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**Unformatted text preview: **1.1. Solve: 1.2. Solve: 1.3. Solve: 1.4. Solve: (a) The basic idea of the particle model is that we will treat an object as if all its mass is concentrated into a single point. The size and shape of the object will not be considered. This is a reasonable approximation of reality if (i) the distance traveled by the object is large in comparison to the size of the object, and (ii) rotations and internal motions are not significant features of the objects motion. The particle model is important in that it allows us to simplify a problem. Complete realitywhich would have to include the motion of every single atom in the objectis too complicated to analyze. By treating an object as a particle, we can focus on the most important aspects of its motion while neglecting minor and unobservable details. (b) The particle model is valid for understanding the motion of a satellite or a car traveling a large distance. (c) The particle model is not valid for understanding how a car engine operates, how a person walks, how a bird flies, or how water flows through a pipe. 1.5. Solve: (a) An operational definition defines a concept or an idea in terms of a procedure, or a set of operations, that is used to identify or measure the concept. (b) The displacement .r.. of an object is a vector found by drawing an arrow from the objects initial location to its final location. Mathematically, f i.r.. = r.. - r... The average velocity v.. of an object is a vector that points in the same direction as the displacement .r.. and has length, or magnitude, .r../.t, where f i .t = t - t is the time interval during which the object moves from its initial location to its final location. 1.6. Solve: The player starts from rest and moves faster and faster (accelerates). 1.7. Solve: The particle starts with an initial velocity but as it slides it moves slower and slower till coming to rest. This is a case of negative acceleration because it is an acceleration opposite to the positive direction of motion. 1.8. Solve: The acceleration of an object is a vector formed by finding the ratio of .v.. , the change in the objects velocity, to .t, the time in which the change occurs. The acceleration vector a.. points in the direction of .v.., which is found by vector subtraction. 1.9. Solve: (a) Acceleration is found by the method of Tactics Box 1.3. Let 0 v.. be the velocity vector between points 0 and 1 and 1 v.. be the velocity vector between points 1 and 2. (b) Speed 1 v is greater than speed 0 v because more distance is covered in the same interval of time. 1.10. Solve: (a) Acceleration is found by the method of Tactics Box 1.3. Let 0 v.. be the velocity vector between points 0 and 1 and 1 v.. be the velocity vector between points 1 and 2. (b) Speed 1 v is greater than speed 0 v because more distance is covered in the same interval of time. 1.11. Solve: (a) (b) 1.12. Solve: (a) (b) 1.13. Model: Represent the car as a particle. Visualize: The dots are equally spaced until brakes are applied to the car. Equidistant dots indicate constant average speed. On braking, the dots get closer as the average speed decreases. 1.14. Model: Represent the (child + sled) system as a particle. Visualize: The dots in the figure are equally spaced until the sled encounters a rocky patch. Equidistant dots indicate constant average speed. On encountering a rocky patch, the average speed decreases and the sled comes to a stop. This part of the motion is indicated by a separation between the dots that becomes smaller and smaller. 1.15. Model: Represent the tile as a particle. Visualize: The tile falls from the roof with an acceleration equal to a = g = 9.8 m/s2. Starting from rest, its velocity increases until the tile hits the water surface. This part of the motion is represented by dots with increasing separation, indicating increasing average velocity. After the tile enters the water, it settles to the bottom at roughly constant speed. 1.16. Model: Represent the tennis ball as a particle. Visualize: The particle falls freely for the three stories under the acceleration of gravity. It strikes the ground and very quickly decelerates to zero (while decompresses) and finally travels upward with negative acceleration under gravity to zero velocity at a height of two stories. The downward and upward motions of the ball are shown in the figure. The increasing length between the dots during downward motion indicates increasing average velocity or downward acceleration. On the other hand, the decreasing length between the dots during upward motion indicates acceleration in a direction opposite to its motion; that is, in the downward direction. Assess: For a free-fall motion, acceleration due to gravity is always vertically downward. 1.17. Model: Represent the toy car as a particle. Visualize: As the toy car rolls down the ramp, its average speed increases. This is indicated by the increasing length of the velocity arrows. That is, motion down the ramp is under an acceleration a... At the bottom of the ramp, the toy car continues with the speed obtained with no change in velocity. 1.18. Solve: Dot Time (s) x (m) 1 0 0 2 2 30 3 4 95 4 6 215 5 8 400 6 10 510 7 12 600 8 14 670 (a) 9 16 720 (b) 1.19. Solve: A forgetful physics professor goes for a walk on a straight country road. Walking at a constant speed, he covers a distance of 300 m in 300 s. He then stops and watches the sunset for 100 s. Finding that it was getting dark, he walks faster back to his house covering the same distance in 200 s. 1.20. Solve: Forty miles into a car trip north from his home in El Dorado, an absent-minded English professor stopped at a rest area one Saturday. After staying there for one hour, he headed back home thinking that he was supposed to go on this trip on Sunday. Absent-mindedly he missed his exit and stopped after one hour of driving at another rest area 20 miles south of El Dorado. After waiting there for one hour, he drove back very slowly, confused and tired as he was, and reached El Dorado in two hours. 1.21. Visualize: The bicycle is moving with an acceleration of 1.5 m/s2. Thus, the velocity will increase by 1.5 m/s each second of motion. 1.22. Visualize: The particle moves upward with a constant acceleration a.. . The final velocity is 200 m/s and is reached at a height of 1000 m. 1.23. Solve: (a) 6 9.12 s (9.12 s) 10 s 9.12 10 6 s 1 s - - . . = . . = . . (b) 3 3.42 km (3.42 km) 10 m 3.42 103 m 1 km . . = . . = . . (c) 2 2 3 44 cm/ms 44 cm 10 m 1 ms 4.4 10 m/s ms 1 cm 10 s - - . .. .. . = . .. .. . = . .. .. . (d) km 103 m 1 hour 80 km/hour 80 22 m/s hour 1 km 3600 s . .. .. . = . .. .. . = . .. .. . 1.24. Solve: (a) 2.54 cm 10 2 m 8.0 inches 8.0 (inch) 0.20 m 1 inch 1 cm . .. - . = . .. . = . .. . (b) 66 feet/s 66 feet 12 inch 1 m 20 m/s s 1 foot 39.37 inch = . .. .. . = . .. .. . . .. .. . (c) miles 1.609 km 103 m 1 hour 60 mph 60 27 m/s hour 1 mile 1 km 3600 s . .. .. .. . = . .. .. .. . = . .. .. .. . (d) 2 14 square inches 14 (inches)2 1 m 9.0 10 3 square meter 39.37 inches = .. .. = - . . 1.25. Solve: (a) 1 hour 1(hour) 3600 s 3600 s 3.60 103 s 1 hour = . . = = . . . . (b) 1 day 1 (day) 24 hours 3600 s 8.64 104 s 1 day 1 hour . .. . = . .. . = . .. . (c) 4 1 year 1 (year) 365.25 days 8.64 10 s 3.16 107 s 1 year 1 day . .. . = . .. . = . .. . (d) 2 2 2 32 ft/s 32 ft 12 inch 1 m 9.75 m/s s 1 ft 39.37 inch = . .. .. . = . .. .. . . .. .. . 1.26. Solve: (a) 20 ft 20(ft) 1 m 7.0 m 3 ft = . . = . . . . (b) 60 miles 60(miles) 1 km 1000 m 1.0 105 m 0.6 miles 1 km = . .. . = . .. . . .. . (c) 60 mph 60(mph) 1 m/s 30 m/s 2 mph . . = . . = . . (d) 1 cm 10 2 m 8 in 8(in) 0.16 m 1/2 in 1 cm . .. - . = . .. . = . .. . 1.27. Solve: (a) (30 cm) 4 in 12 in 10 cm . . = . . . . (b) (25 m/s) 2 mph 50 mph 1 m/s . . = . . . . (c) (5 km) 0.6 mi 3 mi 1 km . . = . . . . (d) 1 in 1 cm 2 1 in 2 1 cm 4 . . . .. . . .. . = . .. . 1.28. Solve: (a) 33.3 25.4 = 846 (b) 33.3 - 25.4 = 7.9 (c) 33.3 = 5.77 (d) 333.3 25.4 =13.1 1.29. Solve: (a) (33.3)2 =1.109103. For numbers starting with 1 an extra digit is kept. (b) 33.3 45.1 =1.50103 Scientific notation is an easy way to establish significance. (c) 22.2 -1.2 = 3.5 (d) 1/ 44.4 = 0.0225 1.30. Solve: The length of a typical car is 15 ft. Or 15(ft) 12 inch 1 m 4.6 m 1 ft 39.37 inch . .. .= . .. . . .. . This length of 15 ft is approximately two-and-a-half times my height. 1.31. Solve: The height of a telephone pole is estimated to be around 50 ft or 15 m. This height is approximately 8 times my height. 1.32. Solve: I typically take 15 minutes in my car to cover a distance of approximately 6 miles from home to campus. My average speed is 6 miles 60 min 24 mph 24(mph) 0.447 m/s 11 m/s 15 min 1 hour 1 mph . . . . . . = = . . = . . . . 1.33. Solve: My barber trims about an inch of hair when I visit him every month for a haircut. The rate of hair growth is 2 9 6 9 1(inch) 2.54 cm 10 m 1 month 1 day 1 h 9.8 10 m/s (month) 1 inch 1 cm 30 days 24 h 3600 s 9.8 10 m 10 m 3600 s 35 m/h s 1 m 1 h - - - . .. .. .. .. .. .. .. .. .. . = . .. .. .. .. . . .. .. . = . .. .. . = . .. .. . 1.34. Model: Represent the Porsche as a particle for the motion diagram. Visualize: 1.35. Model: Represent the watermelon as a particle for the motion diagram. Visualize: 1.36. Model: Represent (Sam + car) as a particle for the motion diagram. Visualize: 1.37. Model: Represent the speed skater as a particle for the motion diagram. Visualize: 1.38. Model: Represent the wad as a particle for the motion diagram. Visualize: 1.39. Model: Represent the ball as a particle for the motion diagram. Visualize: 1.40. Model: Represent the ball as a particle for the motion diagram. Visualize: 1.41. Model: Represent the motorist as a particle for the motion diagram. Visualize: 1.42. Model: Represent Bruce and the puck as particles for the motion diagram. Visualize: 1.43. Model: Represent Fred and yourself as particles for the motion diagram. Visualize: 1.44. Solve: Rahul was coasting on interstate highway I-35 from Wichita to Kansas City at 65 mph. Seeing an accident at a distance of 200 feet in front of him, he braked his car to a stop with steady deceleration. 1.45. Solve: A car starts coasting at an initial speed of 30.0 m/s up a 10 incline. 230 m up the incline the road levels out to a flat road and the car continues coasting at a reduced speed along the road. 1.46. Solve: A skier starts from rest down a 25 slope with very little friction. At the bottom of the 100 m slope the skier moves to a flat area and continues at constant velocity. 1.47. Solve: A ball is dropped from a height to check its rebound properties. It rebounds to 80% of its original height. 1.48. Solve: Two boards lean against each other at equal angles to the vertical direction. A ball rolls up the incline, over the peak, and down the other side. 1.49. Solve: (a) (b) Sue passes 3rd Street doing 40 mph, slows steadily to the stop sign at 4th Street, stops for 1 s, then speeds up and reaches her original speed as she passes 5th Street. If the blocks are 50 m long, how long does it take Sue to drive from 3rd Street to 5th Street? (c) 1.50. Solve: (a) (b) A train moving at 100 km/hour slows down in 10 s to a speed of 60 km/hour as it enters a tunnel. The driver maintains this constant speed for the entire length of the tunnel that takes the train a time of 20 s to traverse. Find the length of the tunnel. (c) 1.51. Solve: (a) (b) Jeremy has perfected the art of steady acceleration and deceleration. From a speed of 60 mph he brakes his car to rest in 10 seconds with a constant deceleration. Then he turns into an adjoining street. Starting from rest, Jeremy accelerates with exactly the same magnitude as his earlier deceleration and reaches the same speed of 60 mph over the same distance in exactly the same time. Find the cars acceleration or deceleration. (c) 1.52. Solve: (a) (b) A coyote (A) sees a rabbit and begins to run toward it with an acceleration of 3.0 m/s2. At the same instant, the rabbit (B) begins to run away from the coyote with an acceleration of 2.0 m/s2. The coyote catches the rabbit after running 40 m. How far away was the rabbit when the coyote first saw it? (c) 1.53. Solve: Since area equals length width, the smallest area will correspond to the smaller length and the smaller width. Similarly, the largest area will correspond to the larger length and the larger width. Therefore, the smallest area is (64 m)(100 m) = 6.4 103 m2 and the largest area is (75 m)(110 m) = 8.3 103 m2. 1.54. Solve: (a) We need kg/m3. There are 100 cm in 1 m. If we multiply by 3 100 cm (1)3 1 m . . = . . . . we do not change the size of the quantity, but only the number in terms of the new unit. Thus, the mass density of aluminum is 3 3 3 3 3 2.7 10 kg 100 cm 2.7 10 kg cm 1 m m - .. .... .. = . .. . (b) Likewise, the mass density of alcohol is 3 3 3 0.81 g 100 cm 1 kg 810 kg cm 1 m 1000 g m . .. . . . . .. . . . = . .. . . . 1.55. Model: The car is represented by the particle model as a dot. Solve: (a) Time t (s) Position x (m) 0 1200 10 975 20 825 30 750 40 700 50 650 60 600 70 500 80 300 90 0 (b) 1.56. Solve: Susan enters a classroom, sees a seat 40 m directly ahead, and begins walking toward it at a constant leisurely pace, covering the first 10 m in 10 seconds. But then Susan notices that Ella is heading toward the same seat, so Susan walks more quickly to cover the remaining 30 m in another 20 seconds, beating Ella to the seat. Susan stands next to the seat for 10 seconds to remove her backpack. 1-1 1.57. Solve: A crane operator holds a ton of bricks 30 m above the ground. Four seconds after he is told to lower the bricks, he takes four seconds to lower them 15 m at a constant rate before stopping the bricks to make an eight-second safety check. He then continues lowering the bricks the remaining 15 m, taking four more seconds. 2.1. Model: We will consider the car to be a particle that occupies a single point in space. Visualize: Solve: Since the velocity is constant, we have xf = xi + vx.t. Using the above values, we get 1 x = 0 m+ (10 m/s)(45 s) = 450 m Assess: 10 m/s 22 mph and implies a speed of 0.4 miles per minute. A displacement of 450 m in 45 s is reasonable and expected. 2.2. Model: We will consider Larry to be a particle. Visualize: Solve: Since Larrys speed is constant, we can use the following equation to calculate the velocities: f i s f i v s s t t - = - (a) For the interval from the house to the lamppost: 1 200 yd 600 yd 200 yd/min 9:07 9:05 v - = =- - For the interval from the lamppost to the tree: 2 1200 yd 200 yd 333 yd/min 9:10 9:07 v - = =+ - (b) For the average velocity for the entire run: avg 1200 yd 600 yd 120 yd/min 9:10 9:05 v - = =+ - 2.3. Model: Cars will be treated by the particle model. Visualize: Solve: Beth and Alan are moving at a constant speed, so we can calculate the time of arrival as follows: 1 0 1 0 1 0 1 0 v x x x t t x x t t t v . - - = = . = + . - Using the known values identified in the pictorial representation, we find: Alan 1 Alan 0 Alan 1 Alan 0 Beth 1 Beth 0 Beth 1 Beth 0 8:00 AM 400 mile 8:00 AM 8 hr 4:00 PM 50 miles/hour 9:00 AM 400 mile 9:00 AM 6.67 hr 3:40 PM 60 miles/hour t t x x v t t x x v - = + = + = + = - = + = + = + = (a) Beth arrives first. (b) Beth has to wait Alan 1 Beth 1 t - t = 20 minutes for Alan. Assess: Times of the order of 7 or 8 hours are reasonable in the present problem. 2.4. Solve: (a) The time for each segment is .t1 = 50 mi/40 mph = 5/4 hr and 2 .t = 50 mi/60 mph = 5/6hr. The average speed to the house is 100 mi 48 mph 5/6 h 5/4 h = + (b) Julie drives the distance 1 .x in time 1 .t at 40 mph. She then drives the distance 2 .x in time 2 .t at 60 mph. She spends the same amount of time at each speed, thus 1 2 1 2 1 2 .t = .t ..x /40 mph = .x /60 mph..x = (2/3).x But 1 2 .x + .x =100 miles, so 2 2 (2/3).x + .x =100 miles. This means 2 .x = 60 miles and 1 .x = 40 miles. Thus, the times spent at each speed are 1 .t = 40 mi/40 mph =1.00 h and 2 .t = 60 mi/60 mph =1.00 h. The total time for her return trip is 1 2 .t + .t = 2.00 h. So, her average speed is 100 mi/2 h = 50 mph. 2.5. Model: The bicyclist is a particle. Visualize: Please refer to Figure EX2.5. Solve: The slope of the position-versus-time graph at every point gives the velocity at that point. The slope at t = 10 s is 100 m 50 m 2.5 m/s 20 s v s t . - = = = . The slope at t = 25 s is 100 m 100 m 0 m/s 10 s v - = = The slope at t = 35 s is 0 m 100 m 10 m/s 10 s v - = =- 2.6. Visualize: Please refer to Figure EX2.6. Solve: (a) We can obtain the values for the velocity-versus-time graph from the equation v = .s/.t. (b) There is only one turning point. At t = 3 s the velocity changes from +5 m/s to -20 m/s, thus reversing the direction of motion. At t =1 s, there is an abrupt change in motion from rest to +5 m/s, but there is no reversal in motion. 2.7. Visualize: Please refer to Figure EX2.7. The particle starts at x0 =10 m at 0 t = 0. Its velocity is initially in the x direction. The speed decreases as time increases during the first second, is zero at t =1 s, and then increases after the particle reverses direction. Solve: (a) The particle reverses direction at t =1 s, when vx changes sign. (b) Using the equation f 0 x = x + area of the velocity graph between 1 t and f t , 2 s 3 s 10 m (area of triangle between 0 s and 1 s) + (area of triangle between 1 s and 2 s) 10 m 1 (4 m/s)(1 s) + 1 (4 m/s)(1 s) 10 m 2 2 10 m area of trapazoid between 2 s and 3 s 10 m 1 (4 m/s + 8 m/s 2 x x = - = - = = + = + 4 s 3 s )(3 s 2 s) 16 m area between 3 s and 4 s 16 m 1 (8 m/s + 12 m/s)(1 s) 26 m 2 x x - = = + = + = 2.8. Visualize: Please refer to Figure EX2.8. Solve: A constant velocity from t = 0 s to t = 2 s means zero acceleration. On the other hand, a linear increase in velocity between t = 2 s and t = 4 s implies a constant positive acceleration. 2.9. Visualize: Please refer to Figure EX2.9. Solve: (a) The acceleration of the train at t = 3.0 s is the slope of the v vs t graph at t = 3 s. Thus a = (2 m/s - ( - 2 m/s)) (8 s) = 0.5 m/s2 . (b) 2.10. Visualize: Please refer to Figure EX2.10. Solve: (a) At t = 2.0 s, the position of the particle is 2 s 2.0 m area under velocity graph from 0 s to 2.0 s 2.0 m 1 (4 m/s)(2.0 s) 6 m 2 x = + t = t = = + = (b) From the graph itself at t = 2.0 s, v = 4 m/s. (c) The acceleration is f i 2 6 m/s 0 m/s 2 m/s 3 s x x x x a v v v t t . - - = = = = . . 2.11. Visualize: Please refer to Figure EX2.11. Solve: (a) Using the equation xf = xi + area under the velocity-versus-time graph between i t and f t we have x(at t =1 s) = x(at t = 0 s) + area between t = 0 s and t =1 s = 2.0 m + (4 m/s)(1 s) = 6 m Reading from the velocity-versus-time graph, (at 1 s) 4 m/s. xv t = = Also, slope / 0 m/s2. xa = = .v .t = (b) x(at t = 3.0 s) = x(at t = 0 s) + area between t = 0 s and t = 3 s = 2.0 m + 4 m/s 2 s + 2 m/s 1 s + (1/2) 2 m/s 1 s =13.0 m Reading from the graph, ( 3 s) 2 m/s. x v t= = The acceleration is ( 3 s) slope (at 4 s) (at 2 s) 2 m/s2 2 s x x x a t v t v t = - = = = = =- 2.12. Model: Represent the jet plane as a particle. Visualize: Solve: (a) Since we dont know the time of acceleration, we will use 2 2 1 0 1 0 2 2 2 2 1 0 2 1 2 ( ) (400 m/s) (300 m/s) 8.75 m/s 2 2(4000 m) v v ax x a v v x = + - - - . = = = (b) The acceleration of the jet is approximately equal to g, the acceleration due to gravity. 2.13. Model: We are using the particle model for the skater and the kinematics model of motion under constant acceleration. Solve: Since we dont know the time of acceleration we will use 2 2 f i f i 2 2 2 2 f i 2 f i 2 ( ) (6.0 m/s) (8.0 m/s) 2.8 m/s 2( ) 2(5.0 m) v v ax x a v v x x = + - - - . = = = - - Assess: A deceleration of 2.8 m/s2 is reasonable. 2.14. Model: We are assuming both cars are particles. Visualize: Solve: The Porsches time to finish the race is determined from the position equation 2 P1 P0 P0 P1 P0 P P1 P0 2 2 P1 P1 ( ) 1 ( ) 2 400 m 0 m 0 m 1 (3.5 m/s )( 0 s) 15.1 s 2 x x v t t a t t t t = + - + - . = + + ...

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