north the boat traelled: Required: Analysis: d T d x + d y d y d T d y d T d y d T ( ) + ( 750 m) 1100 m d y 800 m Statement: The boat traelled 800 m north. Determine the direction the boat traelled: Required: θ Analysis: sin d T Let θ represent the angle d T with the y-axis. sin d T 750 m 1100 m 43 Statement: The boat traelled in the direction [N 43 E] 5. Gien: 13 m; θ [N 3 W] Required: d T Analysis: sin d T d T sin d T sin 13 m sin3 d T 5 m Statement: He needs to kick the ball at least 5 m in order to make it through the centre of the posts. 53. Gien: d 1 11 m [N]; t s; d 6 m [W 4 N]; t 1. s Required: a Analysis: a d t Determine the total x-component and y-component of d T : d Tx d 1x + d x m ( )( cos4 ) [W] 19.3 m [W] d Tx 19 m [W] d Ty d 1y + d y 11 m [N] + 6 m ( )( sin4 ) [N] 11 m [N] m [N] 8.40 m [N] d Ty 8 m [N] Determine the magnitude of d T : d T d Tx + d Ty d T d Tx + d Ty ( ) + ( 8.40 m) (two extra digits carried) 19.3 m d T 34 m Let φ represent the angle d T with the y-axis. tan d Tx d Ty tan 19.3 m 8.40 m 34 (one extra digit carried) Copyright 011 Nelson Education Ltd. Chapter : Motion in Two Dimensions -11

11 Determine the aerage elocity: a d t d T t 1 + t m [N 34 W] 0.55 s + 1. s m [N 34 W] 1.75 s a 0 m/s [N 34 W] Statement: The puck s aerage elocity is 0 m/s [N 34 W]. 54. Gien: x 5. m/s; 35 m; d y 5 m Required: y Analysis: y d y t Determine the time to cross the rier: x t t x 35 m 5. m s s t 6.7 s Determine the speed of the rier current: y d y t 5 m s y 3.7 m/s (two extra digits carried) Statement: The speed of the current is 3.7 m/s. 55. (a) The student is on the south bank of an eastwest rier, so the resulting direction of the elocity of the boat must be north. (b) The student should point the boat north, then turn it west into the current at an angle that will result in the boat traelling directly across the rier. (c) Gien: d y 50 m; x 1.1 m/s; T 3.8 m/s Required: θ Analysis: sin y T sin y T 1.1 m s 3.8 m s 17 Statement: The student should point the boat [N 17 W]. (d) Gien: d y 50 m; T 3.8 m/s Required: t Analysis: d t t d t d 50 m $ # cos17 % 3.8 m s t 14 s Statement: It takes the student 14 s to cross the rier. 56. Gien: d y 1. m; a y 9.8 m/s ; x 1.5 m/s; y 0 m/s Determine the time of flight: Required: t Analysis: d y y t + 1 a y t t d y a y t t d y a y t d y a y d y a y ( ) 1. m 9.8 m % # $ s ' s t 0.49 s Statement: The hockey puck is in flight for 0.49 s. Copyright 011 Nelson Education Ltd. Chapter : Motion in Two Dimensions -1

12 Determine the range: Required: Analysis: x t x t 1.5 m $ # s % s 0.74 m ( ) (two extra digits carried) Statement: The range of the hockey puck is 0.74 m. (b) Gien: d y 1. m; a y 9.8 m/s ; x 1.5 m/s; y 0 m/s Required: f Analysis: f fx + fy fy a y t ( )( s) (two extra digits carried) 9.8 m/s [down] fy 4.85 m/s [down] Use the Pythagorean theorem: f fx + fy f fx + fy f ( ) + ( 4.85 m/s) (one extra digit carried) 1.5 m/s 5.1 m/s Let φ represent the angle f with the x-axis. tan fy fx 4.85 m s 1.5 m (one extra digit carried) s 73 Statement: The puck s final elocity is 5.1 m/s [73 aboe horizontal]. 57. (a) Gien: a y 9.8 m/s ; i 16.5 m/s; θ 35 Required: t Analysis: d y y t + 1 a y t d y y t + 1 a y t i ( sin )t + 1 a y t 0 ( m/s) ( 4.9 m/s )t ( t 0) m s t 4.9 m s s t 1.9 s Statement: The soccer ball s time of flight is 1.9 s. (b) Gien: i 16.5 m/s; θ 35 Required: Analysis: x t x t i cos t 16.5 m % # $ s ' cos35 6 m ( )( s ) (two extra digits carried) Statement: The soccer ball s range is 6 m. (c) Gien: a y 9.8 m/s ; i 16.5 m/s; θ 35 ; fy 0 m/s Required: d y Analysis: fy iy + a y d y d y fy iy a y d y fy iy a y ( ) ( ) ( )( sin35 ) 0 sin35 i 9.8 m/s m/s # 19.6 m/s m s 19.6 m s d y 4.6 m Statement: The soccer ball reached a maximum height of 4.6 m. 58. (a) Gien: a y 9.8 m/s ; t. s; 17 m; d y 5. m Required: i Analysis: i ix + iy $ % 0 ( 16.5 m/s) ( sin35 )t + 1 (9.8 m/s )t 0 ( m/s)t ( 4.9 m/s )t Copyright 011 Nelson Education Ltd. Chapter : Motion in Two Dimensions -13

13 Determine the x-component: ix t 17 m. s 7.77 m/s ix 7.7 m/s Determine the y-component: fy iy + a y d y iy fy a y d y iy fy a y d y ( )( 5. m) m/s m /s m/s iy 10 m/s Use the Pythagorean theorem: f fx + fy f fx + fy f ( ) + ( m/s) (two extra digits carried) 7.77 m/s 13 m/s Statement: The soccer ball is kicked with an initial elocity of 13 m/s. (b) Gien: a y 9.8 m/s ; t. s; 17 m; d y 5. m Required: θ Analysis: tan iy ix tan iy ix m s 7.77 m (two extra digits carried) s 53 Statement: The soccer ball is kicked at an angle of 53. Ealuation 59. Answers may ary. Sample answer: Scale drawings require accuracy in scale, drawing, and measurement in order to get the correct answer. It can also be time consuming. It is less effectie than calculating a precise answer. 60. Answers may ary. Sample answer: Time is a scalar because it always has the same direction: forward in time. The analogy in the question relates time to distance and ordering magnitudes only. 61. (a) The ertical distance is being manipulated to obsere differences in the horizontal displacement (range) and time of flight. Initial elocity and angle of launch are being controlled, while acceleration due to graity is constant. (b) Student B s data will be the most alid because she is repeating her launches ten times to aoid any errors due to measurement or malfunction. (c) Answers may ary. Sample answer: Errors can happen due to bad measuring or problems keeping the initial elocity and angle controlled. Haing more than one person measuring time and distance can preent errors, as well as making sure the launch mechanism is working consistently. Reflect on Your Learning 6. (a) When drawing two ectors that are added together, draw the first ector then draw the second ector, keeping its size and direction, starting at the tip of the first. The resultant ector starts at the tail of the first ector and ends at the tip of the second. (b) Answers may ary. Sample answer: To subtract two ectors, you could use the rules for ector addition in reerse. Rearrange your ector subtraction to look like an addition problem and apply the steps of addition to find the missing ector. First, draw the resultant ector. You know that this connects the tail of one ector to the tip of the other. Then draw the other known ector starting its tail at the same point as the tail of the resultant ector. The missing ector is the ector connecting the tip of the second ector to the tip of the resultant ector. (c) Answers may ary. Sample answer: I prefer the algebraic method of ector addition. Sometimes it seems like a lot more work since ectors must be first broken down into components, then added, then the resultant ector must be determined, but this method is more accurate. As long as you are careful with your algebra there are fewer errors and mistakes than using scale diagrams. 63. Students should discuss any gaps in their understanding of motion problems and how they could learn more about soling motion problems. Copyright 011 Nelson Education Ltd. Chapter : Motion in Two Dimensions -14

14 Research 64. Students answers should be a few paragraphs that discuss the origin and deelopment of the compass rose. 65. Students papers should discuss the Cartesian coordinate system and its deelopment. Descartes should be mentioned as well as a reference to at least one other type of coordinate system such as a three-dimensional rectilinear, polar, or spherical coordinate system. 66. Students reports should discuss the uses of accelerometers in nature. The examples gien include the study of shark mating behaiour, the study of gliding behaiour of lemurs, and turning laptops into earthquake sensors. 67. Students should describe both radar and laser deices. The physics behind each technology should be briefly outlined in one or two paragraphs. Copyright 011 Nelson Education Ltd. Chapter : Motion in Two Dimensions -15