Chapter Solutions Kinematic Vocabulary One of the difficulties in studying mechanics is that many common words are used with highly specific technical meanings, among them velocity, acceleratio n, position, speed, and displacement. The series of questions in this problem is designed to get you to try to think of these quantities like a physicist. Answer the questions in this problem using words from the following list: A. position B. direction C. displacement D. coordinates E. velocity F. acceleration G. distance H. magnitude I. vector J. scalar K. components Part A Velocity differs from speed in that velocity indicates a particle's of motion. B Also accepted: direction Part B Unlike speed, velocity is a quantity. I Also accepted: vector Part C A vector has, by definition, both and direction. H Also accepted: magnitude Part D
Once you have selected a coordinate system, you can epress a two-dimensional vector using a pair of quantities known collectively as. K Also accepted: components, D, coordinates Part E Speed differs from velocity in the same way that differs from displacement. Hint 1. Definition of displacement Displacement is the vector that indicates the difference of two positions (e.g., the final position from the initial position). Being a vector, it is independent of the coordinate system used to describe it (although its vector components depend on the coordinate system). G Also accepted: distance Part F Consider a physical situation in which a particle moves from point A to point B. This process is described from two coordinate systems that are identical ecept that they have different origins. The of the particle at point A differ(s) as epressed in one coordinate system compared to the other, but the from A to B is/are the same as epressed in both coordinate systems. Type the letters from the list given in the problem introduction that best complete the sentence. Separate the letters with commas. There is more than one correct answer, but you should only enter one pair of comma-separated letters. For eample, if the words "vector" and "scalar" fit best in the blanks, enter I,J. A,C Also accepted: position,c, D,C, coordinates,c, A,displacement, position,displacement, D,displacement, coordinates,displacement, A,G, position,g, D,G, coordinates,g, A,distance, position,distance, D,distance, coordinates,distance, A,E, position,e, D,E, coordinates,e, A,velocity, position,velocity, D,velocity, coordinates,velocity, A,B, position,b, D,B, coordinates,b, A,direction, position,direction, D,direction, coordinates,direction The coordinates of a point will depend on the coordinate system that is chosen, but there are several other quantities that are independent of the choice of origin for a coordinate system: in particular, distance, displacement, direction, and velocity. In working physics problems, unless you are interested in the position of an object or event relative to a specific origin, you can usually choose the coordinate system origin to be wherever is most convenient or intuitive. Note that the vector indicating a displacement from A to B is usually represented as.
Part G Identify the following physical quantities as scalars or vectors. Direction of Velocity and Acceleration Vector Quantities Conceptual Question For each of the motions described below, determine the algebraic sign (,, or ) of the velocity and acceleration of the object at the time specified. For all of the motions, the positive y ais is upward. Part A An elevator is moving downward when someone presses the emergency stop button. The elevator comes to rest a short time later. Give the signs for the velocity and the acceleration of the elevator after the button has been pressed but before the elevator has stopped. Enter the correct sign for the elevator's velocity and the correct sign for the elevator's acceleration, separated by a comma. For eample, if you think that the velocity is positive and the acceleration is negative, then you would enter +,-. If you think that both are zero, then you would enter 0,0. Hint 1. Algebraic sign of velocity The algebraic sign of velocity is determined solely by comparing the direction in which the object is moving with the direction that is defined to be positive. In this eample, upward is defined to be positive. Therefore, any object moving upward, whether speeding up, slowing down, or traveling at constant speed, has positive velocity. Hint. Algebraic sign of acceleration The algebraic sign of acceleration is more difficult to determine than the algebraic sign of velocity. The acceleration of an object points in the same direction as the change in the velocity of an object. If an object is speeding up, the change in the velocity points in the same direction as the velocity: If an object is slowing down, the change in velocity points in the opposite direction to that of the velocity:
Once you know the direction of the acceleration, you can determine its sign by comparing it to the defined positive direction, in this case, upward. -,+ Part B A child throws a baseball directly upward. What are the signs of the velocity and acceleration of the ball immediately after the ball leaves the child's hand? Enter the correct sign for the baseball's velocity and the correct sign for the baseball's acceleration, separated by a comma. For eample, if you think that the velocity is positive and the acceleration is negative, then you would enter +,-. If you think that both are zero, then you would enter 0,0. Hint 1. Algebraic sign of velocity The algebraic sign of velocity is determined solely by comparing the direction in which the object is moving with the direction that is defined to be positive. In this eample, upward is defined to be positive. Therefore, any object moving upward, whether speeding up, slowing down, or traveling at constant speed, has positive velocity. Hint. Algebraic sign of acceleration The algebraic sign of acceleration is more difficult to determine than the algebraic sign of velocity. The acceleration of an object points in the same direction as the change in the velocity of an object. If an object is speeding up, the change in the velocity points in the same direction as the velocity:
If an object is slowing down, the change in velocity points in the opposite direction to that of the velocity: Once you know the direction of the acceleration, you can determine its sign by comparing it to the defined positive direction, in this case, upward. +,- Part C A child throws a baseball directly upward. What are the signs of the velocity and acceleration of the ball at the very top of the ball's motion (i.e., the point of maimum height)? Enter the correct sign for the baseball's velocity and the correct sign for the baseball's acceleration, separated by a comma. For eample, if you think that the velocity is positive and the acceleration is negative, then you would enter +,-. If you think that both are zero, then you would enter 0,0. Hint 1. Algebraic sign of velocity
The algebraic sign of velocity is determined solely by comparing the direction in which the object is moving with the direction that is defined to be positive. In this eample, upward is defined to be positive. Therefore, any object moving upward, whether speeding up, slowing down, or traveling at constant speed, has positive velocity. Hint. Algebraic sign of acceleration The algebraic sign of acceleration is more difficult to determine than the algebraic sign of velocity. The acceleration of an object points in the same direction as the change in the velocity of an object. If an object is speeding up, the change in the velocity points in the same direction as the velocity: If an object is slowing down, the change in velocity points in the opposite direction as the velocity: Once you know the direction of the acceleration, you can determine its sign by comparing it to the defined positive direction, in this case, upward.
0,- Solutions to Problems.. Set Up: From the graph the position t at each time t is: 1 1 0 m, 0, 3 1 0 m, 4 0, 8 6 0 m, and 10 60 m Solve: (a) The displacement is 4 0 (i) 10 1 5 0 m; (ii) 10 3 7 0 m; (iii) 3 1 0 m; (iv) (b) (i) 30 m 10 m 4 0 m; 90 (ii) 10 m 10 m 0 m; (iii) zero (stays at 6 0 m ) *.11. Set Up: 10 century 100 yr 5 1 km 10 cm Solve: (a) d t (50 cm/yr)(100 yr) 500 cm 5 0 m d 550 10 cm 7 (b) t 1 1 10 yr t 50 cm/yr 5.1. Set Up: The distance around the circular track is d (400 m) 16 m For a half-lap, d 63 m Use coordinates for which the origin is at her starting point and the -ais is along a diameter, as shown in the figure below. Solve: (a) After one lap she has returned to her starting point. Thus, 0 and av, 0 (b) 40 0 m and av, 16 m average speed d 01 m/s t 65 s 400 m 63 m 1 39 m/s; average speed 0 m/s t 87 s d t 87 s.13. Set Up: Since sound travels at a constant speed, v t; also, from the appendi we find that 1 mile is 1.609 km. Solve: 1 mi (344 m/s)(75 s) 1 6 mi 3 1609 10 m 1 mi 1 Reflect: The speed of sound is (344 m/s) mi/s 3 1609 10 m 5 d 185 m 185 m.14. Solve: (a) t touch: t 0 043 s; pain: t 303 s 76 m/s 0610 m/s (b) The difference between the two times in (a) is 3.01 s.
.31. Set Up: Assume the ball starts from rest and moves in the -direction We may use the equations for constant acceleration. Solve: (a) 0 150 m, v 450 m/s and v 0 0. Using v v a ( ) gives v0 (450 m/s) v a 675 m/s. ( ) (150 m) 0 0 0 v0 v (b) Using 0 t gives ( 0) (150 m) t 00667 s v v 450 m/s 0 v 450 m/s Reflect: We could also use v v0 at to find t 0 0667 s, which agrees with our previous result. a 675 m/s The acceleration of the ball is very large..34. Set Up: Take the direction to be the direction of motion of the boulder. Solve: (a) Use the motion during the first second to find the acceleration. 0 0, 0 0, 00 m, and t 100 s 1 (00 m) 0 0t a t and a 400 m/s t (100 s) For the second second, 0 4 00 m/s, 0 at (400 m/s )(1 00 s) 4 00 m/s a 4 00 m/s, and t 100 s 1 1 0 0 t a t (400 m/s)(1 00 s) (400 m/s )(1 00 s) 6 00 m We can also solve for the location at t 00 s, starting at t 0: 1 1 0 0 t a t (400 m/s )(00 s) 8 00 m, which agrees with.00 m in the first second and 6.00 m in the second second. The boulder speeds up so it travels farther in each successive second. (b) We have already found 4 00 m/s after the first second. After the second second, 0 at 400 m/s (400 m/s )(1 00 s) 8 00 m/s *.37. Set Up: Let be the direction the car is moving. We can use the equations for constant acceleration. Solve: (a) From Eq. (.13), with v0 0, a (0m/s) v 167m/s ( ) (10 m) (b) Using Eq. (.14), t ( 0)/ v (10 m)/(0 m/s) 1s (c) (1 s)(0 m/s) 40 m 0 Reflect: The average velocity of the car is half the constant speed of the traffic, so the traffic travels twice as far..38. Set Up: 1 mi/h 1466 ft/s The car travels at constant speed during the reaction time. Let be the direction the car is traveling, so a 1 0 ft/s after the brakes are applied. Solve: (a) 0 (07 s) 15 4 ft. 1466 ft/s (150 mi/h) 0 ft/s 1 mi/h For the motion after the brakes are applied, 0 0 ft/s, 0 0 (0 ft/s) 0 a ( 10 ft/s ) ( ) 0 ft The total distance is 154 ft 0 ft 356 ft During the reaction time the car travels a distance of ( 0 ft/s) a 1 0 ft/s, and 0 0 a 0 ( ) gives 1466 ft/s (b) 0 (550 mi/h) 806 ft/s A calculation similar to that of part (a) gives a total stopping 1 mi/h distance of ( 0) 564 ft 707 ft 37 ft
.40. Set Up: Let be the direction the train is traveling. Find 0 for each segment of the motion. Solve: t 0 to 14.0 s: 1 1 0 0 t a t (160 m/s )(14 0 s) 157 m At t 14 0 s, the speed is 0 a (1 60 m/s )(14 0 s) 4 m/s In the net 70.0 s, a 0 and 0 0 t (4 m/s)(70 0 s) 1568 m For the interval during which the train is slowing down, 0 4 m/s, a 3 50 m/s and 0 0 a 0 0 0 (4 m/s) 0 7 m a ( 3 50 m/s ) ( ) gives The total distance traveled is 157 m 1568 m 7 m 1800 m