1.1.
Solve:
1.2.
Solve:
1.3.
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 object’s motion. The particle model is important in that it allows us to simplify a problem. Complete reality—which would have to include the motion of every single atom in the object—is 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.
1.4.
Solve: Position refers to the location of an object at a given time relative to a coordinate system. Displacement, on the other hand, is the difference between the object’s final position at time t1 and the initial position at time t0. Displacement is a vector, whose direction is from the initial position toward the final position. An airplane at rest relative to a runway lamp, serving as the origin of our coordinate system, will have a position, called the initial position. The location of the airplane as it takes off may be labeled as the final position. The difference between the two positions, final minus initial, is displacement.
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 tor identify or measure the concept. (b) The displacement ∆r of an object is a vector found by drawingr an arrow from the object’s initial location to its r r r final location. Mathematically, ∆r = rf − ri . The average velocity v of an object is a vector that points in the same r r direction as the displacement ∆r and has length, or magnitude, ∆r / ∆t, where ∆t = t f − t i 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.
r Solve: The acceleration of an object is a vector formed by finding the ratio rof ∆v , the change in the object’s velocity, to ∆t, the time in which the change occurs. The acceleration vector a points in the direction of r ∆v, which is found by vector subtraction.
1.8.
r Solve: (a) Acceleration is found by the method of Tactics Box 1.3. Let v0 be the velocity vector between r points 0 and 1 and v1 be the velocity vector between points 1 and 2.
1.9.
(b) Speed v1 is greater than speed v0 because more distance is covered in the same interval of time.
r Solve: (a) Acceleration is found by the method of Tactics Box 1.3. Let v0 be the velocity vector between r points 0 and 1 and v1 be the velocity vector between points 1 and 2.
1.10.
(b) Speed v1 is greater than speed v0 because more distance is covered in the same interval of time.
1.11. Solve: The acceleration vector at each location points directly toward the center of the Ferris wheel’s circular motion.
1.12. Solve: The acceleration vectors of the runner at the top and at the bottom of the circular track are directly toward the center.
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.
Model: Represent the (child + sled) system as a particle. Visualize:
1.14.
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 compressing the ball) then quickly accelerates upward (while the ball 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 car as a particle. Visualize: The car (particle) moves at a constant speed v so the distance rbetween the dots is constant. While r turning v remains constant, but the direction of v changes. There will be a ∆v during the turn. Therefore, there is an acceleration during the turn.
1.18. 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 r 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.
1.19. 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.
r 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.
1.20.
1.21.
Solve:
(a)
10 −6 s −6 9.12 µs = (9.12 µs) = 9.12 × 10 s 1 µs
(b)
10 3 m 3 3.42 km = (3.42 km) = 3.42 × 10 m 1 km
(c)
cm 10 −2 m 1 ms 44 cm/ ms = 44 = 440 m/s ms 1 cm 10 −3 s
(d)
km 10 3 m 1 hour 80 km / hour = 80 = 22.2 m /s hour 1 km 3600 s
1.22. (b) (c)
Solve:
(a)
−2 2.54 cm 10 m 8 inches = 8 (inch) = 0.203 m 1 inch 1 cm
feet 12 inch 1m 66 feet /s = 66 = 20.1 m /s s 1 foot 39.37 inch miles 1.609 km 10 3 m 1 hour 60 mph = 60 = 26.8 m /s hour 1 mile 1 km 3600 s 2
(d)
1m = 9.03 × 10 −3 square meter 14 square inches = 14 (inches)2 39.37 inches
1.23. (b)
Solve:
(a)
3600 s 1 hour = 1( hour) = 3600 s = 3.60 × 10 3 s 1 hour
24 hours 3600 s 1 day = 1 (day) = 8.64 × 10 4 s 1 day 1 hour
365.25 days 8.64 × 10 4 s 7 1 year = 1 ( year) = 3.16 × 10 s 1 year 1 day ft 12 inch 1m (d) 32 ft /s 2 = 32 2 = 9.75 m /s 2 s 1 ft 39.37 inch (c)
1.24. (b)
1 m = 7.0 m 20 ft = 20(ft ) 3 ft 1 km 1000 m = 1.0 × 10 5 m 60 miles = 60( miles) 0.6 miles 1 km Solve:
(a)
(c)
1 m /s 60 mph = 60( mph) = 30 m /s 2 mph
(d)
1 cm 10 −2 m 8 in = 8(in) = 0.16 m 1/2 in 1 cm
1.25. Solve: Since area equals length Ă— 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) = 6.40 Ă— 103 m2 and the largest area is (75 m)(110 m) = 8.25 Ă— 103 m2.
1.26.
Solve:
(a) We need kg/m3. There are 100 cm in 1 m. If we multiply by 3
100 cm = (1) 3 1m 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
kg 100 cm kg 2.7 × 10 −3 = 2.7 × 10 3 3 cm 3 1 m m (b) Likewise, the mass density of alcohol is 3 g 100 cm 1 kg kg = 810 3 0.81 3 cm 1 m 1000 g m
1.27. Solve: (a) The number 6.21 has three significant figures. (b) The number 62.1 has three significant figures. (c) The number 0.620 has three significant figures. The final zero is significant because it is expressed. (d) The number 0.062 has two significant figures. The second zero places only the decimal point.
1.28. Solve: (a) The number 6200 has two significant figures. The zeros place the decimal point. (b) The number 0.006200 has four significant figures. The initial two zeroes place the decimal point. The last two zeroes do not have to be there, but when they are they are significant. (c) The number 1.0621 has five significant figures. (d) The number 6.21 Ă— 103 has three significant figures.
1.29. (b) (c) (d)
Solve: (a) 33.3 × 25.4 = 846 33.3 – 25.4 = 7.9 33.3 = 5.77 333.3 ÷ 25.4 = 13.1
1.30. Solve: (a) (33.3)2 = 1.11 × 103 (b) 33.3 × 45.1 = 1.50 × 103 Scientific notation is an easy way to establish significance. (c) (d)
22.2 − 1.2 = 3.5 1/44.4 = 0.0225
1.31.
Solve:
The length of a typical car is 15 ft. Or
12 inch 1m 15(ft ) = 4.6 m 1 ft 39.37 inch This length of 15 ft is approximately two and a half times my height.
1.32. 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.
1.33. 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 0.447 m /s 6 miles 60 min = 24 mph = 24( mph ) = 11 m /s 15 min 1 hour 1 mph
1.34. Solve: growth is
My barber trims about an inch of hair when I visit him every month for a haircut. The rate of hair
1(inch ) 2.54 cm 10 −2 m 1 month 1 day 1 hr = 9.8 × 10 −9 m /s ( month ) 1 inch 1 cm 30 days 24 hr 3600 s m 10 6 µm 3600 s = 35 µm / hr = 9.8 × 10 −9 s 1 m 1 hr
1.35. Model: Represent the Porsche as a particle for the motion diagram. Visualize:
1.36. Model: Represent the watermelon as a particle for the motion diagram. Visualize:
1.37. Model: Represent (Sam + car) as a particle for the motion diagram. Visualize:
1.38. Model: Represent the speed skater as a particle for the motion diagram. Visualize:
1.39. Model: Represent the wad 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 ball as a particle for the motion diagram. Visualize:
1.42. Model: Represent the motorist as a particle for the motion diagram. Visualize:
1.43. Model: Represent Bruce and the puck as particles for the motion diagram. Visualize:
1.44. Model: Represent Fred and yourself as particles for the motion diagram. Visualize:
1.45. 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.
1.46.
Solve: A car starts coasting at an initial speed of 30.0 m/s up a 10째 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.
1.47.
Solve: A skier starts from rest down a 25째 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.
1.48. Solve: A race car slows from an initial speed of 100 mph to 50 mph in order to negotiate a tight turn. After making the 90째 turn the car accelerates back up to 100 mph in the same time it took to slow down.
1.49. height.
Solve:
A ball is dropped from a height to check its rebound properties. It rebounds to 80% of its original
1.50.
Solve:
A car drives up a hill, over the top, and down the other side at constant speed.
1.51.
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)
1.52.
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)
1.53.
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 car’s acceleration or deceleration. (c)
1.54.
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)
1.55.
Solve:
(a)
(b) A ball rolls along a level table at 3 m/s. It rolls over the edge and falls 1 m to the floor. How far away from the edge of the table does it land? (c)
1.56.
Solve:
(a)
(b) Phil is standing on a third floor balcony, 10 m above the ground. He throws a ball with a speed of 20 m/s at an upward angle of 30째. Where does the ball hit the ground? (c)
1.57. Solve: (a) The figure shows a motion diagram of a pendulum as it swings from one side to the other. It’s clear that the velocity at the lowest point is not zero. The velocity vector at this point is tangent to the circle. We can use the method of Tactics Box 1.3 to find the acceleration at the lowest point. The acceleration is not zero. Instead, you can see that the acceleration vector points toward the center of the circle.
(b) The end of the arc is like the highest point of a ball tossed straight up. The velocity is zero for an instant as the vector changes from pointing outward to pointing inward. However, the acceleration is not zero at this point. r The velocity is changing at the end point, and this requires an acceleration. The motion diagram shows that ∆v, and r thus a, is tangent to the circle at the end of the arc.