Pearson Physics Level 20 Unit I Kinematics: Chapter 2 Solutions

Transcription

1 Pearson Phsics Leel 0 Unit I Kinematics: Chapter Solutions Student Book page 71 Skills Practice Students answers will ar but ma consist of: (a) scale 1 cm : 1 m; ector will be 5 cm long scale 1 m forward 5 m [forward] (b) scale 1 cm : 5 m; ector will be 4 cm long scale 5 m up 0 m [down] (c) scale 1 cm : 10 km; ector will be 3 cm long scale N 10 km 30 km [north] (d) scale 1 cm : 50 km; ector will be 3 cm long scale 50 km 150 km [left] Student Book page 73 Eample.1 Practice Problems 1. (a) dtotal = 5.0 m + 10 m + 10 m + 10 m + 0 m + 10 m + 40 m + 10 m = 115 m (b) Consider forward to be positie. d = +5.0 m + ( 10 m) + (+10 m) + ( 10 m) + (+0 m) + ( 10 m) + (+40 m) + ( 10 m) = +35 m = 35 m [forward] Pearson Phsics Solutions Unit I Chapter 1 Copright 009 Pearson Education Canada

2 Student Book page 74 Eample. Practice Problems 1. Gien d m [N] d 0.0 m [N] d m [N] displacement ( d ) Analsis and Solution Choose an appropriate scale and reference direction. Scale 1.00 cm : 10.0 m, north is positie. N 40.0 m 0.0 m m 10.0 m d = cm [N] 1.00 cm = m [N] Sprinter s displacement is m [N].. Gien d m [right] d 3.50 m [left] displacement ( d ) Analsis and Solution Choose an appropriate scale and reference direction. Scale 1.00 cm : 1.00 m, right is positie. d 0.75 cm 0.75 m 3.50 m 1.00 m d =.75 cm [left] 1.00 cm =.75 m [left] Plaer s displacement is.75 m [left]. 3. Gien d m [back] d 0.85 m [forth] distance ( d ) displacement ( d ) Analsis and Solution Choose an appropriate scale and reference direction. Pearson Phsics Solutions Unit I Chapter Copright 009 Pearson Education Canada

3 Scale 1.00 cm : 1.00 m, back or left is positie m d 0, since the ectors cancel out m The bricklaer s hand moes 1.70 m back and forth four times, so Δ d 4( d1 d). d 4(0.85 m 0.85 m) 6.80 m Since the plaer starts and finishes in the same spot, displacement is zero. d 0 m Total distance is 6.80 m. Total displacement is zero. Student Book page 75.1 Check and Reflect Knowledge 1. The ectors hae the same magnitude (equal length) but opposite directions.. The ector 5 m [N] should be half the length of, and point in the opposite direction to, the ector 10 m [S]. 5.0 m [S] 0 m [S] cm 6.0 cm 0 m [S] 5.0 m [S] 4 cm 5.0 cm 1.0 cm km 100 km 1.0 cm 5.0 cm 0 km scale = 1.0 cm : 0 km Applications 5. Gien scale = 1.0 cm : 50 km map distance = 4.0 cm separation in kilometres () Analsis and Solution Use equal scale ratios. Pearson Phsics Solutions Unit I Chapter 3 Copright 009 Pearson Education Canada

4 50 km = 4.0 cm 1.0 cm 50 km = 4.0 cm 1.0 cm = km = 080 km 3 =.1 10 km and Verif km 50 km The separation is.1 10 km. Check: = 4.0 cm 1.0 cm 6. (a) d d1 d d3 d4 d5 d6 d7 d8 d9 d10 d1 ( d1) d ( d) d3 ( d3) d4 ( d4) d5 ( d5) d6 ( d6) d7 ( d7) d8 ( d8) d9 ( d9) d10 d 10 (b) d = +100 ards = 100 ards [down field] (c) d = (10) + (0) + (30) + (40) + (50) + (60) + (70) + (80) + (90) = 1000 ards 7. Let represent each displacement south. Since the car s final position is 50 km [N], its total distance traelled south is 450 km. + (50 + ) + (100 + ) = 450 km = 450 km 3 = 300 km = 100 km Therefore, the car s three displacements are 100 km [S], 150 km [S], and 00 km [S]. 8. Vectors A and B are equal. The hae the same magnitude and point in the same direction. 9. (1) The distance the ball traels is d = 10.0 m m m m m +.0 m = 36.0 m () If down is negatie, the ball s displacement from its drop point is d df di.0 m 10.0 m 8.0 m 8.0 m [down] Pearson Phsics Solutions Unit I Chapter 4 Copright 009 Pearson Education Canada

5 Student Book page 78 Skills Practice 1. (a) polar coordinates (b) polar coordinates (c) polar coordinates (d) naigator Pearson Phsics Solutions Unit I Chapter 5 Copright 009 Pearson Education Canada

6 (e) polar coordinates (f) naigator. (a) 3 m [E] (b) 17 m/s [65 S of W] Pearson Phsics Solutions Unit I Chapter 6 Copright 009 Pearson Education Canada

7 (c) 7 m [65 N of E] (d) 8 m/s [15] (e) m [8 W of N] (f) 1 m/s [330] Student Book page 79 Concept Check [60 S of W] ma also be written [30 W of S]. Pearson Phsics Solutions Unit I Chapter 7 Copright 009 Pearson Education Canada

8 Eample.3 Practice Problems 1. Student Book page 8 d1 d3 d dr 1.30 km [67 E of S] d d3 d1 dr 1.30 km [3 S of E] 1.30 km [67 E of S]. The student s position relatie to her starting point is 7 m [60 S of E]. Pearson Phsics Solutions Unit I Chapter 8 Copright 009 Pearson Education Canada

9 Skills Practice Distance d (m) Final position d (m) [direction] reference point Displacement d (m) [direction] AB 560 m 448 m [W] 448 m [W] BC 630 m 4 m [N] 504 m [7 N of E] CD 600 m 704 m [N] 480 m [N] AC 1190 m 4 m [N] 4 m [N] AD 1790 m 704 m [N] 704 m [N] Eample.4 Practice Problems 1. d = 15 km [40 E of N] Student Book page 84 The angle is gien with respect to the -ais (E of N), so use the cosine function to calculate the north component: d = (15 km)(cos 40) = 11 km [N]. = 10 m/s [45] = (10 m/s)(cos 45) = 4. m/s = (10 m/s)(sin 45) = 9.1 m/s Pearson Phsics Solutions Unit I Chapter 9 Copright 009 Pearson Education Canada

10 3. d = 65 km [37 E of S] d = (65 km)(sin 37) = 39 km [E] Concept Check Designate up and right as positie directions. Quadrant R R I + + II + III IV + Skills Practice 1. (a) R 15 km/h [45 N of W] R (15 km/h)(cos 45 ) 11 km/h R (15 km/h)(sin 45 ) 11 km/h (b) R 00 km/h [5 E of S] (c) R (00 km/h)(sin 5 ) 84.5 km/h R (00 km/h)(cos 5 ) 181 km/h R 10 km/h [N] R 0 R 10 km/h Student Book page 85 Pearson Phsics Solutions Unit I Chapter 10 Copright 009 Pearson Education Canada

11 . (a) R = 1 m, R = 7 m R R R 1 m 7 m 14 m opposite tan adjacent 7 m 1 m tan (0.583) 30 R 14 m [30 ] (b) R = 40 km/h, R R R R = 55 km/h 40 km/h 55 km/h 68 km/h opposite tan adjacent 55 km/h 40 km/h tan (1.375) 54 R 68 km/h [54 ] (c) R = 30 cm, R = 10 cm R R R 30 cm 10 cm 3 cm opposite tan adjacent 10 cm 30 cm tan (0.333) 18 R 3 cm [18 ] Pearson Phsics Solutions Unit I Chapter 11 Copright 009 Pearson Education Canada

12 Student Book page 88 Eample.5 Practice Problems 1. Gien d m [0 ] d 60.0 m [335 ] displacement ( d ) Analsis and Solution Resole both ectors in their components and then perform ector addition. Determine the resultant displacement using the Pthagorean theorem. d d 1 1 = 80.0 m = 0 d = d cos d = (60.0 m)(cos 5 ) = m = d sin =-(60.0 m)(sin 5 ) =-5.36 m d = d + d d 1 = 80.0 m m = m d = d + d d 1 = 0 + (-5.36 m) =-5.36 m Pearson Phsics Solutions Unit I Chapter 1 Copright 009 Pearson Education Canada

13 ( ) ( ) d = d + d = ( m ) + (-5.36 m) = 137 m opposite tan = adjacent 5.36 m = m = = tan (0.1887) = Since the angle is with respect to the positie -ais in the negatie direction, the angle with respect to the positie -ais in the positie direction is The farmer s displacement was 137 m [349 o ].. Gien o d1 15 m [15 N of E] o d 13 m [5 W of N] displacement ( d ) Analsis and Solution Resole each position ector into its and components and then find the resultant using trigonometr. d d 1 1 = (15 m)(cos 15 ) = m 1 1 = d cos = d sin = (15 m)(sin 15 ) = 3.88 m Pearson Phsics Solutions Unit I Chapter 13 Copright 009 Pearson Education Canada

14 d d =-(13 m)(sin 5 ) =-1.13 m =-d sin (since in second quadrant) = d cos = (13 m)(cos5 ) = 1.95 m d = d + d d 1 = m + (-1.13 m ) = m d = d + d 1 = 3.88 m m = m ( ) ( ) d = d + d = ( m ) + (16.83 m) = 1.49 m opposite tan = adjacent m = m = = tan (1.60) = 5 The soccer plaer s displacement is 1 m [5 o N of E]. 3. Gien o d1 300 m [S] d 550 m [75 N of E] displacement ( d ) Pearson Phsics Solutions Unit I Chapter 14 Copright 009 Pearson Education Canada

15 Analsis and Solution Resole each position ector into its and components and then add them using trigonometr. d d d d 1 1 = 0 =-300 m = (550 m)(cos75 ) = 14.4 m = d cos = d sin = (550 m)(sin 75 ) = m d = d + d d 1 = m = 14.4 m d = d + d d 1 =- 300 m m = 31.3 m Pearson Phsics Solutions Unit I Chapter 15 Copright 009 Pearson Education Canada

16 ( ) ( ) d = d + d = ( 14.4 m ) + (31.3 m) = 7 m opposite tan = adjacent 31.3 m = 14.4 m = = tan (1.64) = 58 The wildlife biologist s displacement is 7 m [58 o N of E]. Student Book page 90. Check and Reflect Knowledge 1. The cosine function, R = R cos, is used if the angle is with respect to the -ais. R θ R R = R cos θ R The sine function, R = R sin, is used if the angle is with respect to the -ais. R θ R R R = R sin θ. (a) False. The order in which ectors are added is not important because ectors are drawn to scale, and the magnitude and direction of a resultant should be the same regardless of how the ectors are added. Pearson Phsics Solutions Unit I Chapter 16 Copright 009 Pearson Education Canada

17 (b) False. Displacement is not alwas equal to distance. Displacement is a measure of the straight line directl joining a starting point and an endpoint, whereas distance is a measure of the path traelled between the starting point and the endpoint. Distance ma be greater than or equal to displacement. 3. The naigator method is used when the compass directions, north, south, east and west, are inoled. Applications 4. The components of this ector are d = (55 cm)(cos 30) = 48 cm d = (55 cm)(sin 30) = 8 cm 5. (a) Gien d1 5.0 km [W] d 3.0 km [N] d3.0 km [E] d4 1.5 km [S] distance ( d ) displacement ( d ) Analsis and Solution For distance, add the magnitudes of the four ectors. For displacement, add the collinear ectors and then use the Pthagorean theorem and the tangent function. d 5.0 km 3.0 km.0 km 1.5 km 11.5 km d = 5.0 km [W] km [N] +.0 km [E] km [S] = 5.0 km [W] + (-.0 km [W]) km [N] + (-1.5 km [N]) = 3.0 km [W] km [N] R = ( 3.0 km) + ( 1.5 km) = 3.4 km -1æ1.5 km ö = tan ç çè3.0 km ø = 7 d = 3.4 km [7 N of W] Pearson Phsics Solutions Unit I Chapter 17 Copright 009 Pearson Education Canada

18 The distance is 11.5 km and the displacement is 3.4 km [7 o N of W]. (b) Gien d1 500 m [N] l 150 m [E] distance ( d ) displacement ( d ) Analsis and Solution Attempting to cross the rier, the swimmer is moed off course. Her distance is how far she actuall swam in the water. d 500 m 150 m 5 m d = 500 m [N] m [E] To find the direction for displacement, use of the tangent function m = tan æ ç ö çè500 m ø = 17 d = 5 m [17 E of N] The swimmer swam a distance of 5 m and her displacement was 5 m [17 o E of N]. (c) Gien o d1 750 m [90 ] o d.0 km [70 ] distance ( d ) displacement ( d ) Analsis and Solution To determine distance, add the magnitudes of the position ectors. For displacement, add the position ectors once the units and angles are the same. Pearson Phsics Solutions Unit I Chapter 18 Copright 009 Pearson Education Canada

19 d 750 m.0 km 0.75 km.0 km.95 km d = 750 m [90 ] +.0 km [70 ] =-0.75 km [70 ] +.0 km [70 ] = 1.45 km [70 ] The distance traelled is.95 km and the displacement is 1.45 km [70 o ]. 6. Gien d 1 = 5.0 km [45 o W of N] d = 7.0 km [45 o S of E] displacement ( d ) Analsis and Solution The ectors are in opposite directions. So, the magnitude of the displacement of the boat is.0 km. The direction is [45 o S of E]. The boat s displacement is.0 km [45 o S of E]. 7. Gien o 355 m/s [30 ] magnitude of the ertical elocit component ( ) Pearson Phsics Solutions Unit I Chapter 19 Copright 009 Pearson Education Canada

20 Analsis and Solution Resole the ector into its ertical component using sin. sin (355 m/s)(sin 30 ) 178 m/s The magnitude of the ertical component of the pellet s elocit is 178 m/s. 8. Gien d1 1.0 km [55 N of E] d 3.15 km [70 S of E] displacement ( d ) Analsis and Solution Resole each ector into its components and perform ector addition. d1 = d1 cos = (1.0 km)(cos 55 ) = km d1 = d1 sin = (1.0 m)(sin 55 ) = km Pearson Phsics Solutions Unit I Chapter 0 Copright 009 Pearson Education Canada

21 d d = d sin = (3.15 km)(sin 0 ) = km =-d cos =-(3.15 km)(cos 0 ) =-.960 km d = d + d d 1 = km km = km d = d + d d 1 = km km = km ( ) ( ) d = d + d = ( km ) + ( km) =.651 km opposite tan = adjacent km = km = = tan (1.10) = 48 From the diagram, this angle is with respect to the positie -ais in the clockwise direction, or S of E. The jet ski s displacement is.65 km [48 o S of E]. 9. Gien o km/h [5 N of W] o 4.5 km/h [65 E of N] t 1 35 min t 0 min Pearson Phsics Solutions Unit I Chapter 1 Copright 009 Pearson Education Canada

22 displacement ( d ) aerage elocit ( ae ) Analsis and Solution Using the uniform motion equation, determine the displacement in both directions. km 1 h d1 6.0 [5 N of W] 35 min h 60 min 3.5 km [5 N of W] km 1 h d 4.5 [65 E of N] 0 min h 60 min 1.5 km [65 E of N] Resole each displacement ector into its and components and then perform ector addition. d =-d d d d cos =-(3.5 km)(cos 5 ) =-3.17 km = d sin = (3.5 m)(sin 5 ) = 1.48 km = d sin = (1.5 km)(sin 65 ) = 1.36 km = d cos = (1.5 km)(cos65 ) = km d = d + d 1 = km km =-1.81 km d = d + d 1 = 1.48 km km =.11 km Pearson Phsics Solutions Unit I Chapter Copright 009 Pearson Education Canada

23 ( ) ( ) d = d + d = ( km ) + (.11 km) =.78 km opposite tan = adjacent.11 km = 1.81 km = = tan (1.17) = 49 From the diagram aboe, this angle is 49 o N of W. Dd Find aerage elocit using the equation ae =. D t Dd ae = D t.78 km [49 N of W] 60 min = (35 + 0) min 1 h = 3.0 km/h [49 N of W] The jogger s total displacement is.8 km [49 o N of W]. The jogger s aerage elocit is 3.0 km/h [49 o N of W]. 10. Gien d 38 m [90 ] (a) components of displacement ( d, d ) (b) distance to second base ( d ) Analsis and Solution Since the baseball diamond is square in shape, the two sides of the triangle formed from home base to first base, and from first base to second base, must be equal. The angle must be 45 o because the triangle is isosceles. Pearson Phsics Solutions Unit I Chapter 3 Copright 009 Pearson Education Canada

24 Use the sine and cosine functions to determine the components of the plaer s displacement. (a) d dcos 38 m (cos 45 ) 7 m d dsin 38 m (sin 45 ) 7 m (b) The distance is not 38 m because the plaer does not run in a straight line but to first then to second base, or 7 m 7 m 54 m. (a) The components of the baseball plaer s displacement are 7 m in both the and directions. (b) The plaer s distance traelled is 54 m. 11. Gien o d m [310 ] o d 35.0 m [135 ] displacement ( d ) Analsis and Solution Resole each ector into its and components and then perform ector addition. Pearson Phsics Solutions Unit I Chapter 4 Copright 009 Pearson Education Canada

25 d d d d d cos 45.0 m (cos 50 ) 8.95 m d sin 45.0 m (sin 50 ) m d cos 35.0 m (cos 45 ) 4.75 m d sin 4.75 m d d d 35.0 m (sin 45 ) m ( 4.75 m) m d d d d m 4.75 m d 9.7 m d d d m 9.7 m 10.6 m opposite tan adjacent 9.7 m m.38 1 tan (.38) 66.8 Pearson Phsics Solutions Unit I Chapter 5 Copright 009 Pearson Education Canada

26 From the diagram, this angle is with respect to the positie -ais in the clockwise direction. Using the polar coordinates method, this angle is The skateboarder s displacement is 10.6 m [93 o ]. Student Book page 91 Concept Check Since the obserer on the train has the same elocit as the stationar ball and the train (5 m/s [forward]), the obserer sees the ball roll at a elocit of 5 m/s [forward]. The obserer on the ground will see the ball rolling at the combined elocities of the ball and train, or 5 m/s [forward] + 5 m/s [forward] = 50 m/s [forward]. If the ball rolls 5 m/s [backward], its elocit is the same for an obserer on the train. For an obserer on the ground, its elocit is 5 m/s [forward] + 5 m/s [backward] = 5 m/s [forward] 5 m/s [forward] = 0 m/s. Therefore, the ball appears stationar. Eample.6 Practice Problems 1. Gien swimmer = 1.8 m/s [N] d = 00 m current = 1. m/s [W] (a) ground elocit ( ground ) (b) time ( t ) Analsis and Solution (a) ground = swimmer + current Student Book page 95 ( ) ( ) ( ) ground swimmer current ground (1.8 m/s) (1. m/s). m/s Pearson Phsics Solutions Unit I Chapter 6 Copright 009 Pearson Education Canada

27 opposite tan adjacent 1 1. m/s tan 1.8 m/s 34 From the diagram, this angle is W of N. d (b) Use the equation and the swimmer s speed. Use the scalar form of the t equation because ou are diiding b a ector. d t 00 m m 1.8 s s (a) The swimmer s ground elocit is. m/s [34º W of N]. (b) It takes the swimmer s to cross the rier.. Gien current = 1. m/s [W] swimmer = 1.8 m/s direction of ground elocit = N (a) direction (b) time ( t ) Analsis and Solution ground (a) = swimmer + current opposite sin hpotenuse 1 1. m/s sin 1.8 m/s 4 From the diagram, this angle is E of N. Pearson Phsics Solutions Unit I Chapter 7 Copright 009 Pearson Education Canada

28 d (b) Use the equation. Since displacement and elocit must be in the same t direction, first find the ground speed. ( ) ( ) ( ) swimmer ground current ( ground ) ( swimmer ) ( current ) ground (1.8 m/s) (1. m/s) 1.34 m/s Use the scalar form of the equation because ou are diiding b a ector. d t 00 m m 1.34 s s (a) The swimmer must direct herself at an angle of [4 E of N]. (b) It takes the swimmer s to cross the rier. Student Book page 97 Eample.7 Practice Problems 1. [Note: Practice Problem 1 on page 97 in the 4 th edition should be corrected to read as follows: An airplane flies with an air elocit of 750 km/h [N]. If the wind elocit is 60 km/h [15 E of N], what is the plane s ground elocit? The correct answer to this question is km/h [1 E of N] and the solution is as follows.] Gien o wind 60 km/h [15 E of N] air 750 km/h [N] ground elocit ( ground ) Analsis and Solution Determine the ground elocit of the airplane b resoling each ector into its east and north components and then using the Pthagorean theorem and the tangent function. Consider north and east positie. Pearson Phsics Solutions Unit I Chapter 8 Copright 009 Pearson Education Canada

29 wind wind air wind wind sin 60 km/h (sin 15 ) 15.5 km/h cos 60 km/h (cos 15 ) km/h 0 air 750 km/h ground air wind ground air wind km/h 15.5 km/h wind wind 15 wind Pearson Phsics Solutions Unit I Chapter 9 Copright 009 Pearson Education Canada

30 ground air wind 750 km/h km/h 808 km/h ground ground ground 15.5 km/h 808 km/h 808 km/h = km/h opposite tan adjacent ground ground 15.5 km/h 808 km/h tan (0.019) 1 Pearson Phsics Solutions Unit I Chapter 30 Copright 009 Pearson Education Canada

31 The angle is therefore 1 o E of N. The airplane s ground elocit is km/h [1 E of N].. Gien o ground 856 km/h [5.0 W of S] wind 65.0 km/h [S] air elocit ( air ) Analsis and Solution Perform ector subtraction to determine the components of the air elocit. Using the Pthagorean theorem, find the magnitude of the air elocit and then use the tangent function to determine the air elocit direction. N ground ground wind wind ground ground sin 856 km/h (sin 5.0 ) km/h cos 856 km/h (cos 5.0 ) km/h km/h Pearson Phsics Solutions Unit I Chapter 31 Copright 009 Pearson Education Canada

32 ground air wind air ground wind air ground wind km/h km/h air ground wind km/h ( 65.0 km/h) km/h air air air km/h km/h 798 km/h opposite tan adjacent km/h km/h tan (0.509) 7.0 From the diagram, the angle is 7.0 o W of S. The air elocit of the jetliner is 798 km/h [7.0 o W of S]. 3. Gien o ground 795 km/h [5 W of N] d 100 km [N] time ( t ) Pearson Phsics Solutions Unit I Chapter 3 Copright 009 Pearson Education Canada

33 Analsis and Solution Calculate the north component of the ground elocit of the plane. d Use to determine the time. t ground = (795 km/h)(cos5 ) = 70.5 km/h d = t d t = 100 km = km 70.5 h = h It will take the plane h to trael 100 km north. Student Book page 99 Eample.8 Practice Problems 1. Gien boat 4.50 m/s [W] current.0 m/s [0 W of N] ground elocit ( ground ) Analsis and Solution Resole the boat and current ectors into their west and north components, then perform ector addition. Pearson Phsics Solutions Unit I Chapter 33 Copright 009 Pearson Education Canada

34 boat boat current current 4.50 m/s 0 current current sin.0 m/s (sin 0 ) ground boat current ground boat current m/s cos 1.88 m/s.0 m/s (cos 0 ) 4.50 m/s ( m/s) 5.18 m/s ground boat current m/s 1.88 m/s ground ground ground 5.18 m/s 1.88 m/s 5.5 m/s opposite tan adjacent 1.88 m/s 5.18 m/s tan (0.363) 0 From the diagram, the angle is 0 o N of W. Pearson Phsics Solutions Unit I Chapter 34 Copright 009 Pearson Education Canada

35 The boat s ground elocit is 5.5 m/s [0 o N of W].. Gien boat 13 m/s [N] jogger 3.75 m/s [0 N of E] ground elocit ( ground ) Analsis and Solution Determine the north and east components of the jogger s elocit and the ship. Perform ector addition on each component of the resultant elocit and then calculate the magnitude of the resultant using the Pthagorean theorem. Use the tangent function to determine the resultant direction. boat boat jogger jogger 0 13 m/s jogger jogger cos 3.75 m/s (cos 0 ) 3.5 m/s sin 3.75 m/s (sin 0 ) 1.8 m/s ground boat jogger ground boat jogger ground boat jogger 03.5 m/s 3.5 m/s 13 m/s 1.8 m/s 14.8 m/s Pearson Phsics Solutions Unit I Chapter 35 Copright 009 Pearson Education Canada

36 ground ground ground 3.5 m/s 14.8 m/s 15 m/s opposite tan adjacent 14.8 m/s 3.5 m/s tan (4.05) 76 From the diagram, this angle is 76 o N of E. The resultant elocit of the jogger is 15 m/s [76 o N of E]. 3. Gien d 65.0 km [N] t 3.0 h direction = 55 o [W of N] elocit of ship ( ship ) Analsis and Solution Determine the northern component of the ship s elocit using the equation Use the result to determine the ship s elocit. d. t Pearson Phsics Solutions Unit I Chapter 36 Copright 009 Pearson Education Canada

37 d t 65.0 km 3.0 h 1.67 km/h cos cos 1.67 km/h cos km/h The ship s elocit is 38 km/h [55 o W of N]. Student Book page 100 Eample.9 Practice Problems 1. Gien current = 3.0 m/s [W] d = 80 m boat = 4.0 m/s (a) time ( t ) if the boat points directl north and moes at an angle downstream (b) time ( t ) if the boat points at an angle upstream and moes directl north Analsis and Solution (a) Draw a ector diagram. Pearson Phsics Solutions Unit I Chapter 37 Copright 009 Pearson Education Canada

38 d Use the equation to find the time taken to cross. Since the boat s elocit is t parallel to the width of the rier, use the boat s elocit to calculate the time. Use the scalar form of the equation because ou are diiding b a ector. d t 80 m m 4.0 s 0 s (b) Draw a ector diagram. Since the ground elocit is parallel to the width of the rier, first calculate the ground elocit. ground boat current boat ground current ground boat current ground boat current 4.0 m/s 3.0 m/s.65 m/s d Use the equation to find the time taken to cross. Use the scalar form of the t equation because ou are diiding b a ector. d t 80 m m.65 s 30 s (a) The time taken to cross the rier if the boat points directl north is 0 s. (b) The time taken to cross the rier if the boat points at an angle upstream and moes directl north is 30 s. Pearson Phsics Solutions Unit I Chapter 38 Copright 009 Pearson Education Canada

39 Student Book page Check and Reflect Knowledge 1. Wind or current will increase an object s ground speed when the wind or current is in the same direction, or at least has a component in the same direction as the object s motion. For eample, a north wind will increase an object s ground speed if the object is moing in the northerl direction.. Wind or current will change an object s direction (or heading) with respect to the ground, but not the object s speed in the original, intended direction, onl if its elocit is perpendicular to the elocit of the object with respect to the medium itself. For eample, a rier s current that is perpendicular to a swimmer s motion will not alter a swimmer s swimming speed in the water but will alter the swimmer s direction, pushing the swimmer downstream. The swimmer s ground elocit increases, but the time taken to cross the rier will remain the same. 3. Zero displacement occurs when the current or wind has the same speed as the moing object, but in the opposite direction. For eample, in a training pool, the current is set at the same speed, but in the opposite direction to the swimmer s motion. As a result, the swimmer remains in the same spot relatie to the swimming pool. 4. Perpendicular components of motion are independent of one another. An eample is a bullet fired horizontall. It will continue its horizontal motion at muzzle elocit, but will fall erticall in step with an object that is dropped from rest. Applications 5. Gien current 0.60 m/s d m swimmer 1.35 m/s (a) time ( t ) when swimmer is angled upstream and ends up on the far bank directl across from where he started (b) time ( t ) and distance ( d ) when swimmer is pointed directl across the rier and is carried downstream Analsis and Solution (a) Draw a ector diagram current swimmer ground Pearson Phsics Solutions Unit I Chapter 39 Copright 009 Pearson Education Canada

40 ( ) ( ) ( ) swimmer ground current ( ) ( ) ( ) ground swimmer current ( ) ( ) ground swimmer current ground (1.35 m/s) (0.60 m/s) 1.09 m/s d Use the scalar equation to sole for time. t d t m m 1.09 s 88 s It will take the swimmer 88 s to cross the rier. (b) Draw a ector diagram current swimmer ground Use the scalar equation d t d t to sole for time m m 1.35 s 79 s Now, calculate how far downstream from his starting point the swimmer ends up. d t m s s 47 m (a) It will take the swimmer 88 s to cross the rier. (b) It will take the swimmer 79 s to cross the rier and he will end up 47 m downstream from where he started. Pearson Phsics Solutions Unit I Chapter 40 Copright 009 Pearson Education Canada

41 6. Gien air 65 km/h [N] headwind 3.0 km/h [S] tailwind 3.0 km/h [N] crosswind 3.0 km/h [W] ground elocit ( ground ) Analsis and Solution (a) A headwind will decrease the plane s elocit. Use ector addition. ground = air + wind = 65 km/h [N] km/h [S] = 65 km/h [N] + (-3.0 km/h [N]) = 65 km/h [N]-3.0 km/h [N] = 33 km/h [N] (b) A tailwind will increase the plane s elocit. Use ector addition. ground = air + wind = 65 km/h [N] km/h [N] = 97 km/h [N] (c) A crosswind will alter the plane s direction. Use the Pthagorean theorem. = + ground air wind ground = ( 65 km/h) + ( 3.0 km/h) = 67 km/h opposite tan = adjacent 3.0 km/h = 65 km/h -1 = tan (0.11) = 6.9 From the diagram, this angle is 6.9 W of N or 83.1 o N of W. Pearson Phsics Solutions Unit I Chapter 41 Copright 009 Pearson Education Canada

42 The plane s resultant elocit is (a) 33 km/h [N] (b) 97 km/h [N] (c) 67 km/h [83.1 o N of W] 7. Gien dupstream 300 m ddownstream 300 m current 1.0 m/s swimmer 1.5 m/s time ( t ) Analsis and Solution Calculate the time to go downstream and the time to go upstream using the equation d. Then add the two times. The current will increase the swimmer s speed going t downstream and will decrease the swimmer s speed going upstream. downstream swimmer current 1.5 m/s 1.0 m/s.5 m/s d t ddownstream tdownstream downstream 300 m m.5 s 10 s upstream swimmer current 1.5 m/s 1.0 m/s 0.5 m/s dupstream tupstream upstream 300 m m 0.5 s 600 s t tdownstream tupstream 10 s 600 s 70 s = 7.10 s The swimmer takes s to complete the round trip. Pearson Phsics Solutions Unit I Chapter 4 Copright 009 Pearson Education Canada

43 8. Gien air 800 km/h [E] wind 60 km/h [4 E of N] ground elocit ( ground ) Analsis and Solution Resole the air and wind elocities into their components and then perform ector addition. Consider east and north positie. air air wind 800 km/h 0 sin wind 60 km/h (sin 4 ) 40.1 km/h wind wind cos 60 km/h (cos 4 ) 44.6 km/h ground air wind ground air wind ground air wind 800 km/h 40.1 km/h km/h km/h 44.6 km/h ground ground ground km/h 44.6 km/h km/h opposite tan adjacent km/h km/h tan (0.0531) 3 ground θ ground ground Pearson Phsics Solutions Unit I Chapter 43 Copright 009 Pearson Education Canada

44 From the diagram, the angle is 3 o N of E. The ground elocit of the airplane is km/h [3 o N of E]. 9. Gien air 3.0 m/s [E] d 3. km o wind 1.75 m/s [5 W of S] ground elocit ( ground ) time ( t ) Analsis and Solution Draw a diagram Resole the air and wind elocities into their and components. Perform ector addition to determine the plane s ground elocit. Consider south and west negatie. d Use the ground speed and the equation to determine the time taken to trael t 3. km. air air wind 3.0 m/s 0 sin wind 1.75 m/s (sin 5 ) m/s Pearson Phsics Solutions Unit I Chapter 44 Copright 009 Pearson Education Canada

45 wind wind cos 1.75 m/s (cos 5 ) 1.59 m/s ground air wind ground air wind 3.0 m/s ( m/s).6 m/s ground air wind 0 ( 1.59 m/s) 1.59 m/s ground ground ground.6 m/s 1.59 m/s.76 m/s opposite tan adjacent ground 1.59 m/s θ.6 m/s ground tan (0.70) 35 From the diagram, the angle is 35 o S of E. d = t d t = ground ground 300 m = m.76 s 3 = s = 19 min The plane s ground elocit is.8 m/s [35 o S of E]. It would take the plane 19 minutes to trael 3. km. 10. Gien air = 600 km/h [W] wind = 40 km/h [45 o W of S] ground elocit ( ground ) Pearson Phsics Solutions Unit I Chapter 45 Copright 009 Pearson Education Canada

46 Analsis and Solution Resole the air and wind elocities into their and components. Perform ector addition to determine the plane s ground elocit. Consider west and south negatie. air air wind 600 km/h 0 sin wind 40 km/h (sin 45 ) 8.3 km/h wind wind cos 40 km/h (cos 45 ) 8.3 m/s ground wind air ground wind air 8.3 km/h ( 600 km/h) 68.3 km/h ground wind air 8.3 km/h km/h ground ground ground 68.3 km/h 8.3 km/h 68.9 km/h = km/h opposite tan adjacent 8.3 km/h 68.3 km/h ground ground tan (0.0450) 3 From the diagram, the angle is 3 o S of W. The plane s ground elocit is km/h [3 o S of W]. 11. Gien boat = 4.0 m/s [N] current =.5 m/s [W] d = 0.80 km = 800 m θ ground Pearson Phsics Solutions Unit I Chapter 46 Copright 009 Pearson Education Canada

47 (a) ground elocit ( ground ) (b) time ( t ) Analsis and Solution (a) The ectors are perpendicular, so use the Pthagorean theorem to determine the magnitude of the ground elocit. ground boat current ground boat current 4.0 m/s.5 m/s m/s Use the tangent function to find the angle. opposite tan adjacent.5 m/s 4.0 m/s 0.65 tan (0.65) 3 From the diagram, the angle is 3 o W of N. d (b) To find the time, use the equation and the magnitude of the boat s t elocit because it is parallel to the width of the rier. Use the scalar form of the equation because ou are diiding b a ector. d t 800 m m 4.0 s.010 s Pearson Phsics Solutions Unit I Chapter 47 Copright 009 Pearson Education Canada

48 (a) The canoe s ground elocit is 4.7 m/s [3 o W of N]. (b) It takes the canoe.0 10 s to cross the rier. Student Book page 104 Concept Check position uniform Horizontal elocit uniform acceleration uniform (= 0) position non-uniform Vertical elocit non-uniform acceleration uniform (= 9.8 m/s ) Student Book page 105 Concept Check (a) Air resistance (friction) and spin are being neglected. (b) The projectile finall stops when it hits the ground. Thus, friction in the horizontal direction brings the object to rest. (c) The projectile s path would still be parabolic, if its initial elocit had a ertical component, because its motion is still affected b acceleration due to grait. Student Book page 107 Eample.10 Practice Problems 1. Gien 30 cm/s d 1.5 m i 0 m/s range ( d ) Analsis and Solution Choose down to be positie. The coin accelerates due to grait once it leaes the table. Its initial ertical elocit is zero. Determine the time to fall from a height of m using the equation d i t a. t Pearson Phsics Solutions Unit I Chapter 48 Copright 009 Pearson Education Canada

49 1 d i t a t d t a (1.5 m ) m 9.81 s s d Then determine the range b using the equation. t d t cm 30 (0.505 s ) s 15 cm The coin lands 15 cm from the base of the table.. Gien 5.0 m/s t.50 s h m ertical distance ( ) d Analsis and Solution Choose down to be positie. The distance dropped is a measure of ertical displacement. Since the object is fired horizontall from the top of the cliff, its initial 1 ertical elocit is zero. Use the equation d i t a. t 1 d i t a t 1 m (.50 s ) s 30.7 m The arrow will fall 30.7 m in.50 s. Pearson Phsics Solutions Unit I Chapter 49 Copright 009 Pearson Education Canada

50 3. Gien d 40.0 m d 100 m horizontal speed ( ) Analsis and Solution Choose down to be positie. Determine the amount of time it takes the object to fall m using the equation d i t a t, where i = 0. 1 d i t a t d t a (100 m) 9.81 m/s s Calculate the horizontal elocit b using the equation d t 40.0 m s 8.86 m/s The horizontal speed of the object is 8.86 m/s. d. t Eample.11 Practice Problems 1. (a) Gien i = 10.0 m/s t = 3.0 s (a) range ( d ) (b) height ( d ) Student Book page 109 Pearson Phsics Solutions Unit I Chapter 50 Copright 009 Pearson Education Canada

51 Analsis and Solution (a) Since there is no acceleration in the horizontal direction, use the equation d to find the range. t d t (10.0 m/s)(3.0 s) 30 m 1 (b) Use the equation d i t a t, where i = 0 because there is no initial ertical elocit component. 1 d 0 (9.81 m/s ) 3.0 s 44 m (a) The ball traels a horizontal distance of 30 m. (b) The ball was thrown from a height of 44 m.. (a) Gien i = 0.0 m/s [30 o ] t = 3.0 s (a) range ( d ) (b) height ( d ) (c) maimum height ( d ) Analsis and Solution Choose up and forward to be positie. (a) Find the component of the ball s initial elocit. i i cos (0.0 m/s)(cos 30 ) 17.3 m/s d Since there is no acceleration in the horizontal direction, use the equation to t find the range. d t (17.3 m/s)(3.0 s) 5 m (b) Find the component of the ball s initial elocit. i i sin (0.0 m/s)(sin 30 ) 10.0 m/s 1 Use the equation d i t a t to find the height from which the ball was thrown. Pearson Phsics Solutions Unit I Chapter 51 Copright 009 Pearson Education Canada

52 1 d (10.0 m/s) 3.0 s ( 9.81 m/s ) 3.0 s 14 m The negatie sign means that the ball was thrown from a height 14 m aboe a reference point. (c) Since the projectile s final ertical elocit component is zero the instant it reaches maimum height, use the equation f i a d to calculate maimum height. a d f i f i d a 0 (10.0 m/s) ( 9.81 m/s ) 5.1 m Since the ball was thrown from a height of 14 m, the total maimum height reached b the ball is 14 m m = 19 m (a) The ball traels a horizontal distance of 5 m. (b) The ball was thrown from a height of 14 m. (c) The ball s maimum height is 19 m. Eample.1 Practice Problems 1. Gien i = 17.0 m/s [0 o ] maimum height ( ) d Student Book page 111 Analsis and Solution Choose up and forward to be positie. Calculate the component of the elocit. i (17.0 m/s)(sin 0 ) 5.81 m/s Since the projectile s final ertical elocit component is zero the instant it reaches maimum height, use the equation a d to calculate maimum height. f i Pearson Phsics Solutions Unit I Chapter 5 Copright 009 Pearson Education Canada

53 f i f i d a a d 0 (5.81 m/s) ( 9.81 m/s ) 1.7 m The projectile reaches a maimum height of 1.7 m.. Gien d 300 km [forward] m [forward] d 100 km [up] m [up] maimum initial elocit ( i ) Analsis and Solution Choose up and forward to be positie. Determine the amount of time it takes for 1 the rocket to fall from maimum height. Use the equation d i t a, t where i (at maimum height) equals zero. 1 d i t a t d t a ( m ) m 9.81 s 14.8 s This time is the time taken to fall down from maimum height. The total time of flight is therefore 14.8 s 85.6 s. d Using the equation =, diide the range of the rocket b the total time to t determine the initial horizontal elocit component. Pearson Phsics Solutions Unit I Chapter 53 Copright 009 Pearson Education Canada

54 i = = d t m 85.6 s 1050 m/s = f i Find the initial ertical elocit component using the equation a, since t ertical motion is uniforml accelerated and the final ertical elocit component at maimum height is zero. direction: f - i a = t = -a t i f æ m ö = s ç è s ( ) ø =1401 m/s Use the Pthagorean theorem to determine the magnitude of the initial elocit. ( ) ( ) = + i i i = ( 1050 m/s) + ( 1401 m/s) = 1751 m/s 3 = m/s Use the tangent function to determine the direction of the initial elocit. opposite tan = adjacent = i i 1401 m/s = 1050 m/s -1 = tan (1.333) = The rocket had a maimum initial elocit of m/s.4 Check and Reflect Student Book page 11 [ 53.1 ]. Knowledge 1. Both athletes hae ertical and horizontal components to their dies into the water. Platform diers must ensure that the do not hit the diing board on the wa down but want to minimize their horizontal moement. Speed swimmers want to get into the pool as far as possible from the starting block, so the maimize the horizontal Pearson Phsics Solutions Unit I Chapter 54 Copright 009 Pearson Education Canada

55 elocit component while tring not to add too much to the minimum ertical elocit that is determined b the ertical distance between the starting block and the water surface.. The range increases with angle up to a launch angle of 45, then decreases back to zero for ertical projection. 3. (a) Yes, there is a location on its trajector where the acceleration and elocit ectors are perpendicular. At maimum height, acceleration is straight down and the elocit ector has onl the horizontal component at this point. (b) No, there isn t a location on its trajector where the acceleration and elocit ectors are parallel. The elocit is initiall at an angle and it has a horizontal component. If the ectors were to be parallel, the horizontal elocit would hae to become zero, but it remains constant throughout the whole trajector. 4. A noice swimmer who jumps straight up risks injur to the spine because he or she will land on the deck of the pool. Most swim instructors teach students to put their toes oer the edge of the pool and jump out, giing them horizontal uniform motion, so the land in the water, not on the deck. Applications 5. Gien 6.0 m/s 0 m/s i d 0.50 m range ( d ) Analsis and Solution Choose down to be positie. The cup has an initial horizontal elocit component of 6.0 m/s, whereas its initial ertical elocit component is zero. Use the equation 1 d i t a, t where i = 0. direction: Find t. Pearson Phsics Solutions Unit I Chapter 55 Copright 009 Pearson Education Canada

56 1 d i t a t d t a (0.50 m ) m 9.81 s s The runner must drop the cup in front of the garbage can so that it will land inside the can and not on the ground. The amount of time taken to fall 0.50 m is equal to the amount of time the cup has to trael horizontall at a uniform rate of 6.0 m/s. direction: Calculate the range. d t m 6.0 (0.319 s ) s.0 m The runner should release his cup when he is.0 m from the garbage can. 6. Gien o i 7.0 m/s [35 ] initial horizontal elocit component ( i ) initial ertical elocit component ( i ) range ( d ) maimum height ( d ) Analsis and Solution Use cos and sin to determine horizontal () and ertical () elocit components. i (7.0 m/s)(cos 35 ).1 m/s i (7.0 m/s)(sin 35 ) 15.5 m/s Calculate the maimum height using the equation f i ad, where f = 0. Pearson Phsics Solutions Unit I Chapter 56 Copright 009 Pearson Education Canada

57 direction: a d f d i f i a m/s 9.81 m/s 1. m Calculate the time the ball is in the air using the ertical elocit component and the f i equation a. The initial and final ertical elocit components hae the same t magnitude but opposite direction, so f i a t f i t a 15.5 m/s (15.5 m/s) 9.81 m/s m 31.0 s m 9.81 s 3.16 s d Then calculate how far the ball will trael using the equation. t direction: The ball s range is therefore d t m.1 (3.16 s ) s 69.8 m The initial horizontal elocit component is.1 m/s. The initial ertical elocit component is 15.5 m/s. The ball traels up ( d ) 1. m and out ( d ) 69.8 m. 7. Gien o i 17.0 m/s [5 ] receier s speed ( receier ) Analsis and Solution Choose up and forward to be positie. Pearson Phsics Solutions Unit I Chapter 57 Copright 009 Pearson Education Canada

58 Resole the football s elocit into its and components. i icos (17.0 m/s)(cos 5 ) m/s sin i i (17.0 m/s)(sin 5 ) m/s f i Calculate the amount of time the ball is in the air using the equation a. t Since the football lands at the same height from which it is thrown, the magnitudes of the initial and final ertical elocit components are equal but their directions are opposite. So, f i a t f i t a m/s (7.185 m/s) 9.81 m/s m s m 9.81 s s d Determine the distance the ball traels using the equation. t The ball is in the air for s. During this time, the range of the ball is d t m s s.57 m Then subtract the receier s initial position from this alue to determine how far the receier must run. The receier must run a distance of drun d dreceier.57 m 1.0 m m Pearson Phsics Solutions Unit I Chapter 58 Copright 009 Pearson Education Canada

59 Diide the distance the receier must run b the time the ball is in flight to determine her speed. drun receier t m s 7.15 m/s The receier must hae a speed of 7.1 m/s to catch the ball. 8. Gien 3 d 8.59 m takeoff speed ( i ) Analsis and Solution Choose up and forward to be positie. i i cos i sin direction: f i a t f i t a i isin 3 ( isin 3 ) 9.81 m/s i sin m/s i sin m/s direction: d t Pearson Phsics Solutions Unit I Chapter 59 Copright 009 Pearson Education Canada

60 t d i 8.59 m i cos 3 Time traelled in the horizontal direction equals time traelled in the ertical direction. Sole for takeoff speed: t t i sin m 9.81 m/s i cos m 9.81 m/s i cos 3 sin 3 i 10.8 m/s Phillips takeoff speed was 10.8 m/s. 9. Gien i 10 m/s [55.0 ] d 50.0 m (a) time to reach maimum height ( ) tma height (b) maimum height with respect to the ground ( ) (c) total time in the air ( t ) (d) range ( d ) (e) final elocit components ( f, ) f Analsis and Solution d ma Choose up and forward to be positie. (a) Calculate the horizontal and ertical components of the initial elocit. i i cos Pearson Phsics Solutions Unit I Chapter 60 Copright 009 Pearson Education Canada

61 i (10 m/s)(cos 55.0 ) m/s sin i i (10 m/s)(sin 55.0 ) m/s f i At maimum height, ertical elocit is zero. Use the equation a to find t time. f i a t f i tup a m s m 9.81 s 10.0 s (b) Calculate the maimum height using the equation a d, where f = 0. f i d a m/s 9.81 m/s m s m 19.6 s f i 49.5 m Add the maimum height to the cliff height to determine the ertical displacement. d 49.5 m 50.0 m ma 54.5 m 1 d i t a t, where (at maimum height) equals zero and is negatie because motion is downward. (c) To find the time taken to fall down, use the equation i t down d a 54.5 m m 9.81 s 10.5 s Pearson Phsics Solutions Unit I Chapter 61 Copright 009 Pearson Education Canada

62 Total time in the air is the time taken to reach maimum height plus the time taken to fall down to the ground. t t t up down 10.0 s 10.5 s 0.54 s (d) The projectile eperiences uniform motion in the horizontal direction. d Use the equation. t d t m s s 1414 m = 1.41 km (e) Since motion in the horizontal direction is uniform, the horizontal component of the final elocit is the same as the horizontal component of the initial elocit. For the f i ertical component of final elocit, use the equation a. t f i a t a t f i m s m s s 103 m/s Since up and forward are positie, the negatie sign indicates that motion is downward. (a) The projectile reaches maimum height in 10.0 s. (b) Maimum height with respect to the ground is 54 m. (c) The total time in the air 0.5 s. (d) The range is 1.41 km. (e) The horizontal component of the final elocit is 68.8 m/s and the ertical component is 103 m/s. Pearson Phsics Solutions Unit I Chapter 6 Copright 009 Pearson Education Canada

63 Etension 10. Eplanation: i (m/s) i (m/s) d (m) d (m) Angle ( ) To find i and i, use i = i sin θ and i = i cos θ, where i = 0.0 m/s. To find d, first find the time (Δt) taken to trael the entire range using the equation d and d = i Δt + 1 a (Δt), where d = 0 and a = 9.81 m/s. Substitute the alue of Δt into the equation d = i Δt to sole for the range, d. To find maimum height,, diide the time (Δt) in half and substitute this alue into the equation d d = i Δt + 1 a (Δt). Pearson Phsics Solutions Unit I Chapter 63 Copright 009 Pearson Education Canada

64 Student Book pages Chapter Reiew Knowledge 1. Final position d (km) Displacement Distance [direction] d (km) d (km) reference point [direction] AB 3.0 km 3.0 km [E] 3.0 km [E] BC.0 km 3.6 km [34 N of E].0 km [N] CD 5.0 km 8. km [14 N of E] 5.0 km [E] AC 5.0 km 3.6 km [34 o N of E] 3.6 km [34 o N of E] AD 10.0 km 8. km [14 o N of E] 8. km [14 o N of E]. d dcos (55 m)(cos ) 41 m d dsin (55 m)(sin ) 37 m 3. When a projectile reaches maimum height, its ertical elocit component is zero. 4. Grait has an effect on the ertical component. As the football rises, the ertical component of its elocit decreases. When the football reaches maimum height, its ertical elocit component is zero. 5. The ector should be 10.0 cm long and point north. scale 50.0 km N 6. Since a ector s length represents its magnitude, ou can onl compare ectors when the are drawn to scale. 7. (a) d 5.0 m [S] 10.0 m [N] scale d N 5.0 m d 5.0 m [N] (b) d 65.0 cm [E] 75.0 cm [E] d 1 d Pearson Phsics Solutions Unit I Chapter 64 Copright 009 Pearson Education Canada

65 scale 10.0 cm d d 1 d d cm [E] (c) d 1.0 km [forward] 3.5 km [backward] scale 1.0 km d d d 1 d.5 km [backward] (d) d 35.0 km [right] 45.0 km [left] 35.0 km [right] 45.0 km [right] scale 10.0 km d d 1 d d 80.0 km [right] 8. An object thrown erticall upwards has an initial horizontal elocit component of zero. Applications 9. Gien = 50 knots D t= 50 min distance ( D d ) Analsis and Solution Conert knots to km/h and min to h. Manipulate the elocit equation: Dd = D t D d = Dt km 1 h min h 60 min D d = 50 knots 1 knot = km The plane traels km. 10. Gien i 30.0 m/s [55 ] range ( d ) maimum height ( ) d Pearson Phsics Solutions Unit I Chapter 65 Copright 009 Pearson Education Canada

66 Analsis and Solution Choose up and forward to be positie. Calculate the horizontal and ertical components of elocit as an epression of the initial elocit. i (30.0 m/s)(cos 55 ) 17. m/s i (30.0 m/s)(sin 55 ) 4.6 m/s f i Determine the amount of time the ball remains in the air using the equation a. t f i a t f i t a 4.6 m/s 4.6 m/s 9.81 m/s m 49. s m 9.81 s s d Use the uniform motion equation to determine the range. t d t m s s 86. m The golf ball reaches maimum height in half the time it is in the air or s s. Use the equation d i t a t to find maimum height. 1 d i t a t m s 1 m s s s 30.8 m Pearson Phsics Solutions Unit I Chapter 66 Copright 009 Pearson Education Canada

67 The ball s range is 86. m and its maimum height is 30.8 m. 11. Gien i 80.0 m/s [10 ] range ( d ) time ( t ) Analsis and Solution Choose up and forward to be positie. Calculate the ertical and horizontal components of the initial elocit. i (80.0 m/s)(cos 10 ) 78.8 m/s i (80.0 m/s)(sin 10 ) 13.9 m/s f i Determine the time the ball remains airborne using the equation a. t f i a t f i t a 13.9 m/s (13.9 m/s) 9.81 m/s m 7.8 s m 9.81 s.83 s d Calculate the range using the uniform motion equation. t d t m 78.8 (.83 s ) s.310 m The golf ball traels.3 10 m horizontall and is in the air for.83 s. 1. Gien = 4.0 m/s [N] boat Pearson Phsics Solutions Unit I Chapter 67 Copright 009 Pearson Education Canada

68 d = 10 m current = 5.0 m/s [E] (a) ground elocit ( ground ) (b) time ( t ) Analsis and Solution (a) ground = boat + current Use the Pthagorean theorem to calculate the ground elocit. ground boat current 4.0 m/s 5.0 m/s ground 6.4 m/s Use the tangent function to determine the angle. opposite tan adjacent m/s tan 4.0 m/s 51 From the diagram, this angle is 51 E of N. (b) To find the time, substitute distance and the canoe s speed into the uniform d motion equation,. Use the scalar form of the equation because ou are t diiding b a ector. d t 10 m 4.0 m/s 30 s (a) The ground elocit of the canoeist is 6.4 m/s [51 E of N]. (b) It takes the canoeist 30 s to cross the rier. Pearson Phsics Solutions Unit I Chapter 68 Copright 009 Pearson Education Canada

69 13. Gien 7.50 m/s t 3.0 s i 0 m/s (a) final ertical elocit component ( f ) (b) range ( d ) (c) horizontal elocit component at t = 1.50 s ( (1.50 s) ) Analsis and Solution Choose down to be positie. f i (a) Use the equation a, where i 0. t a t f i m s s 9 m/s d (b) Use the equation. t d t m 7.50 (3.0 s ) s 3 m (c) Because horizontal motion is uniform, the elocit at 1.50 s is the same as the initial horizontal elocit. (1.50 s) 7.50 m/s Pearson Phsics Solutions Unit I Chapter 69 Copright 009 Pearson Education Canada

70 (a) The object s ertical elocit component when it reaches the ground is 9 m/s [down]. (b) The object lands 3 m from the base of the cliff. (c) The object s horizontal elocit remains constant throughout, at 7.50 m/s [awa from cliff]. 14. Gien d.0 m d.0 m initial elocit ( i ) Analsis and Solution Choose up and forward to be positie. Determine the initial ertical elocit component using the equation f i a d, where = 0 (at maimum height). f a d f i i f a d m m s 6.6 m/s Then determine the amount of time it takes to trael.0 m erticall using the f i equation a. t f i t a 6.6 m/s 6.6 m/s 9.81 m/s m 1.5 s m s s Pearson Phsics Solutions Unit I Chapter 70 Copright 009 Pearson Education Canada