G ∆v G aav = ∆t 10.2 m/s [45° W of S] = 2.0s G 2 aav = 5.1 m/s [45° W of S] 19. (c) constant vix = 5.5 m/s CHAPTER 1 REVIEW (Pages 64–67) Understanding Concepts 1× 103 m 1h = 27.8 m/s 1. (a) 100 km/h 3 1km × 3.6 10 s Thus, 100 km/h is 27.8 m/s. 3 1km 3.6 × 10 s 2 (b) 97 m/s = 3.5 × 10 km/h 1h 3 × 1 10 m The peregrine falcon can dive at speeds of 3.5 × 102 km/h. (c) To convert km/h to m/s, divide by 3.6. To convert m/s to km/h, multiply by 3.6. 2. (a) L × T–1 results in speed. L L (b) 3 ¯ T = 2 results in acceleration. T T L (c) 2 ¯ T ¯ T = L results in length. T G G 3. (a) ∆d = a (∆t ) 2 Therefore: ? L L = 2 × T2 T ? 4. 5. 6. 7. L=L The equation is dimensionally correct because both sides of the equation have the same dimensions. (b) An equation may be dimensionally correct yet still be wrong. In (a), for example, the equation could be one of the following: G G 1G ∆d = v ∆t + a ∆t 2 i 2 G 1G 2 or ∆d = a ∆t 2 (a) The instantaneous speed is equal to the average speed throughout the motion. (b) The instantaneous velocity is equal to the average velocity throughout the motion. (c) The instantaneous speed is equal to the magnitude of the average velocity throughout the motion. (a) The displacement is equal to the area under a velocity-time graph. (b) The instantaneous acceleration is the slope of the line (for constant acceleration) or the slope of the tangent to the curve (for nonconstant acceleration) on a velocity-time graph. No, a component of a vector cannot have a magnitude greater than the vector’s magnitude. A vector can be broken into components using sine and cosine, which have a maximum of one. (a) Yes, two vectors having the same magnitude can be combined to give a zero resultant vector by adding two vectors that have opposite directions; for example, 5 m [S] + 5 m [N] = 0. (b) No, two vectors having different magnitudes cannot be combined to give a zero resultant vector. (c) Yes, three vectors having different magnitudes can be combined to give zero. Adding three vectors in two dimensions head-to-tail, can total zero. Even in one dimension three vectors can add to zero; for example, 2 m [S] + 5 m [N] + 3 m [S] = 0. Copyright © 2003 Nelson Chapter 1 Kinematics 71 8. G d1 = 214 m [E] G d 2 = 96 m [28° N of E] G d3 = 12 m [25° S of E] (a) G (b) ∆d = ? Let the +x direction be to the east and the +y direction be to the north. G G G G ∆d = ∆d1 + ∆d 2 + ∆d 3 ∆d y = ∆d1, y + ∆d 2, y + ∆d 3, y ∆d x = ∆d1, x + ∆d 2, x + ∆d 3, x = (96 m) sin 28° − (12 m) sin 25° = 214 m + (96 m) cos 28° + (12 m) cos 25° ∆d y = 40 m ∆d x = 310 m G ∆d = ∆d x 2 + ∆d y 2 = (310 m/s)2 + (40 m/s) 2 G ∆d = 312 m/s ∆d y tan θ = ∆d x 9. 40 m/s θ = tan −1 310 m/s θ = 7.4° The displacement from the tee needed to get a hole-in-one using components is 312 m/s, or 3.1 × 102 m [7.4° N of E]. Let the +x direction be to the east and the +y direction be north. G Let C be the addition of the two vectors. G G G C = A+ B C x = Ax + Bx C y = Ay + By = (5.1 km) cos 38° + (6.8 km) sin 19° C x = 6.2 km C y = −3.3 km = (5.1 km) sin 38° − (6.8 km) cos19° G (a) The vector that would add to the vector C to give a resultant displacement of zero would be equal in magnitude, but G G opposite in direction to the vector C . Thus, vector C is 72 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson G C = Cx 2 + C y 2 = (6.2 km)2 + (−3.3km) 2 G C = 7.0 km Cy tan θ = Cx 3.3 km θ = tan −1 6.2 km θ = 28° Thus, the vector that must be added is 7.0 km [28° N of W]. G (b) Let the vector D represent the desired vector. G G G 4.0 km [W] = A + B + D G G 4.0 km [W] = C + D G G D = 4.0 km [W] − C Dx = −4.0 km − 6.2 km Dy = −C y Dx = −10.2 km Dy = −3.3 km G D = Dx 2 + Dy 2 = (−10.2 km) 2 + (−3.3 km)2 G D = 11 km Dy tan θ = Dx 3.3km θ = tan −1 10.2 km θ = 18° The vector is 11 km [18° N of W]. G 10. (a) At 5 P.M., assume that people are sitting down to dinner. If we assume an average nose-to-toes ∆d value of 1 m [down] (when person is sitting), for a town of 2000, the resultant displacement vector of the sum of all the nose-to-toes vectors is about 2000 m or about 2 × 103 m [down]. (b) At 5 A.M., assume that everyone is sleeping lying down. Because the beds face different directions, we can assume that the vectors point in random directions and average to 0 m. Thus the resultant displacement averages to 0 m. 304.29 km 11. d = = 4.4100 km 69 laps ∆t = 84.118 s vav = ? d vav = ∆t 4.4100 × 103 m = 84.118 s vav = 52.426 m/s The average speed for the lap is 52.426 m/s. Copyright © 2003 Nelson Chapter 1 Kinematics 73 12. (a) ∆t = 2.0 s vav = 115 km/h = 31.9 m/s d=? d ∆t d = vav ∆t vav = = (31.9 m/s)(2.0 s) d = 64 m The distance in metres if the speed of your car is 115 km/h is 64 m. (b) Assuming that the average length of a car is 5.0 m, the number of car lengths is 13. (a) v1 = 24 m/s ∆d1 = 1.2 × 103 m v2 = 18 m/s ∆d2 = 1.2 × 103 m ∆t = ? 64 m = 13 car lengths. 5.0 m ∆t = ∆t1 + ∆t2 = ∆d1 ∆d 2 + v1 v2 = 1.2 ×103 m 1.2 × 103 m + 24 m/s 18 m/s ∆t = 1.2 × 102 s The time interval is 1.2 × 102 s. (b) vav = ? d vav = ∆t 1.2 × 103 m + 1.2 × 103 m = 1.2 × 102 s vav = 21 m/s The eagle’s average speed during this motion is 21 m/s. 14. (a) Starting from a positive position and a low velocity toward the zero position, the speed increases gradually at first and then dramatically. Upon reaching the zero position, the velocity changes direction. The motion now is a high velocity followed by decreasing velocity back to the initial position. The last part of the motion takes about half as long as the first part. (b) Starting from rest, the object undergoes uniform acceleration in one direction, followed by a higher uniform acceleration in the opposite direction for a shorter period of time until coming to a stop. 15. d1 = 4.5 m d2 = 6.8 m ∆t = 5.0 s (a) vav = ? d + d2 vav = 1 ∆t 4.5 m + 6.8 m = 5.0 s vav = 2.3 m/s The firefighter’s average speed is 2.3 m/s. (b) Let the +x direction be horizontal and the +y direction be down. Thus, ∆dx = 6.8 m ∆dy = 4.5 m G ∆d = ? 74 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson G ∆d = ∆d x 2 + ∆d y 2 = (6.8 m)2 + (4.5 m)2 G ∆d = 1.6 m ∆d y tan θ = ∆d x 4.5 m θ = tan −1 6.8 m θ = 33° The firefighter’s average velocity is 1.6 m/s [33° below the horizontal]. G 16. ∆d1 = 16 m [35° S of W] G ∆d 2 = 22 m [15° S of E] ∆t = 6.4 s G (a) ∆d = ? Solve using components. Let the +x direction be to the east and the +y direction be south. G G G ∆d = ∆d1 + ∆d 2 ∆d y = ∆d1 y + ∆d 2 y ∆d x = ∆d1 x + ∆d 2 x = ( −16 m) cos 35° + (22 m) cos 15° ∆d x = 8.1 m = (16 m) sin 35° + (22 m) sin 15° ∆d y = 15 m G ∆d = ∆d x 2 + ∆d y 2 = (8.1m)2 + (15 m)2 G ∆d = 17 m ∆d y tan θ = ∆d x 15 m θ = tan −1 8.1m θ = 61° G ∆ d = 17 m [61° S of E] or 17 m [29° E of S] , thus, the player’s resultant displacement is 17 m [29° E of S]. G (b) vav = ? G ∆d G vav = ∆t 17 m [29° E of S] = 6.4 s G vav = 2.7 m/s [29° E of S] Thus, the player’s average velocity is 2.7 m/s [29° E of S]. 17. vav = 100 km/h The directions are found by estimating the directions of the tangents at the positions D, E, and F. The instantaneous velocity at position D is 100 km/h [10° S of E]. The instantaneous velocity at position E is 100 km/h [10° N of E]. The instantaneous velocity at position F is 100 km/h [75° N of E]. Copyright © 2003 Nelson Chapter 1 Kinematics 75 18. (a) vi = 42 km/h vf = 105 km/h ∆t = 26 s = 7.2 × 10–3 h ∆d = ? ∆d = vav ( ∆t ) 1 (105 km/h + 42 km/h)(7.2 × 10−3 h) 2 ∆d = 0.53km The car travels 0.53 km in the time interval. G (b) aav = ? G G vf − vi G aav = ∆t 105 km/h [fwd] − 42 km/h [fwd] = 26 s G aav = 2.4 (km/h)/s [fwd] The magnitude of the car’s average acceleration is 2.4 (km/h)/s. G 19. vi = 0 G ∆d = 15 m [fwd] ∆t = 1.2 s G (a) a = ? G G 1G 2 ∆d = vi ∆t + a ( ∆t ) 2 G 1G 2 ∆d = 0 + a ( ∆t ) 2 G G 2∆d a= 2 ( ∆t ) = = 2(15 m [fwd]) (1.2 s) 2 G a = 21m/s 2 [fwd] The average acceleration of the cars is 21 m/s2 [fwd]. G (b) vf = ? G G G vf = vi + a ∆t = 0 + (20.8 m/s 2 [fwd])(1.2 s) G vf = 25 m/s [fwd] The velocity of the cars at 1.2 s is 25 m/s [fwd]. G (c) The magnitude of the acceleration in terms of g can be determined as: G a 20.8 m/s 2 G = g 9.8 m/s 2 G G a = 2.1 g G 20. vf = 4.0 × 102 m/s [fwd] G vi = 0 G ∆d = 0.80 m [fwd] G a=? 76 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson G 2 G2 G G vf = vi + 2a ∆d G G G vf 2 = 0 + 2a ∆d G G v2 a= f G 2 ∆d (4.0 × 102 m/s[fwd])2 = 2(0.8 m[fwd]) G a = 1.0 × 105 m/s 2 [fwd] The constant acceleration needed by the bullet is 1.0 × 105 m/s2 [fwd]. G 21. vi = 2.28 × 102 m/s [fwd] G a = 6.25 × 101 m/s2 [fwd] G ∆d = 1.86 × 103 m [fwd] G vf = ? G 2 G2 G G vf = vi + 2a ∆d G vf = (2.28 × 10 2 m/s[fwd]) 2 + 2(6.25 × 101 m/s 2 [fwd])(1.86 × 103 m[fwd]) G vf = 5.33 × 102 m/s[fwd] Thus, the rocket’s velocity is 533 m/s [fwd]. G 22. vf = 0 G a = 9.8 m/s2 [down] G ∆d =1.9 m [up] G vi = ? G 2 G2 G G vf = vi + 2a ∆d G G G 0 = vi 2 + 2 a ∆ d G G G vi = −2a ∆d = −2( −9.8 m/s 2 [up])(1.9 m [up] ) G vi = 6.1 m/s [up] The minimum vertical velocity needed by the salmon to jump to the top of the waterfall is 6.1 m/s [up]. G 23. ∆d = 2.0 × 102 m [fwd] G a =1.6 m/s2 [fwd] G (a) vi = 0.0 m/s ∆t = ? G G 1G 2 ∆d = vi ∆ t + a ( ∆ t ) 2 G 1G 2 ∆d = 0 + a ( ∆t ) 2 G 2 ∆d ∆t = G a = 2(2.0 × 102 m[fwd]) 1.6 m/s 2 [fwd] ∆t = 16s The motion takes 16 s. Copyright © 2003 Nelson Chapter 1 Kinematics 77 G (b) vi = 8.0 m/s [fwd] ∆t = ? G G 1G 2 ∆d = vi ∆t + a ( ∆t ) 2 G G 1G 2 0 = vi ∆t + a ( ∆t ) − ∆d 2 G G 1G 2 0 = a ( ∆ t ) + vi ∆ t − ∆ d 2 This is a quadratic equation with solution: ∆t = −b ± b 2 − 4ac 2a G G G G a −vi ± vi 2 − 4 ( −∆d ) 2 = G a 2 2 = −8.0 m/s[fwd] ± (8.0 m/s [fwd]) 2 + 2(1.6 m/s 2 [fwd])(2.0 × 102 m [fwd]) 1.6 m/s 2 [fwd] ∆t = 12 s The motion takes 12 s. 24. Let the +x direction be east and the +y direction be south. G vi = 240 m/s [28° S of E] G vf = 220 m/s [28° E of S] ∆t = 35 s G aav = ? ∆v y = vfy − viy ∆vx = vfx − vix = (220 m/s) sin 28° − (240 m/s) cos 28° ∆vx = −109 m/s = (220 m/s) cos 28° − (240 m/s) sin 28° ∆v y = 82 m/s G 2 2 ∆v = ∆v x + ∆v y = (−109 m/s)2 + (82 m/s) 2 G ∆v = 136 m/s G G ∆v a= ∆t 136 m/s = 35 s G a = 3.9 m/s 2 tan θ = vx vy 109 m/s θ = tan −1 82 m/s θ = 53° The airplane’s average acceleration during this time interval is 3.9 m/s2 [53° W of S]. 78 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson G 25. vf = 54 m/s [N] G aav =19 m/s 2 [45° W of N] ∆t = 4.0 s G vi = ? To find the initial velocity: G G G vf = vi + aav ∆t G G G vi = vf − aav ∆t G G = vf + ( − aav ∆t ) ( = 54 m/s [N] + −19 m/s 2 [45° W of N](4.0 s = 54 m/s [N] + ( −76 m/s [45° W of N] ) G vi = 54 m/s [N] + 76 m/s [45° E of S] ) Using components with +x east and +y north: vix = 0 + 76 m/s ( sin 45° ) vix = 54 m/s viy = 54 m/s − 76 m/s ( cos 45° ) = 54 m/s − 54 m/s viy = 0 m/s G vi = 54 m/s [E] The car should enter the curve with a velocity of 54 m/s [E]. G 26. vi = 17 m/s [up] G ∆d = 5.2 m G a = 9.8 m/s2 [down] ∆t = ? G G 1G 2 ∆d = vi ∆t + a ( ∆t ) 2 G G 1G 2 0 = vi ∆t + a ( ∆t ) − ∆d 2 G G 1G 2 0 = a ( ∆ t ) + vi ∆ t − ∆ d 2 This is a quadratic equation with solution: ∆t = −b ± b 2 − 4ac 2a G G G G a −vi ± vi 2 − 4 ( −∆d ) 2 = G a 2 2 = −17 m/s[up] ± (17 m/s[up]) 2 + 2( −9.8 m/s 2 [up])(5.2 m[up]) −9.8 m/s 2 [up] = 1.7 s ± 1.4 s ∆t = 0.34 s or 3.1 s The ball passes the camera at 0.34 s on the way up and at 3.1 s on the way down. Copyright © 2003 Nelson Chapter 1 Kinematics 79 27. (a) The slopes of the line segments on the velocity-time graph indicate the accelerations. (b) The area between the line and the x-axis up to the specific times indicate the total displacements up to those times. 28. rVenus = 1.08 × 1011 m TVenus = 1.94 × 107 s (a) v = ? 2π r v= T 2π (1.08 ×1011 m) = 1.94 × 107 s v = 3.50 × 10 4 m/s 1 km 3600 s = 3.50 × 10 4 m/s 1000 m 1 h v = 1.26 ×105 km/h Venus has an average speed of 3.50 × 104 m/s, or 1.26 × 105 km/h. G (b) vav = ? G Determine the average velocity after half a revolution using ∆d = 2r. Thus, G ∆d G vav = ∆t 2r = T 2 4r = T 4(1.08 × 1011 m) = 1.94 × 107 s G vav = 2.23 × 104 m/s The magnitude of the average velocity of Venus after it has completed half a revolution around the Sun is 2.23 × 104 m/s. 80 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson (c) Use the initial and final velocities to determine the average acceleration after a quarter revolution. Choose directions such as east and south that are perpendicular to define the directions at these two positions. G a =? G G G v − vi a= f ∆t G G vf − vi = T 4 = 4(3.50 × 10 4 m/s[E] − 3.50 × 104 m/s[S]) 1.94 × 107 s G a =7.22 × 10 −3 m/s 2 [E] − 7.22 ×10−3 m/s 2 [S] G a = ax 2 + a y 2 = 2(7.22 × 10−3 m/s 2 ) 2 G a = 1.02 × 10−2 m/s 2 The magnitude of its average acceleration during a quarter of a revolution around the Sun is 1.02 × 10–2 m/s2. 29. (a) The horizontal component of the acceleration of a projectile is zero. The vertical component of the acceleration of a projectile is 9.8 m/s2 toward Earth. (b) With air resistance affecting both components, the horizontal acceleration would be negative (assuming the direction of the horizontal component of the velocity is defined as positive), and the vertical component of the acceleration would be less than 9.8 m/s2. 30. (a) Let +y be down. G vi = 18 m/s [horizontal] ay = 9.8 m/s2 ∆x = 9.0 m ∆t = ? Find the horizontal component of the initial velocity: Horizontally (constant vix ): vix = 18.0 m/s ∆t = ∆x vix 9.0 m 18.0 m/s ∆t = 0.50 s The snowball will hit the tree after 0.50 s. (b) Find the vertical component of the initial velocity: = Vertically (constant a y ): viy = 0 m/s 1 ∆y = viy ∆t + a y (∆t ) 2 2 1 = a y ( ∆t ) 2 2 (9.8 m/s 2 )(0.50 s) 2 = 2 ∆y = 1.2 m Since the snowball’s original location was 1.5 m above ground, the height that the snowball will hit the tree is 1.5 m – 1.2 m = 0.3 m. Copyright © 2003 Nelson Chapter 1 Kinematics 81 G (c) vf = ? 2 2 vfy = viy + 2a y ∆y vfy 2 = 0 + 2a y ∆y vfy = 2a y ∆y = 2(9.8 m/s 2 )(1.2 m) vfy = 4.8 m/s Therefore, G 2 2 2 vf = vfx + vfy G vf = (18 m/s) 2 + (4.8 m/s) 2 G vf = 19 m/s θ = tan −1 = tan −1 vfy vfx 4.8 m/s 18 m/s θ = 15° The snowball’s velocity as it strikes the tree is 19 m/s [15° below the horizontal]. 31. (a) Let +y be down. ∆y = 1.5 may = 9.8 m/s2 ∆x = 16 m viy = 0 m/s vix = ? 1 ∆y = viy ∆t + a y (∆t ) 2 2 1 ∆y = a y ( ∆t ) 2 2 2 ∆y ∆t = ay 2(1.5 m) 9.8 m/s 2 ∆t = 0.55 s = ∆x ∆t 16 m = 0.55 s vix = 29 m/s Thus, the initial velocity of a ball projected horizontally is 29 m/s [horizontally]. 32. (a) In the train’s frame of reference, the path of the ball is straight downward from the release position. (b) In the other frame of reference (actually Earth’s frame), the path of the ball is parabolic and is shaped just like any projectile whose initial velocity is horizontal. (The magnitude of the initial velocity relative to Earth is equal in magnitude to the magnitude of the train’s velocity relative to Earth.) vix = 82 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson 33. Use the subscripts P for the plane, A for the wind (air), and G for the ground. Let the +x direction be east and the +y direction be south. G vPA = 285 km/s [45° S of E] G vAG = 75 km/h [22° E of N] G vPG = ? G G G vPG = vPA + vAG vPG,x = (285 km/h)cos 45° + (75 km/h)sin 22° vPG,y = (285 km/h)sin 45° − (75 km/h)cos 22° vPG,x = 230 km/h vPG,y = 132 km/h G 2 2 vPG = vPG,x + vPG,y = (230 km/h)2 + (132 km/h)2 G vPG = 2.6 ×10 2 km/h vPG,x tan θ = vPG,y 230 km/h θ = tan −1 132 km/h θ = 60° The velocity of the plane relative to the ground is 2.6 × 102 km/h [60° E of S]. 34. Use the subscripts S for the swimmer, W for the water, and G for the ground. Let the +x direction be downstream from the initial shore and the +y direction be across the river. G vSW = 0.80 m/s [across the river] ∆y = 86 m ∆x = 54 m G =? (a) v WG ∆y ∆t = G vSW = 86 m 0.80 m/s [across the river] ∆t = 108s G ∆x vWG = ∆t 54 m = 108s G vWG = 0.50 m/s The speed of the river current is 0.50 m/s. G (b) vSG = ? G G G vSG = vSW + vWG vSG,x = vWG = 0.50 m/s Copyright © 2003 Nelson vSG,y = vSW = 0.80 m/s Chapter 1 Kinematics 83 G vSG = vSG,x 2 + vSG,y 2 = (0.50 m/s)2 + (0.80 m/s)2 G vSG = 0.94 m/s vSG,y tan θ = v SG,x 0.80 m/s θ = tan −1 0.50 m/s θ = 58° The swimmer’s velocity relative to the shore is 0.94 m/s [downstream, 58° from the initial shore]. (c) θ = ? G G G vSG = vSW + vWG G 2 G 2 G 2 vSW = vSG + vWG G 2 G 2 G 2 vSG = vSW − vWG G vSG = (0.80 m/s)2 − (0.50 m/s) 2 G vSG = 0.62 m/s G vSG tan θ = G v WG 0.62 m/s θ = tan −1 0.50 m/s θ = 51° The direction in which the swimmer should have aimed to land directly across from the departure position is upstream, 51° from the near shore. 35. Use the subscripts P for the plane, A for the wind (air), and G for the ground. Let the +x direction be south and the +y direction be east. G ∆ d = 2.5 × 103 km [18° S of W] ∆t = 5.3 h G vAG = 85 km/h [E] G vPA = ? First we find the velocity of the plane relative to the ground that allows it to reach its destination on schedule. G ∆d PG G vPG = ∆t 2.5 × 103 km [18° S of W] = 5.3h G 2 vPG = 4.7 × 10 km/h [18° S of W] G G G vPG = vPA + vAG G G G vPA = vPG − vAG Using the relative velocity equation: G G G vPG = vPA + vAG G G G vPA = vPG − vAG 84 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson Using components with +x west and +y south: vPA,x = (4.7 × 102 km/h) cos18° − ( −85 km/h) vPA,y = (4.7 × 102 km/h) sin18° − 0 vPA,x = 5.34 × 102 km/h vPA,y = 1.45 × 102 km/h G vPA = vPA,x 2 + vPA,y 2 = (5.34 × 102 km/h)2 + (1.45 × 102 km/h)2 G vPA = 5.5 × 102 km/h vPA,y tan θ = v PA,x 1.45 × 10 2 km/h θ = tan −1 2 5.34 × 10 km/h θ = 15° The velocity of the plane relative to the air is 5.5 × 102 km/h [15° S of W]. 36. Let +y be up. G vi = 21.0 m/s [47° above the horizontal] ay = 9.8 m/s2 ∆x = 25 m ∆y = ? Find the horizontal and vertical components of the initial velocity: Horizontally (constant G vix = vi cos θ vix ) = (21.0 m/s)(cos 47°) vix = 14.3 m/s Vertically (constant ay) G viy = vi sin θ = (21.0 m/s)(sin 47°) viy = 15.4 m/s First we must determine the change in time: ∆x ∆t = vix 25 m 14.3 m/s ∆t = 1.7 s To calculate the vertical height of the football over the bar: 1 ∆y = viy ∆t + a y (∆t ) 2 2 1 = (15.4 m/s)(1.7 s) + (9.8 m/s 2 )(1.7 s)2 2 ∆y = 12 m Since the horizontal bar of the goal post is 3.0 above the field, the football will pass above the bar at a height of 12 m – 3.0 m = 8.9 m. = Copyright © 2003 Nelson Chapter 1 Kinematics 85 Applying Inquiry Skills 37. (a) Since the only instrument allowed is a metre stick or a measuring tape, the student could measure the horizontal range of the ball that is observed to have the maximum range. The angle (θ) of the ball’s initial velocity must be estimated. The equation derived for the maximum horizontal range for a projectile that lands at the same level as its starting position can be applied. (See the text, page 49.) ∆x = vi 2 = vi 2 g sin 2θ g ∆x sin 2θ vi = g ∆x sin 2θ (b) The biggest source of random error is in estimating the launch angle of the ball. Another source of random error is in measuring the horizontal range. A systematic error may occur if the horizontal range is measured to a position below the level at launch. A source of error that could be either random or systematic, depending on the situation, is the effect of the wind and/or air resistance. 38. L = 62.0 cm dU = 9.9 cm dL = 4.3 cm (a) θ = ? d − dL sin θ = U L 9.9 cm − 4.3 cm θ = sin −1 62 cm θ = 5.2° The angle of incline of the air table is 5.2° above the horizontal. (b) a = ? Refer to Investigation 1.4.1, questions (c) and (j), for background information. a = g sin θ = (9.8 m/s 2 )(sin 5.2°) a = 0.89 m/s 2 The magnitude of the acceleration of the puck parallel to the inclined plane is 0.89 m/s2. (c) The main random sources of error occur in using a ruler to measure the required distances. A systematic source of error may occur if the ruler or metre stick has a worn end or if the calibration does not begin at 0.0 cm. (Notice that the sources of error mentioned here relate to the way in which the data were found and applied in this question. In a complete experiment, the motion of the puck would be analyzed and there would be many more sources of error.) 39. No matter what values of the launch angle, horizontal range, and initial velocity students choose for their example, their calculations will show that the distance the target falls equals the difference between its height above the launch level and ∆y of the projectile. A specific example follows. ∆x = 8.50 m G vi = 38.0 m/s [33.0° above the horizontal] To satisfy the condition that the dart launcher is aimed at the stationary target, tan θ = target above the launch level: h ∆x , where h is the level of the h = ∆x tan θ = (8.50 m )( tan 33.0° ) h = 5.52 m 86 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson The time of flight can be found by using the horizontal component of the dart’s motion: ∆x vix = ∆t ∆x ∆t = vix = = ∆x vi cos θ 8.50 m (38.0 m )(cos 33.0° ) ∆t = 0.267 s In that time interval, the target falls by an amount ∆yT: 1 ∆yT = viy ∆t + a y ∆t 2 2 ( = 0 + 4.9 m/s 2 )(0.267 s ) 2 ∆yT = 0.349 m The target is thus 5.52 m − 0.349 m = 5.17 m above the launch level. Finally, consider the vertical component of the dart’s displacement, ∆yT, using +y up: 1 ∆yD = viy ∆t + a y ∆t 2 2 1 = (vi sin θ ) ∆t + a y ∆t 2 2 ( = (38.0 m/s )(sin 33.0° )(0.267 s ) − 4.9 m/s 2 )(0.267 s ) 2 ∆yD = 5.18 m Thus, in the time interval that the target falls to a specific height above the launch level, the dart’s motion results in the dart striking the target. (Rounding error resulted in a slight discrepancy in this example.) If different students choose different initial conditions, the final result will be the same. This problem can also be solved for general cases. If you have your students try this, be sure they use a consistent +y direction throughout the entire solution. Making Connections 40. v1 = 125 km/h v2 = 100 km/h d = 17 km ∆t = ? (a) ∆t = t2 − t1 = d d − v2 v1 17 km 17 km − 100 km/h 125 km/h = 0.17 h − 0.14 h = ∆t = 0.03 h or 2.0 min The driver saves 2.0 minutes by breaking the speed limit. (b) At a higher speed there is more air resistance and more heat produced by the higher amount of friction in the moving parts of the car and its engines. More fuel per unit distance moved is required to compensate for these effects. Copyright © 2003 Nelson Chapter 1 Kinematics 87 41. Let ∆t represent the time it takes for the signal to travel from Earth to the satellite and tD represent the delay time. d = 4.8 × 107 m c = 3.0 × 108 m/s tD = 0.55 s ttotal = ? (a) First we must calculate the change in time: d ∆t = c 4.8 ×10 7 m = 3.0 × 108 m/s ∆t = 0.16 s Now we can determine the total time: ttotal =∆t + tD + ∆t = 2(0.16 s) + 0.55 s ttotal = 0.87 s The total time interval between sending the signal and receiving the return signal on Earth is 0.87 s. (b) The time delays mentioned are obvious in live television broadcasts involving large distances between interviewers. (Time delays are less when transmission is via cable rather than via satellite.) 42. Three-dimensional positions, velocities, and accelerations of the past and present could be analyzed and used to predict the motion of an asteroid or other body in the future. This applies to both Earth and the asteroid. 43. Typical estimated speed and stopping distance are 5.0 m/s and 0.50 m, respectively, both of which are conservative values. Using these values, we calculate the acceleration using magnitudes: vf 2 = vi 2 + 2a∆d 0 = vi 2 + 2 a∆d a= = −vi 2 2 ∆d −(5.0 m/s) 2 2(0.50 m) a = −25 m/s 2 G Since this acceleration is more than 2.5 g , it is recommended that the patient avoid a rigorous racket sport. (Likely the true acceleration would be even greater than indicated here because most sports enthusiasts can run faster than 5.0 m/s, and often the stopping distance is less than 50 cm.) 44. At this stage, students are expected to confine their answers to concepts related to Chapter 1. (a) The main principles are velocity and relative velocity of a flowing liquid (i.e., blood). The equations are: d G G G vblood/artery = vblood/sensor + vsensor/artery v = blood and blood ∆t G (b) Answers will vary. One design could be a tiny device that sends signals outside the body so its motion ( vsensor/artery ) could be monitored by a receiver at the same time that a sensor within the device measures the speed of the blood G ( vblood/sensor ). Both sets of data could be stored in a computer for later analysis. 45. (a) The explosive was fired vertically upward and detonated right at its maximum height. (b) The explosive was fired vertically upward and detonated while still rising fairly quickly. (c) The explosive was beginning to fall back downward and was moving slightly to the right when detonation occurred. (d) The explosive was moving upward and fairly quickly to the left when detonation occurred. Extension 46. Let d1 represent the distance the helicopter is from the cliff, the subscript S represent the sonar signal, and the subscript H represent the helicopter. ∆d1 = 7.0 × 102 m ∆tS = 3.4 s vS = 3.5 × 102 m/s vH = ? 88 Unit 1 Forces and Motion: Dynamics Copyright © 2003 Nelson First we must calculate the distance the sonar signal travels: ∆dS = vS ∆tS = (3.5 ×10 2 m/s)(3.4s) ∆dS = 1.2 × 103 m Next we calculate the distance the helicopter travels: ∆d H = 2∆d1 − ∆dS = 2(7.0 ×10 2 m) − 1.2 × 103 m ∆d H = 2.1 × 10 2 m Finally, we can determine the speed of the helicopter: ∆d H vH = ∆tS = 2.1 × 10 2 m 3.4 s vH = 62 m/s The speed of the helicopter is 62 m/s 47. Let the subscript T represent the truck travelling at constant speed, the subscript P represent the police cruiser. vT = 18 m/s aP = 2.2 m/s2 viP = 0 m/s (a) To calculate how far the cruiser travels before catching the truck, ∆dP = ∆dT. ∆d T = vT ∆t ∆t = ∆d T vT ∆d P = viP ∆t + ∆d P = 1 a P ( ∆t ) 2 2 1 a P ( ∆t ) 2 2 Substitute for ∆t: 2 1 ∆d ∆d P = aP T which is equal to 2 vT ∆d P = 1 ∆d P aP 2 vT 2 2 = 2 vT aP = 2(18 m/s) 2 2.2 m/s2 ∆d P = 2.9 × 102 m The cruiser travels 2.9 × 102 m before catching the truck. (b) ∆t = ? ∆d T ∆t = vT = 2.9 ×10 2 m 18 m/s ∆t = 16 s The pursuit lasts 16 s. Copyright © 2003 Nelson Chapter 1 Kinematics 89 48. Let the +x direction be east and the +y direction be south. vi = 80 km/h [E] vf = 100 km/h [45° S of E] a = 5.0 (km/h)/s (a) vfx =? vfy = ? G vfy = vf sin θ G vfx = vf cos θ = (100 km/h)(sin 45°) = (100 km/h)(cos 45°) vfy = 71 km/h vfx = 71 km/h The direction of the acceleration is the same as the direction of the change of velocity. G G G ∆v = vf − vi ∆vx = vfx − vix 1 = 70.7 km/h − 8.0 × 10 km/h ∆vx = −9.3 km/h tan θ = ∆v x ∆v y θ = tan −1 (9.3 km/h ) (70.7 km/h ) θ = 7.5° The direction of the acceleration is [7.5° W of S]. (b) ∆t = ? From (a) above, we can calculate the change in velocity: ∆v = = ( ∆vx )2 + ( ∆v y ) 2 ( −9.3 km/h )2 + (70.7 km/h )2 ∆v = 71 km/h We can now calculate the time interval: G G ∆v a= ∆t G ∆v ∆t = G a 71 km/h [7.5° W of S] = 5.0 (km/h)/s [7.5° W of S] ∆t = 14 s The time interval of the acceleration is 14 s. 49. Let +y be up, then ay = −g = −9.8 m/s2. Begin by determining an expression for ∆t using the vertical component of the motion: 1 ∆y = viy ∆t + a y ∆t 2 2 ( ) ∆y = vi sin θ∆t − 4.9 m/s 2 ∆t 2 (4.9 m/s ) ∆t 2 90 Unit 1 Forces and Motion: Dynamics 2 − vi sin θ∆t + ∆y = 0 Copyright © 2003 Nelson Using the quadratic formula to solve for ∆t: ∆t = ∆t = −b ± b 2 − 4 ac 2a vi sin θ ± ( −vi sin θ )2 − (19.9 m/s 2 ) ∆y 9.8 m/s 2 Substituting this equation for ∆t into the equation involving the horizontal component of the motion: ∆x = vix ∆t ∆x = vi cosθ∆t v sin θ ± i ∆x = vi cosθ ( −vi sin θ )2 − (19.9 m/s 2 ) ∆y 9.8 m/s 2 50. Consider the motion in the frame of reference of the flowing river, which moves at a constant velocity downstream relative to the shore. The sunbather swims directly away from the raft for 15 min at a constant speed, so it will take her exactly the same length of time to swim directly back to the raft. Thus, the raft drifts for 30 min or 0.50 h while moving (relative to the shore) 1.0 km downstream. Let R represent the river and S represent the shore. G ∆d G vRS = ∆t 1.0 km [downstream] = 0.50 h G vRS = 2.0 km/h [downstream] The speed of the current in the river is 2.0 km/h. (A more complex solution from the frame of reference of Earth or the shore provides the same answer. Refer to the SIN solutions book, question 75-10.) Copyright © 2003 Nelson Chapter 1 Kinematics 91
