Mecanica

2
Motion in One Dimension
CHAPTER OUTLINE
2.1 2.2 2.3 2.4 2.5 2.6 2.7 Position, Velocity, and Speed Instantaneous Velocity and Speed Acceleration Motion Diagrams One-Dimensional Motion with Constant Acceleration Freely Falling Objects Kinematic Equations Derived from Calculus

ANSWERS TO QUESTIONS
* An asterisk indicates an item new to this edition. *Q2.1 Count spaces (intervals), notdots. Count 5, not 6. The first drop falls at time zero and the last drop at 5 × 5 s = 25 s. The average speed is 600 m 25 s = 24 m s, answer (b). The net displacement must be zero. The object could have moved away from its starting point and back again, but it is at its initial position again at the end of the time interval. Yes. Yes. If the speed of the object varies at all over the interval, theinstantaneous velocity will sometimes be greater than the average velocity and will sometimes be less.

Q2.2

Q2.3

*Q2.4

(a) It speeds up and its acceleration is positive. (b) It slows down overall, since final speed 1 m s is slower than 3 m s. Its acceleration is positive, meaning to the right. (c) It slows down and its acceleration is negative. (d) It speeds up to final speed 7 m s. Itsacceleration is negative, meaning toward the left or towards increasing-magnitude negative numbers on the track. No: Car A might have greater acceleration than B, but they might both have zero acceleration, or otherwise equal accelerations; or the driver of B might have tramped hard on the gas pedal in the recent past to give car B greater acceleration just then. (c) A graph of velocity versustime slopes down steadily from an original positive (northward) value. The graph cuts through zero and goes through increasing-magnitude negative values, all with the same constant acceleration. (i) none. All of the disks are moving. (ii) (b) shows equal spacing, meaning constant nonzero velocity and constant zero acceleration. (iii) (b) This question has the same physical meaning as question (ii).(iv) (c) shows positive acceleration throughout. (v) (a) shows negative (leftward) acceleration in the last three images. Tramping hard on the brake at zero speed on a level road, you do not feel pushed around inside the car. The forces of rolling resistance and air resistance have dropped to zero as the car coasted to a stop, so the car’s acceleration is zero at this moment and afterward.Tramping hard on the brake at zero speed on an uphill slope, you feel thrown backward against your seat. Before, during, and after the zero-speed moment, the car is moving with a downhill acceleration if you do not tramp on the brake. Brian Popp suggested the idea for this question.

Q2.5

*Q2.6

*Q2.7

*Q2.8

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Chapter 2

*Q2.9With original velocity zero, displacement is proportional to the square of time in (1 2)at 2. Making the time one-third as large makes the displacement one-ninth as large, answer (c). No. Constant acceleration only. Yes. Zero is a constant. They are the same. After the first ball reaches its apex and falls back downward past the student, it will have a downward velocity of magnitude vi. Thisvelocity is the same as the velocity of the second ball, so after they fall through equal heights their impact speeds will also be the same. For the release from rest we have (4 m s)2 = 02 + 2 gh. For case (i), we have vf = (3 m s)2 + 2 gh = (3 m s)2 + (4 m s)2. Thus answer (d) is true. For case (ii) the same steps give the same answer (d).
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Q2.10 Q2.11

*Q2.12

*Q2.13 (i) Its speed is zeroat b and e. Its speed is equal at a and c, and somewhat larger at d. On the bounce it is moving somewhat slower at f than at d, and slower at g than at c. The assembled answer is d > f > a = c > g > b = e. (ii) The velocity is positive at a, f, and g, zero at b and e, and negative at c and d, with magnitudes as described in part (i). The assembled answer is f > a > g > b = e > c > d. (iii) The...