Solution Manual for Physics for Scientists and Engineers: A Strategic Approach, 4th Edition

KINEMATICS IN ONE DIMENSION 2 Conceptual Questions 2.1. I left my friend’s house 5 mi east of town and rode my bike east at 4 mi/h for 5 more miles; this took 20 minutes. I then rested and ate a sandwich for 20 minutes and then rode all the way back to town at 4 mi/h. 2.2. With a slow start out of the blocks, a super sprinter reached top speed in about 5 s, having gone only 30 m. He was still able to finish his 100 m in only just over 9 s by running a world record pace for the rest of the race. 2.3. A crane picks up a crate from a platform 50 ft above the ground and then lifts it to 100 ft above the ground in 10 s where it is marked before lowering it all the way to the ground at a faster rate in another 10 s. 2.4. (a) At t  1 s, the slope of the line for object A is greater than that for object B. Therefore, object A’s speed is greater. (Both are positive slopes.) (b) No, the speeds are never the same. Each has a constant speed (constant slope) and A’s speed is always greater. 2.5. (a) A’s speed is greater at t  1 s. The slope of the tangent to B’s curve at t  1 s is smaller than the slope of A’s line. (b) A and B have the same speed at just about t  3 s. At that time, the slope of the tangent to the curve representing B’s motion is equal to the slope of the line representing A. 2.6. (a) B. The object is still moving, but the magnitude of the slope of the position-versus-time curve is smaller than at D. (b) D. The slope is greatest at D. (c) At points A, C, and E the slope of the curve is zero, so the object is not moving. (d) At point B the slope is negative, so the object is moving to the left. 2.7. (a) The slope of the position-versus-time graph is greatest at C, so the object is moving fastest at this point. (b) The slope is negative at point F, meaning the object is moving to the left there. (c) At point F the slope is increasing in magnitude (getting more negative), meaning that the object is speeding up to the left. (d) At point E the object is not moving since the slope is zero. Before point E, the slope is positive, while after E it is negative, so the object is turning around at E. 2.8. (a) The positions of the third dots of both motion diagrams are the same, as are the sixth dots of both, so cars A and B are at the same locations at the time corresponding to dot 3 and again at that of dot 6. (b) The spacing of dots 4 and 5 in both diagrams is the same, so the cars are traveling at the same speeds between times corresponding to dots 4 and 5. 2.9. No, though you have the same position along the road, his velocity is greater because he is passing you. If his velocity were not greater, then he would remain even with the front of your car. 2.10. Yes. The acceleration vector will point west when the bicycle is slowing down while traveling east. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-1 2-2 Chapter 2 2.11. (a) As a ball tossed upward moves upward, its vertical velocity is positive, while its vertical acceleration is negative, opposite the velocity, causing the ball to slow down. (b) The same ball on its way down has downward (negative) velocity. The downward negative acceleration is pointing in the same direction as the velocity, causing the speed to increase. 2.12. For all three of these situations the acceleration is equal to g in the downward direction. The magnitude and direction of the velocity of the ball do not matter. Gravity pulls down at constant acceleration. (Air friction is ignored.) 2.13. (a) The magnitude of the acceleration while in free fall is equal to g at all times, independent of the initial velocity. The acceleration only tells how the velocity is changing. (b) The magnitude of the acceleration is still g because the rock is still in free fall. The speed is increasing at the same rate each instant, that is, by the same v each second. 2.14. (a) The vertical axis of the graph is velocity, not position. The object is speeding up where the velocity is increasing; this is the case at point C. (b) The object is slowing down at point A because the velocity in the x direction is getting smaller. (c) The graph of velocity is always above the t axis, so the velocity is always positive, or in the direction to the right. At none of the points A, B, or C is it moving to the left. (d) The object is moving to the right at all three points because the velocity is positive at all three points. Exercises and Problems Exercises Section 2.1 Uniform Motion 2.1. Model: Cars will be treated by the particle model. Visualize: © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-3 Solve: Beth and Alan are moving at a constant speed, so we can calculate the time of arrival as follows: x x1  x0 x x v   t1  t0  1 0 t t1  t0 v Using the known values identified in the pictorial representation, we find: x x 400 mile tAlan 1  tAlan 0  Alan 1 Alan 0  8:00 AM   8:00 AM  8 hr  4:00 PM v 50 miles/hour x x 400 mile tBeth 1  tBeth 0  Beth 1 Beth 0  9:00 AM   9:00 AM  6.67 hr  3:40 PM v 60 miles/hour (a) Beth arrives first. (b) Beth has to wait tAlan 1  tBeth 1  20 minutes for Alan. Assess: Times of the order of 7 or 8 hours are reasonable in the present problem. 2.2. Model: We will consider Larry to be a particle. Visualize: Solve: Since Larry’s speed is constant, we can use the following equation to calculate the velocities: s s vs  f i tf  ti (a) For the interval from the house to the lamppost: 200 m  600 m v1   200 m/min 9 : 07  9 : 05 For the interval from the lamppost to the tree: 1200 m  200 m v2   333 m/min 9 :10  9 : 07 (b) For the average velocity for the entire run: 1200 m  600 m vavg   120 m/min 9 :10  9 : 05 2.3. Solve: (a) The time for each segment is t1  50 mi/40 mph  5/4 hr and t2  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 x1 in time  t1 at 40 mph. She then drives the distance x2 in time t2 at 60 mph. She spends the same amount of time at each speed, thus t1  t2  x1 /40 mph  x2 /60 mph  x1  (2/3)x2 © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-4 Chapter 2 But x1  x2  100 miles, so (2/3)x2  x2  100 miles. This means x2  60 miles and x1  40 miles. Thus, the times spent at each speed are t1  40 mi/40 mph  1.00 h and t2  60 mi/60 mph  1.00 h. The total time for her return trip is t1  t2  2.00 h. So, her average speed is 100 mi/2 h  50 mph. 2.4. Model: The jogger is a particle. 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 s 50 m  25 m v   1.25 m/s t 20 s The slope at t  25 s is 50 m  50 m v  0.0 m/s 10 s The slope at t  35 s is 0 m  50 m v  5.0 m/s 10 s Section 2.2 Instantaneous Velocity Section 2.3 Finding Position from Velocity 2.5. Solve: (a) We can obtain the values for the velocity-versus-time graph from the equation v  s /t. (b) There are no turning points since the slope doesn’t change from positive to negative (or vice versa) at any points. 2.6. Visualize: Please refer to Figure EX2.6 in the text. The particle starts at x0  10 m at t0  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 xf  x0  area of the velocity graph between t1 and tf , x2s  10 m  (area of triangle between 0 s and 1 s)  (area of triangle between 1 s and 2 s) 1 1  10 m  (4 m/s)(1 s)  (4 m/s)(1 s)  10 m 2 2 x4s  x2s  area between 2 s and 4 s 1  10 m  (4 m/s  12 m/s)(2 s)  26 m 2 2.7. Model: The graph shows the assumption that the blood isn’t moving for the first 0.1 s nor at the end of the beat. Visualize: The graph is a graph of velocity vs. time, so the displacement is the area under the graph—that is, the area of the triangle. The velocity of the blood increases quickly and decreases a bit more slowly. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-5 Solve: Call the distance traveled y. The area of a triangle is 12 BH . y  1 1 BH  (0.20 s)(0.80 m/s)  8.0 cm 2 2 Assess: This distance seems reasonable for one beat. 2.8. Solve: We can calculate the position of the particle at every instant with the equation xf  xi  area under the velocity-versus-time graph between ti and tf The particle starts from the origin at t  0 s, so xi  0 m. Notice that the each square of the grid in Figure EX2.8 has ―area‖ (5 m/s)  (2 s)  10 m. We can find the area under the curve, and thus determine x, by counting squares. You can see that x = 35 m at 6 because there are 3.5 squares under the curve. In addition, x  35 m at t = 10 s because the 5 m represented by the half square between 6 s and 8 s is canceled by the –5 m represented by the half square between 8 s and 10 s. Areas beneath the axis are negative areas. The particle passes through x  35 m at t  6 s and again at t  10 s. Section 2.4 Motion with Constant Acceleration 2.9. Visualize: The object has a constant velocity for 2 s and then speeds up between t  2 and t  4. 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 that is the slope of the velocity line. 2.10. Visualize: The graph is a graph of velocity vs. time, so the acceleration is the slope of the graph. Solve: When the blood is speeding up the acceleration is ay   y t   m/s  m/s2 0.05 s When the blood is slowing down the acceleration is ay   y t  0.80 m/s  5.3 m/s2 0.15 s Assess: 16 m/s2 is an impressive but reasonable acceleration. 2.11. Solve: (a) At t  20 s, the position of the particle is x2 s  2.0 m  area under velocity graph from t  0 s to t  2.0 s 1  2.0 m  (4.0 m/s)(2.0 s)  2.0 m  8.0 m 2 (b) From the graph itself at t  20 s, v  2 m/s (c) The acceleration is ax  vx vfx  vix 0.0 m/s  6.0 m/s    2.0 m/s 2 t t 3s © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-6 Chapter 2 2.12. Solve: (a) Using the equation xf  xi  area under the velocity-versus-time graph between ti and tf 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, vx (at t  1 s)  4 m/s. Also, ax  slope  v/t  0 m/s 2. (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, vx (t  3 s)  2 m/s. The acceleration is ax (t  3 s)  slope  vx (at t  4 s)  vx (at t  2 s)  2 m/s 2 2s 2.13. Model: Represent the car as a particle. Solve: (a) First, we will convert units: 60 miles 1 hour 1610 m    27 m/s hour 3600 s 1 mile The motion is constant acceleration, so v v (27 m/s  0 m/s) v1  v0  at  a  1 0   2.7 m/s2 t 10 s (b) The distance is calculated as follows: 1 1 x1  x0  v0t  a(t )2  a(t )2  1.3 102 m  4.3 102 feet 2 2 2.14. Model: Represent the jet plane as a particle. Visualize: Solve: Since we don’t know the time of acceleration, we will use v12  v02  2a( x1  x0 ) v 2  v02 (400 m/s)2  (300 m/s) 2 a 1   8.75 m/s 2  8.8 m/s 2 2 x1 2(4000 m) Assess: The acceleration of the jet is not quite equal to g, the acceleration due to gravity; this seems reasonable for a jet. 2.15. Model: Represent the spaceship as a particle. Solve: (a) The known information is: x0  0 m, v0  0 m/s, t0  0 s, a  g  9.8 m/s 2 , and v1  3.0 108 m/s. Constant-acceleration kinematics gives v v v1  v0  at  t  t1  1 0  3.06 107 s a © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-7 The problem asks for the answer in days, so we need a conversion: t1  (306 107 s)  1 hour 1 day   3.54 102 days  3.6 102 days 3600 s 24 hour (b) The distance traveled is 1 1 x1  x0  v0t  a(t )2  at12  4.6 1015 m 2 2 (c) The number of seconds in a year is 1 year  365 days  24 hours 3600 s   3.15 107 s 1 day 1 hour In one year light travels a distance 1 light year  (3.0 108 m/s)(315 107 s)  9.47 1015 m The distance traveled by the spaceship is 46 1015 m/946 1015 m  049 of a light year. Assess: Note that x1 gives ―Where is it?‖ rather than ―How far has it traveled?‖ ―How far‖ is represented by x1  x0  They happen to be the same number in this problem, but that isn’t always the case. 2.16. Model: Model the air as a particle. Visualize: Use the definition of acceleration and then convert units. Solve: ax  vx 150 km/h  1000 m   1 h   1 min  2       83 m/s t 0.50 s  1 km   60 min   60 s  Assess: 83 m/s2 is a remarkable acceleration. 2.17. Model: We are using the particle model for the skater and the kinematics model of motion under constant acceleration. Solve: Since we don’t know the time of acceleration we will use vf2  vi2  2a( xf  xi ) a vf2  vi2 (6.0 m/s) 2  (8.0 m/s) 2   2.8 m/s 2 2( xf  xi ) 2(5.0 m) Her acceleration is positive 2.8 m/s2 because it points to the right; not because she is speeding up (she isn’t). Assess: A deceleration of 2.8 m/s 2 is reasonable. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-8 Chapter 2 2.18. Model: We are assuming both cars are particles. Visualize: Solve: The Porsche’s time to finish the race is determined from the position equation 1 xP1  xP0  vP0 (tP1  tP0 )  aP (tP1  tP0 )2 2 1 2  400 m  0 m  0 m  (3.5 m/s )(tP1  0 s) 2  tP1  15.12 s 2 The Honda’s time to finish the race is obtained from Honda’s position equation as 1 xH1  xH0  vH0 (tH1  tH0 )  aH0 (tH1  tH0 )2 2 1 2 400 m  0 m  0 m  (3.0 m/s )(tH1  1 s) 2  tH1  15.33 s 2 So, the Porsche wins by 0.21 s. Assess: The numbers are contrived for the Porsche to win, but the time to go 400 m seems reasonable. 2.19. Model: The car is a particle moving under constant-acceleration kinematic equations. Visualize: Solve: This is a three-part problem. First the car accelerates, then it moves with a constant speed, and then it decelerates. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-9 First, the car accelerates: v1  v0  a0 (t1  t0 )  0 m/s  (4.0 m/s 2 )(6 s  0 s)  24 m/s 1 1 x1  x0  v0 (t1  t0 )  a0 (t1  t0 )2  0 m  (4.0 m/s 2 )(6 s  0 s) 2  72 m 2 2 Second, the car moves at v1: 1 x2  x1  v1(t2  t1)  a1(t2  t1)2  (24 m/s)(8 s  6 s)  0 m  48 m 2 Third, the car decelerates: v3  v2  a2 (t3  t2 )  0 m/s  24 m/s  (3.0 m/s 2 )(t3  t2 )  (t3  t2 )  8 s 1 1 x3  x2  v2 (t3  t2 )  a2 (t3  t2 )2  x3  x2  (24 m/s)(8 s)  (3.0 m/s 2 )(8 s) 2  96 m 2 2 Thus, the total distance between stop signs is: x3  x0  ( x3  x2 )  ( x2  x1)  ( x1  x0 )  96 m  48 m  72 m  216 m Assess: A distance of approximately 600 ft in a time of around 10 s with an acceleration/deceleration of the order of 7 mph/s is reasonable. Section 2.5 Free Fall 2.20. Model: Represent the spherical drop of molten metal as a particle. Visualize: Solve: (a) The shot is in free fall, so we can use free fall kinematics with a   g. The height must be such that the shot takes 4 s to fall, so we choose t1  4 s. Then, y1  y0  v0 (t1  t0 )  1 1 1 g (t1  t0 )2  y0  gt12  (9.8 m/s2 )(4 s) 2  78.4 m 2 2 2 (b) The impact velocity is v1  v0  g (t1  t0 )   gt1  39.2 m/s.  Assess: Note the minus sign. The question asked for velocity, not speed, and the y-component of v is negative because the vector points downward. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-10 Chapter 2 2.21. Model: We model the ball as a particle. Visualize: Solve: Once the ball leaves the student’s hand, the ball is in free fall and its acceleration is equal to the free-fall acceleration g that always acts vertically downward toward the center of the earth. According to the constantacceleration kinematic equations of motion 1  y1  y0  v0 t  a t 2 2 Substituting the known values 2 m  0 m  (15 m/s)t1  (1/2)(9.8 m/s 2 )t12 One solution of this quadratic equation is t1  3.2 s. The other root of this equation yields a negative value for t1, which is not valid for this problem. Assess: A time of 3.2 s is reasonable. 2.22. Model: We will use the particle model and the constant-acceleration kinematic equations. Visualize: © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-11 Solve: (a) Substituting the known values into  y1  y0  v0 t  12 a t 2 , we get 1 10 m  0 m  20 (m/s)t1  (9.8 m/s 2 )t12 2 One of the roots of this equation is negative and is not relevant physically. The other root is t1  453 s, which is the answer to part (b). Using v1  v0  at , we obtain v1  20(m/s)  (9.8 m/s2 )(4.53 s)  24 m/s (b) The time is 4.5 s. Assess: A time of 4.5 s is a reasonable value. The rock’s velocity as it hits the bottom of the hole has a negative sign because of its downward direction. The magnitude of 24 m/s compared to 20 m/s, when the rock was tossed up, is consistent with the fact that the rock travels an additional distance of 10 m into the hole. 2.23. Model: Model the flea as a particle. Both the initial acceleration phase and the free-fall phase have constant acceleration, so use the kinematic equations. Visualize: Solve: We can apply the kinematic equation vf2  vi2  2ay twice, once to find the take-off speed and then again to find the final height. In the first phase the acceleration is up (positive) and v0  0. v12  2a0 ( y1  y0 )  2(1000 m/s2 )(0.50 103 m)v1  1.0 m/s In the free fall phase the acceleration is a1   g and v1  10 m/s and v2  0 m/s. y2  y1  v22  v12 v12 (1.0 m/s)2    5.1 cm 2a1 2( g ) 2(9.8 m/s 2 ) So the final height is y2  51 cm  y1  51 cm  050 mm  52 cm. Assess: This is pretty amazing—about 10–20 times the size of a typical flea. 2.24. Model: The watermelon and Superman will be treated as particles that move according to constantacceleration kinematic equations. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-12 Chapter 2 Visualize: Solve: The watermelon’s and Superman’s position as they meet each other are 1 yW1  yW0  vW0 (tW1  tW0 )  aW0 (tW1  tW0 ) 2 2 1 yS1  yS0  vS0 (tS1  tS0 )  aS0 (tS1  tS0 ) 2 2 1  yW1  320 m  0 m  ( 9.8 m/s 2 )(tW1  0 s) 2 2  yS1  320 m  ( 35 m/s)(tS1  0 s)  0 m Because tS1  tW1, 2 yW1  320 m  (4.9 m/s2 ) tW1 yS1  320 m  (35 m/s) tW1 Since yW1  yS1, 2 320 m  (4.9 m/s2 )tW1  320 m  (35 m/s)tW1  tW1  0 s and 7.1 s Indeed, tW1  0 s corresponds to the situation when Superman arrives just as the watermelon is dropped off the Empire State Building. The other value, tW1  71 s, is the time when the watermelon will catch up with Superman. The speed of the watermelon as it passes Superman is vW1  vW0  aW0 (tW1  tW0 )  0 m/s  (9.8 m/s2 )(7.1 s  0 s)  70 m/s Note that the negative sign implies a downward velocity. Assess: A speed of 157 mph for the watermelon is understandable in view of the significant distance (250 m) involved in the free fall. 2.25. Model: Model the rock as a particle and assume it is in free fall. Visualize: Set up a coordinate system where the origin is at the top of the building, but up is still positive. That is, the location of the bottom of the building is y2 which is a negative number, and downward velocities are also negative. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-13 Solve: Use the third kinematic equation for y2  y1 and y2  y0 . y2  y1  v22  v12 2a y2  y0  v22  v02 2a Now use v0  0 and set up a ratio where the 2a cancels. 0.45  y2  y1 v22  v12 v2   1  12 2 y2  y1 v2 v2 Now the first kinematic equation comes into play. v2  v1  a(t2  t1 ) v2 v12 1  12  1   0.45 v2 (v1  a (t2  t1 )) 2 1  0.45  v12 (v1  a (t2  t1 )) 2  0.55 Take square roots of both sides and solve for v1. v1  0.55  v1  a(t2  t1 ) v1  a(t2  t1) 0.55 (9.8 m/s 2 )(1 s) 0.55   28.13 m/s 1  0.55 1  0.55 Now compute v2. v2  v1  a(t2  t1)  28.13 m/s  (9.8 m/s2 )(1 s)  37.93 m/s Plug back into the third kinematic equation with v0  0 and y0  0. y2  v22 (37.93 m/s) 2   73 m 2a 2(9.8 m/s 2 ) So the building is 73 m tall. Assess: 73 m is a reasonable height of a building. Also, with the information calculated so far we check our work by using kinematic equations to find the other variables: y1  40 m, t1  2.9 s, t2  3.9 s. It is important to keep extra digits in the intermediate calculations and then round to the correct number of significant figures at the end. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-14 Chapter 2 Section 2.6 Motion on an Inclined Plane 2.26. Model: We will model the skier as a particle. Visualize: Note that the skier’s motion on the horizontal, frictionless snow is not of any interest to us. Also note that the acceleration parallel to the incline is equal to g sin10°. Solve: Using the following constant-acceleration kinematic equations, vf2x  vi2x  2ax ( xf  xi )  (15 m/s) 2  (3.0 m/s) 2  2(9.8 m/s 2 )sin10 ( x1  0 m)  x1  64 m vfx  vix  ax (tf  ti )  (15 m/s)  (3.0 m/s)  (9.8 m/s 2 )(sin10 )t  t  7.1 s Assess: A time of 7.1 s to cover 64 m is a reasonable value. 2.27. Model: Represent the car as a particle. Visualize: Solve: Note that the problem ―ends‖ at a turning point, where the car has an instantaneous speed of 0 m/s before rolling back down. The rolling back motion is not part of this problem. If we assume the car rolls without friction, then we have motion on a frictionless inclined plane with an acceleration a   g sin   g sin10  1.7 m/s2. Constant-acceleration kinematics gives v2 (30 m/s) 2 v12  v02  2a( x1  x0 )  0 m2 /s 2  v02  2ax1  x1   0    265 m 2a 2(1.7 m/s 2 ) Notice how the two negatives canceled to give a positive value for x1.  Assess: We must include the minus sign because the a vector points down the slope, which is in the negative x-direction. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-15 2.28. Model: Santa is a particle moving under constant-acceleration kinematic equations. Visualize: Note that our x-axis is positioned along the incline. Solve: Using the following kinematic equation, v12  v02  2a || ( x1  x2 )  (0 m/s)2  2(4.9 m/s 2 )(10 m  0 m)  v1  9.9 m/s Assess: Santa’s speed of 20 mph as he reaches the edge is reasonable. 2.29. Model: The snowboarder is a particle moving under constant-acceleration kinematic equations. The speed does not diminish on the horizontal section. Visualize: Note that our s-axis is positioned along both inclines. Solve: (a) Using the following kinematic equation for the downward slope v12  v02  2a0 (s1  s0 )  (0 m/s)2  2(9.8 m/s2 )(sin15 )(50 m  0 m)  v1  15.93 m/s which is reported as 16 m/s to two significant figures. (b) Use the same equation again on the upward slope (with v2  v1 ). v32  v22  2a2 ( s3  s2 )  (0 m/s) 2  ( s3  s2 )  v12 (15.93 m/s) 2   31 m 2a2 2(9.8 m/s 2 )(sin 25 ) Assess: Because the upward slope is steeper we did not expect the snowboarder to travel as far up the slope. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-16 Chapter 2 2.30. Model: The frog is a particle moving under constant-acceleration kinematic equations. The ramp is frictionless. Visualize: Note that our x-axis is positioned along the incline. Solve: Using the following kinematic equation, 1 v12  v02  2a( x1  x2 )  (5.0 m/s)2  2(9.8 m/s 2 )   (2.0 m  0 m)  v1  2.3 m/s 2 Assess: We knew the speed at the top of the ramp would be smaller than the beginning speed at the bottom. Section 2.7 Instantaneous Acceleration 2.31. Solve: The formula for the particle’s velocity is given by vf  vi  area under the acceleration curve between ti and tf For t  4 s, we get 1 v4 s  8 m/s  (4 m/s 2 )4 s  16 m/s 2 Assess: The acceleration is positive but decreases as a function of time. The initial velocity of 8.0 m/s will therefore increase. A value of 16 m/s is reasonable. 2.32. Visualize: Solve: We will determine the object’s velocity using graphical methods first and then using calculus. Graphically, v(t )  v0  area under the acceleration curve from 0 to t. In this case, v0  0 m/s The area at the time requested is a triangle. t 6 s 1 v(t  6 s)  (6 s)(10 m/s)  30 m/s 2 © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-17 Let us now use calculus. The acceleration function a(t) consists of three pieces and can be written: 0 st 4 s  2.5t  a(t )  t  30 4 s  t  6 s  0 6 s t 8 s  These were determined by the slope and the y-intercept of each of the segments of the graph. The velocity function is found by integration as follows: For 0  t  4 s, t v(t )  v(t  0 s)   a(t )dt  0  2.5 0 t2 2 t  1.25t 2 0 This gives t  4 s v(t  4 s)  20 m/s For 4 s  t  6 s, t  5t 2  v(t )  v(t  4 s)   a(t )dt  20 m/s    30t   2.5t 2  30t  60 4  2  4 t This gives: t  6 s v(t  6 s)  30 m/s Assess: The same velocities are found using calculus and graphs, but the graphical method is easier for simple graphs. 2.33. Solve: x = (2t 3 + 2t +1) m (a) The position t  2 s is x2s  [2(2)3  2(2)  1] m  21 m (b) The velocity is the derivative v  dx /dt and the velocity at t  2 s is calculated as follows: v  (6t 2  2) m/s  v2s  [6(22 )  2] m/s  26 m/s (c) The acceleration is the derivative a  dv /dt and the acceleration at t  2 s is calculated as follows: a = (12t) m/s2 Þ a2s = 24 m/s2 2.34. Solve: The formula for the particle’s position along the x-axis is given by tf xf  xi   vx dt ti Using the expression for vx we get xf  xi  23 [tf3  ti3 ] ax  dvx d  (2t 2 m/s)  4t m/s 2 dt dt (a) The particle’s position at t  1 s is x1 s  1 m  23 m  53 m. (b) The particle’s speed at t  1 s is v1 s  2 m/s. (c) The particle’s acceleration at t  1 s is a1 s  4 m/s2. 2.35. Solve: (a) The velocity-versus-time graph is given by the derivative with respect to time of the position function: dx  (6t 2  18t ) m/s dt For vx  0 m/s, there are two solutions to the quadratic equation: t = 0 s and t = 3 s. vx  © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-18 Chapter 2 (b) At the first of these solutions, x(at t  0 s)  2(0 s)3  9(0 s)2  12  12 m The acceleration is the derivative of the velocity function: dv ax  x  (12t  18) m/s 2  a(at t  0 s)  18 m/s 2 dt At the second solution, x(at t  3 s)  2(3 s)3  9(3 s)2  12  15 m ax (at t  3 s)  12(3 s)  18  18 m/s2 2.36. Solve: First take the derivatives. Every quantity is in SI units. x  2t 3  6t 2  12 v  6t 2  12t a  12t  12 (a) To find the minimum v we take the derivative of the velocity function and set it equal to zero (this is how we dv  12t  12  0  t  1 s. This is the first always minimize things). But we have already taken the derivative of v: dt answer. Now we plug that time back into the equation for v: vmin  6(1)2  12(1)  6 m/s. (b) We have already done this when we minimized v. The acceleration is zero at 0  12t  12  t  1 s. Problems 2.37. Solve: The graph for particle A is a straight line from t  2 s to t  8 s The slope of this line is 10 m/s, which is the velocity at t  70 s The negative sign indicates motion toward lower values on the x-axis. The velocity of particle B at t  70 s can be read directly from its graph. It is 20 m/s. The velocity of particle C can be obtained from the equation vf  vi  area under the acceleration curve between ti and tf This area can be calculated by adding up three sections. The area between t  0 s and t  2 s is 40 m/s, the area between t  2 s and t  5 s is 45 m/s, and the area between t  5 s and t  7 s is 20 m/s. We get (10 m/s)  (60 m/s)  (45 m/s)  (20 m/s)  95 m/s. 2.38. Solve: (a) The velocity-versus-time graph is the derivative with respect to time of the distance-versus-time graph. The velocity is zero when the slope of the position-versus-time graph is zero, the velocity is most positive when the slope is most positive, and the velocity is most negative when the slope is most negative. The slope is zero at t = 0, 1 s, 2 s, 3 s, . . . ; the slope is most positive at t = 0.5 s, 2.5 s, . . ; and the slope is most negative at t = 1.5 s, 3.5 s, . . . (b) © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-19 2.39. Solve: The given function for the velocity is vx  t 2  7t  10. (a) The turning points are when the velocity changes sign. Set vx  0 and check that it actually changes sign at those times. The function factors into the product of two binomials: vx  (t  2)(t  5)  vx  0 when t  2 s and t  5 s Indeed, the function changes sign at those two times. (b) The acceleration is given by the derivative of the velocity. ax  dvx  2t  7 dt Plug in the times from part (a): ax (2 s)  2(2)  7  3 m/s2 and ax (5 s)  2(5)  7  3 m/s2 Assess: This problem does not have constant acceleration so the kinematic equations do not apply, but a  dv /dt always applies. 2.40. Solve: The position is the integral of the velocity. t1 t0 t1 0 x1  x0   vx dt  x0   kt 2dt  x0  13 kt 3 t1 0  x0  13 kt13 We’re given that x0  9.0 m and that the particle is at x1  90 m at t1  30 s Thus 9.0 m  (9.0 m)  13 k (3.0 s)3  (9.0 m)  k (9.0 s3) Solving for k gives k  20 m/s3. 2.41. Solve: (a) The velocity is the integral of the acceleration. t1 t0 t1 0 v1x  v0 x   ax dt  0 m/s   (10  t )dt  (10t  12 t 2 ) The velocity is zero when  t1  10t1  12 t12 0  v1x  0 m/s  10t1  12 t12  (10  12 t1)  t1  t1  0 s or t1  20 s The first solution is the initial condition. Thus the particle’s velocity is again 0 m/s at t1  20 s (b) Position is the integral of the velocity. At t1  20 s, and using x0  0 m at t0  0 s, the position is t1 t0 x1  x0   vx dt  0 m   20 0 10t  t  dt  5t 1 2 2 2 20 0  16 t 3 20 0  667 m 2.42. Solve: (a) The turning point is when v  0, so set the equation for v equal to zero. The velocity is zero when v  (20 m/s)sin( t )  0  t  0 s, 1 s, 2 s, … The first time after t  0 is t  1 s. (b) The acceleration at t = 1 s is found by taking the derivative of the velocity equation and inserting t = 1 s. a dv  (2.0 m/s)  cos( t ) dt a(t  1 s)  (2.0 m/s)  cos( )  2.0 m/s 2 2.43. Model: Represent the ball as a particle. Visualize: Please refer to the figure in the problem. Solve: In the first and third segments the acceleration as is zero. In the second segment the acceleration is negative and constant. This means the velocity vs will be constant in the first two segments and will decrease linearly in the third © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-20 Chapter 2 segment. Because the velocity is constant in the first and third segments, the position s will increase linearly. In the second segment, the position will increase parabolically rather than linearly because the velocity decreases linearly with time. 2.44. Model: Represent the ball as a particle. Visualize: Please refer to Figure P2.44. The ball rolls down the first short track, then up the second short track, and then down the long track. s is the distance along the track measured from the left end (where s = 0). Label t = 0 at the beginning, that is, when the ball starts to roll down the first short track. Solve: Because the incline angle is the same, the magnitude of the acceleration is the same on all of the tracks. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-21 Assess: Note that the derivative of the s versus t graph yields the vs versus t graph. And the derivative of the vs versus t graph gives rise to the as versus t graph. 2.45. Visualize: Please refer to Figure P2.45. Solve: 2.46. Visualize: Please refer to Figure P2.46. Solve: 2.47. Model: The plane is a particle and the constant-acceleration kinematic equations hold. Solve: (a) Using as  v /t , we have, as (t  0 to t  10 s)  23 m/s  0 m/s  2.3 m/s 2 10 s  0 s as (t  20 s to t  30 s)  69 m/s  46 m/s  2.3 m/s 2 30 s  20 s For all time intervals a is 2.3 m/s2 . (b) Using kinematics as follows: vfs  vis  a(tf  ti )  80 m/s  0 m/s  (2.3 m/s 2 )(tf  0 s)  tf  35 s (c) Using the above values, we calculate the takeoff distance as follows: 1 1 sf  si  vis (tf  ti )  as (tf  ti )2  0 m  (0 m/s)(35 s)  (2.3 m/s 2 )(35 s) 2  1410 m 2 2 For safety, the runway should be 3 1410 m  4230 m or 2.6 mi. This is longer than the 2.4 mi long runway, so the takeoff is not safe. 2.48. Model: The car is a particle and constant-acceleration kinematic equations hold. Visualize: Solve: This is a two-part problem. During the reaction time, x1  x0  v0 (t1  t0 )  1/2a0 (t1  t0 )2  0 m  (20 m/s)(0.50 s  0 s)  0 m  10 m © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-22 Chapter 2 After reacting, x2  x1  110 m  10 m  100 m, that is, you are 100 m away from the intersection. To stop successfully, v22  v12  2a1( x2  x1)  (0 m/s)2  (20 m/s)2  2a1(100 m)  a1  2 m/s 2 2.49. Model: We will use the particle model and the constant-acceleration kinematic equations. Visualize: Solve: (a) To find x2 , we first need to determine x1 Using x1  x0  v0 (t1  t0 ), we get x1  0 m  (20 m/s) (050 s  0 s)  10 m. Now, v22  v12  2a1( x2  x1)  0 m2 /s2  (20 m/s)2  2(10 m/s 2 )( x2  10 m)  x2  30 m The distance between you and the deer is ( x3  x2 ) or (35 m  30 m)  5 m. (b) Let us find v0 max such that v2  0 m/s at x2  x3  35 m Using the following equation, v22  v02 max  2a1( x2  x1)  0 m2 /s2  v02 max  2(10 m/s2 )(35 m  x1) Also, x1  x0  v0 max (t1  t0 )  v0 max (0.50 s  0 s)  (0.50 s)v0 max . Substituting this expression for x1 in the above equation yields v02 max  (20 m/s2 )[35 m  (0.50 s) v0 max ]  v02 max  (10 m/s)v0 max  700 m2 /s 2  0 The solution of this quadratic equation yields v0 max  22 m/s (The other root is negative and unphysical for the present situation.) Assess: An increase of speed from 20 m/s to 22 m/s is very reasonable for the car to cover an additional distance of 5 m with a reaction time of 0.50 s and a deceleration of 10 m/s2 . 2.50. Model: We will use the particle model and the constant-acceleration kinematic equations. Visualize: [ © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-23 Solve: Since Car 2 has the same initial speed and same acceleration, then it, too, must stop in 10 m, so it begins braking at the same position that Car 1 did. Compute the time for Car 1 to brake. ( x11  x10 )  1 a1(t1  t0 )2  (t1  t0 )  2 2( x11  x10 ) a The initial speed of Car 1 is the same as the initial speed of Car 2. 2 v10  2a1( x11  x10 )  v10  2a( x11  x10 ) The distance Car 2 travels in time (t1  t0 ) at speed v20  v10 is x21  x 20  2a( x11  x10 ) 2( x11  x10 )  2( x11  x10 )  2(10 m)  20 m a This is the initial separation of the two cars. Assess: A simpler way is to realize that the average speed of Car 1 during breaking is half its initial speed. Car 2 (going at that same initial speed) will cover twice the distance during the same time, so Car 2 travels 20 m while Car 1 brakes for 10 m. 2.51. Model: We will use the particle model and the constant-acceleration kinematic equations. Visualize: The ball can have zero velocity right as it gets to the hole. Solve: Use the kinematic equations at each stage. v32  v22  2a2 ( x3  x2 ) v22  v12  2a1 ( x2  x1 ) v12  v02  2a0 ( x1  x0 ) Solve the last equation for v02 and substitute in from the other equations. v02  v12  2a0 ( x1  x0 )  v22  2a1 ( x2  x1 )   2a0 ( x1  x0 )     v32  2a2 ( x3  x2 )   2a1( x2  x1)   2a0 ( x1  x0 )     0  2(1.0 m/s 2 )(1.0 m)  2(6.0 m/s 2 )(2.0 m)  2( 1.0 m/s 2 )(3.0 m)  5.7 m/s Assess: This seems like a swift putt, but in the reasonable range. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-24 Chapter 2 2.52. Model: The car is represented as a particle. Visualize: Solve: This is a two-part problem. First, we need to use the information given to determine the acceleration during braking. Second, we need to use that acceleration to find the stopping distance for a different initial velocity. First, the car coasts at constant speed before braking: x1  x0  v0 (t1  t0 )  v0t1  (30 m/s)(0.5 s)  15 m Then, the car brakes to a halt. Because we don’t know the time interval, use v22  0  v12  2a1( x2  x1) v12 (30 m/s) 2 2  10 m/s 2 2( x2  x1) 2(60 m  15 m)  We used v1  v0  30 m/s Note the minus sign, because a1 points to the left. We can repeat these steps now with v0   m/s. The coasting distance before braking is  a1   x1  v0t1  (40 m/s)(0.5 s)  20 m The position x2 after braking is found using v22  0  v12  2a1( x2  x1)  x2  x1  v12 (40 m/s) 2  20 m   100 m 2a1 2(10 m/s 2 ) 2.53. Model: Model each of the animals as a particle and use kinematic equations. Assume that the time it takes the cheetah to accelerate to 30 m/s is negligible. Visualize: © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-25 Solve: The cheetah is in uniform motion for the entire duration of the problem, so we can easily solve for its position at t3  25 s: x3C  x1C  (vx )1C t  0 m  (30 m/s)(15 s)  450 m The gazelle’s motion has two phases: one of constant acceleration and then one of constant velocity. We can solve for the position and the velocity at t2 , the end of the first phase. (vx )2G  (vx )1G  (ax )G t  0 m/s  (46 m/s 2 )(5.0 s)  23 m/s 1 1 x2G  x1G  (vx )1G t  (ax )G (t ) 2  170 m  0 m  (4.6 m/s 2 )(5.0 s) 2  227.5 m 2 2 From t2 to t3 the gazelle moves at a constant speed, so we can use the equation for uniform motion to find its final position. x3G  x2G  (vx )2G t  227.5 m  (23 m/s)(10.0 s)  457.5 m  460 m x3C is 450 m; x3G is 460 m. The gazelle is just ten meters ahead of the cheetah when the cheetah has to break off the chase, so the gazelle escapes. Assess: The numbers in the problem statement are realistic, so we expect our results to mirror real life. The speed for the gazelle is close to that of the cheetah, which seems reasonable for two animals known for their speed. And the result is the most common occurrence—the chase is very close, but the gazelle gets away. 2.54. Model: Assume the distance between the front of one car and the back of the car in front of it is negligible. All the cars have the same length. Visualize: Call the length of each car L. Solve: Use the second kinematic equation L 1 a(t1)2 2 where t1  5.0 s We seek to find t7 . 7L  1 a (  t7 ) 2 2 1  1 7  a(t1)2   a(t7 )2  7(t1)2  (t7 )2  t7  7 (t1)  7 (5.0 s)  13 s 2  2 Assess: The numbers in the problem statement are realistic, so we expect our results to mirror real life. The speed for the gazelle is close to that of the cheetah, which seems reasonable for two animals known for their speed. And the result is the most common occurrence—the chase is very close, but the gazelle gets away. 2.55. Model: The rocket is represented as a particle. Visualize: © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-26 Chapter 2 Solve: (a) There are three parts to the motion. Both the second and third parts of the motion are free fall, with a   g. The maximum altitude is y2 . In the acceleration phase: 1 1 1 y1  y0  v0 (t1  t0 )  a(t1  t0 )2  at12  (30 m/s 2 )(30 s) 2  13,500 m 2 2 2 v1  v0  a(t1  t0 )  at1  (30 m/s 2 )(30 s)  900 m/s In the coasting phase, v22  0  v12  2 g ( y2  y1)  y2  y1  v12 (900 m/s)2  13,500 m   54,800 m  54.8 km 2g 2(9.8 m/s 2 ) The maximum altitude is 54.8 km ( 33 miles). (b) The rocket is in the air until time t3  We already know t1  30 s We can find t2 as follows: v v2  0 m/s  v1  g (t2  t1)  t2  t1  1  122 s g Then t3 is found by considering the time needed to fall 54,800 m: y3  0 m  y2  v2 (t3  t2 )  1 1 2 y2 g (t3  t2 )2  y2  g (t3  t2 )2  t3  t2   228 s 2 2 g Assess: In reality, friction due to air resistance would prevent the rocket from reaching such high speeds as it falls, and the acceleration upward would not be constant because the mass changes as the fuel is burned, but that is a more complicated problem. 2.56. Model: We will model the rocket as a particle. Air resistance will be neglected. Visualize: Solve: Using the constant-acceleration kinematic equations, v1  v0  a0 (t1  t0 )  0 m/s  a0 (16 s  0 s)  ao (16 s) 1 1 a0 (t1  t0 ) 2  a0 (16 s  0 s) 2  ao (128 s 2 ) 2 2 1 y2  y1  v1 (t2  t1 )  a1 (t2  t1 ) 2 2 1 2  5100 m  128 s a0  16 s a0 (20 s  16 s)  ( 9.8 m/s 2 )(20 s  16 s) 2  a0  27 m/s 2 2 Assess: This acceleration would produce a final speed (after 16 s) of 400 m/s  900 mph which would be reasonable. y1  y0  v0 (t1  t0 )  © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-27 2.57. Model: We will model the lead ball as a particle and use the constant-acceleration kinematic equations. Visualize: Note that the particle undergoes free fall until it hits the water surface. Solve: The kinematic equation y1  y0  v0 (t1  t0 )  12 a0 (t1  t0 )2 becomes 1 5.0 m  0 m  0 m  (9.8 m/s 2 )(t1  0) 2  t1  1.01 s 2 Now, once again, 1 a1(t2  t1) 2 2  y2  y1  v1(3.0 s  1.01 s)  0 m/s  1.99 v1 y2  y1  v1 (t2  t1)  v1 is easy to determine since the time t1 has been found. Using v1  v0  a0 (t1  t0 ), we get v1  0 m/s  (9.8 m/s2 )(1.01 s  0 s)  9.898 m/s With this value for v1, we go back to: y2  y1  1.99v1  (1.99)(9.898 m/s)  19.7 m y2  y1 is the displacement of the lead ball in the lake and thus corresponds to the depth of the lake, which is 19.7 m. The negative sign shows the direction of the displacement vector. Assess: A depth of about 60 ft for a lake is not unusual. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-28 Chapter 2 2.58. Model: The elevator is a particle moving under constant-acceleration kinematic equations. Visualize: Solve: (a) To calculate the distance to accelerate up: (v1)2  v02  2a0 ( y0  y0 )  (5 m/s)2  (0 m/s)2  2(1 m/s2 )( y1  0 m)  y1  12.5 m (b) To calculate the time to accelerate up: v1  v0  a0 (t1  t0 )  5 m/s  0 m/s  (1 m/s2 )(t1  0 s)  t1  5 s To calculate the distance to decelerate at the top: v32  v22  2a2 ( y3  y2 )  (0 m/s)2  (5 m/s)2  2(1 m/s2 )( y3  y2 )  y3  y2  12.5 m To calculate the time to decelerate at the top: v3  v2  a2 (t3  t2 )  0 m/s  5 m/s  (1 m/s2 )(t3  t2 )  t3  t2  5 s The distance moved up at 5 m/s is y2  y1  ( y3  y0 )  ( y3  y2 )  ( y1  y0 )  200 m  12.5 m  12.5 m  175 m The time to move up 175 m is given by 1 y2  y1  v1(t2  t1)  a1(t2  t1)2  175 m  (5 m/s)(t2  t1)  (t2  t1)  35 s 2 To total time to move to the top is (t1  t0 )  t2  t1   t3  t2   5 s  35 s  5 s  45 s Assess: To cover a distance of 200 m at 5 m/s (ignoring acceleration and deceleration times) will require a time of 40 s. This is comparable to the time of 45 s for the entire trip as obtained above. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-29 2.59. Model: The player is modeled as a particle. Visualize: The player has a constant acceleration during the jump because he is in free fall. However, the elevator is not accelerating. It is another reference frame moving at constant velocity. Solve: If the player can jump 55 cm in one constant velocity reference frame, he can jump 55 cm above the floor in the other constant velocity reference frame. Assess: The player would not even know the elevator was moving (except for accelerations at the start and stop). 2.60. Model: You and the bus are particles; the bus has a constant acceleration and your acceleration is zero. Visualize: Solve: (a) For the bus: 1 xB1  xB0  aBt12 2 For you: xY1  vY0t1 Equations for the bus and you are graphed on the same position-versus-time graph. When you catch the bus xB1  xY1. Then solve the resulting quadratic equation. 1 aBt12  vY0t1  xB0  0  2 t1  2 vY0  vY0  2aB xB0 aB  (4.5 m/s)  (4.5 m/s) 2  2(1.0 m/s 2 )(9.0 m) 1.0 m/s 2  (4.5 s)  (1.5 s)  3.0 s, 6.0 s There are two positive answers because if you don’t jump on the bus when you first catch up to it at 3.0 s then you would pass it and then later it would catch up to you at 6.0 s. You actually have two chances to jump on the bus in this scenario, but the first one comes at 3.0 s. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-30 Chapter 2 (b) The maximum time you could wait before starting to run is best seen in the position-versus-time graph above. The parabola for the bus stays the same, but the straight line for you could move horizontally until it is just tangent to the parabola (touches at one point). The slope of the position graph for the bus is the velocity graph, and we seek the time when its slope is vY0 . vB  aBt  vY0 (4.5 m/s)  (1.0 m/s 2 )t  t  4.5 s When you catch the bus xB  xY . 1 xB (4.5 s)  9  (1.0 m/s 2 )(4.5 s) 2  19.125 m 2 Now we find xY0 . xY  xY0  vY0t 19.125 m  xY0  (4.5 m/s)(4.5 s)  xY0  1.125 m Now we can find the time when xY  0. 0 m  1.125 m  (4.5 m/s)t  t  0.25 s So you could wait ¼ s and still catch the bus by running at the same speed. In this scenario you don’t get two chances to catch the bus. Assess: The data and answers all seem reasonable. When you catch the bus it is not going the same speed you are in part (a), but it is in part (b). 2.61. Model: The cars are represented as particles. Visualize: Solve: (a) Ann and Carol start from different locations at different times and drive at different speeds. But at time t1 they have the same position. It is important in a problem such as this to express information in terms of positions (that is, coordinates) rather than distances. Each drives at a constant velocity, so using constant velocity kinematics gives xA1  xA0  vA (t1  tA0 )  vA (t1  tA0 ) xC1  xC0  vC (t1  tC0 )  xC0  vCt1 The critical piece of information is that Ann and Carol have the same position at t1, so xA1  xC1 Equating these two expressions, we can solve for the time t1 when Ann passes Carol: vA (t1  tA0 )  xC0  vCt1  (vA  vC )t1  xC0  vAtA0  t1  xC0  vAtA0 2.4 mi  (50 mph)(0.5 h)   1.96 h  2.0 h vA  vC 50 mph  36 mph © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-31 (b) Their position is x1  xA1  xC1  xC0  vCt1  7286 miles  73 miles (c) Note that Ann’s graph doesn’t start until t  0.5 hours, but her graph has a steeper slope so it intersects Carol’s graph at t  2.0 hours. 2.62. Model: Both cyclists are modeled as particles obeying the constant-acceleration kinematic equations. Visualize: Solve: Use the same kinematic equation for each cyclist with t0  0.0 s. 1 aAt 2 2 1 xB  xB0  vB0t  aBt 2 2 xA  xA0  vA0t  When they pass at time t1 we have xA  xB. 1 1 xA0  vA0t1  aAt12  xB0  vB0t1  aBt12 2 2 Move all terms to one side and solve the resulting quadratic equation using the quadratic formula. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-32 Chapter 2 1 (aB  aA )t12  (vB0  vA0 )t  ( xB0  xA0 )  0 2 1 (0.40 m/s 2  0.20)t12  ( 1.67 m/s  5.0 m/s)t  (200 m  0.0 m)  0 2  t1  89.1 s, 22.44 s Plug the positive value of t1 back into the position equation for Amir. 1 xA (22.44 s)  (0.0 m)  (5.0 m/ s)(22.44 s)  ( 0.20 m/ s 2 )(22.44 s)2  62 m. 2 This is just before Amir comes to a dead halt. Assess: The answer is shy of halfway up the hill which seems to make sense. 2.63. Model: Model the ice as a particle and use the kinematic equations for constant acceleration. Model the ―very slippery block‖ and ―smooth ramp‖ as frictionless. Set the x-axis parallel to the ramp. Visualize: Note that the distance down the ramp is x  h/sin . Also ax  g sin down a frictionless ramp. Solve: Use vf2  vi2  2ax x, where vi  0. vf2  2ax  vf  2( g sin  ) h  2 gh sin  Assess: We will later learn how to solve this problem in an easier way with energy. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-33 2.64. Model: We will use the particle model and the kinematic equations at constant-acceleration. Visualize: Solve: To find x2 , let us use the kinematic equation v22  v12  2a1( x2  x1)  (0 m/s)2  (50 m/s)2  2(10 m/s 2 )( x2  x1)  x2  x1  125 m Since the nail strip is at a distance of 150 m from the origin, we need to determine x1: x1  x0  v0 (t1  t0 )  0 m  (50 m/s)(0.60 s  0.0 s)  30 m Therefore, we can see that x2  (30  125) m  155 m. That is, he can’t stop within a distance of 150 m so he stops after the nail strip by 5.0 m. He is in jail. Assess: Bob is driving at approximately 100 mph and the stopping distance is of the correct order of magnitude. 2.65. Model: We will use the particle model with constant-acceleration kinematic equations. Visualize: Solve: The acceleration, being the same along the incline, can be found as v12  v02  2a( x1  x0 )  (4.0 m/s)2  (5.0 m/s)2  2a(3.0 m  0 m)  a  1.5 m/s 2 We can also find the total time the puck takes to come to a halt as v2  v0  a(t2  t0 )  0 m/s  (5.0 m/s)  (1.5 m/s2 ) t2  t2  3.3 s Using the above obtained values of a and t2 , we can find x2 as follows: 1 1 x2  x0  v0 (t2  t0 )  a(t2  t0 )2  0 m  (5.0 m/s)(3.3 s)  (1.5 m/s 2 )(3.3 s) 2  8.33 m 2 2 That is, the puck goes through a displacement of 8.3 m. Since the end of the ramp is 8.5 m from the starting position x0 and the puck stops 0.17 m or 17 cm before the ramp ends, you are not a winner. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-34 Chapter 2 2.66. Model: The car is a particle that moves with constant linear acceleration. Visualize: Solve: The reaction time is 1.0 s, and the motion during this time is x1  x0  v0 (t1  t0 )  0 m  (20 m/s)(1.0 s)  20 m During slowing down, 1 a1(t2  t1)2  200 m 2 1  20 m  (20 m/s)(15 s  1.0 s)  a1(15 s  1.0 s) 2  a1  1.02 m/s 2 2 x2  x1  v1(t2  t1)  The final speed v2 can now be obtained as v2  v1  a1(t2  t1)  (20 m/s)  (1.02 m/s2 )(15 s  1 s)  5.7 m/s 2.67. Model: The ball is a particle that exhibits freely falling motion according to the constant-acceleration kinematic equations. Visualize: Solve: Using the known values, we have v12  v02  2a0 ( y1  y0 )  (10 m/s)2  v02  2(9.8 m/s 2 )(5.0 m  0 m)  v0  14 m/s © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-35 2.68. Model: Both cars are particles that move according to the constant-acceleration kinematic equations. Visualize: Solve: (a) David’s and Tina’s motions are given by the following equations: 1 aD (tD1  tD0 ) 2  vD0tD1 2 1 1 2 xT1  xT0  vT0 (tT1  tT0 )  aT (tT1  tT0 )2  0 m  0 m  aTtT1 2 2 xD1  xD0  vD0 (tD1  tD0 )  When Tina passes David the distances are equal and tD1  tT1, so we get xD1  xT1  vD0tD1  1 1 2v 2(30 m/s) 2 aTtT1  vD0  aTtT1  tT1  D0   30 s 2 2 aT 2.0 m/s2 Using Tina’s position equation, xT1  1 1 2 aTtT1  (2.0 m/s 2 )(30 s) 2  900 m 2 2 (b) Tina’s speed vT1 can be obtained from vT1  vT0  aT (tT1  tT0 )  (0 m/s)  (2.0 m/s2 )(30 s  0 s)  60 m/s Assess: This is a high speed for Tina (~134 mph) and so is David’s velocity (~67 mph). Thus the large distance for Tina to catch up with David (~0.6 miles) is reasonable. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-36 Chapter 2 2.69. Model: We will represent the dog and the cat in the particle model. Visualize: Solve: We will first calculate the time tC1 the cat takes to reach the window. The dog has exactly the same time to reach the cat (or the window). Let us therefore first calculate tC1 as follows: xC1  xC0  vC0 (tC1  tC0 )  1 aC (tC1  tC0 )2 2 1 2  3.0 m  1.5 m  0 m  (0.85 m/s 2 )tC1  tC1  1.879 s 2 In the time tD1  1879 s, the dog’s position can be found as follows: 1 aD (tD1  tD0 ) 2 2 1  0 m  (1.50 m/s)(1.879 s)  ( 0.10 m/s 2 )(1.879 s)2  2.64 m 2 xD1  xD0  vD0 (tD1  tD0 )  That is, the dog is shy of reaching the cat by 3.0 m – 2.64 m = 0.36 m. The cat is safe. 2.70. Model: Assume the water drops are particles in free fall and use the constant-acceleration kinematic equations. Visualize: Each drop falls from rest so the initial speed is zero. We will also put the origin of the coordinate system at the level of the roof, so the values of the position will be negative. The acceleration for each drop will be a   g. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension For each drop y  2-37 1 ( g )(t )2 . 2 In the figure the labels show where the different drops are at the instant the first drop hits the ground and the fifth drop starts to fall. y1  h Solve: First we want to calculate the time between drops t. It is important to note that drop 1 was where drop 2 now is one t before. So the first drop has fallen for 4t , the second drop has fallen for 3t , and the third drop has fallen for 2t. 1 y3  y2  ( g ) (2t )2  (3t )2   1.0 m   2 2(1.0 m) 4(t ) 2  9(t ) 2  (9.8 m/s 2 ) (5)(t )2  2(1.0 m) (9.8 m/s ) 2  t  2(1.0 m) (5)(9.8 m/s 2 )  0.202 s Now plug t into the y equation for the first drop. 1 y1  (9.8 m/s 2 )(4(0.202 s)) 2  3.2 m 2 So the height of the building is h   y1  3.2 m. Assess: This is a low roof, so this isn’t a multi-story building. But checking our work gives y3  1.8 m, and y2  0.8 m, which is the proper spacing. 2.71. Model: Treat the car and train in the particle model and use the constant-acceleration kinematic equations. Visualize: © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-38 Chapter 2 Solve: In the particle model the car and train have no physical size, so the car has to reach the crossing at an infinitesimally sooner time than the train. Crossing at the same time corresponds to the minimum a1 necessary to avoid a collision. So the problem is to find a1 such that x2  45 m when y2  60 m. The time it takes the train to reach the intersection can be found by considering its known constant velocity. v0 y  v2 y  30 m/s  y2  y0 60 m   t2  2  0 s t2  t0 t2 Now find the distance traveled by the car during the reaction time of the driver. x1  x0  v0 x (t1  t0 )  0  (20 m/s)(0.50 s)  10 m The kinematic equation for the final position at the intersection can be solved for the minimum acceleration a1 1 a1(t2  t1) 2 2 1  10 m  (20 m/s)(1.5 s)  a1(1.5 s) 2 2 x2  45 m  x1  v1x (t2  t1)   a1  4.4 m/s 2 Assess: The acceleration of 4.4 m/s 2 = 2.0 miles/h/s is reasonable for an automobile to achieve. However, you should not try this yourself! Always pay attention when you drive! Train crossings are dangerous locations, and many people lose their lives at one each year. 2.72. Model: Model the ball as a particle. Since the ball is heavy we ignore air resistance. 1 Visualize: We use the kinematic equation y  v0t  a(t )2 , but we set the origin at the ground so y0  h and 2 y1  0; this means y  y1  y0  h. We release the ball from rest so at t0  0 we have v0  0 and t  t. We also note that a   g where g is the free-fall acceleration on Planet X. Making all these substitutions leaves 1  h   g t2 2  So we expect a graph of h vs. t 2 to produce a straight line whose slope is g /2 and whose intercept is zero. Compare to y  mx  b where y  h, m  g /2, x  t 2 , and b  0. Solve: First look at a graph of height vs. fall time and notice that it is not linear. It would be difficult to analyze. Even though the point (0,0) is not a measured data point, it is valid to add to the data table and graph because it would take zero time to fall zero distance. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-39 However, the theory has guided us to expect that a graph of height vs. fall time squared would be linear and the slope would be g /2. First we use a spreadsheet to square the fall times and then graph the height vs. fall time squared to see if it looks linear and that the intercept is close to zero. It looks linear and R2  0996 tells us the linear fit is very good. We also see that the intercept is a very small negative number that is close to zero, so we have confidence in our model. The fit is not perfect and the intercept is not exactly zero probably because of uncertainties in timing the fall. We now conclude that the slope of the best fit line m  37326 is g /2 in the proper units, so g  2  37326 m/s2  75 m/s2 on Planet X. Assess: The free-fall acceleration on Planet X is a little bit smaller than on earth, but is reasonable. It is customary to put the independent variable on the horizontal axis and the dependent variable along the vertical axis. Had we done so here we would have graphed t 2 vs. h and the slope would have been 2/g. Our answer to the question would be the same. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-40 Chapter 2 2.73. Model: Model the ball as a particle. Ignore air resistance. Visualize: Solve: We can apply the kinematic equation vf2  vi2  2ay twice, once for the launching phase and again for the free fall phase. In the launching phase the acceleration is up (positive), v0  0 and y  y1  y0  d . v12  2a0d In the free fall phase the acceleration is a1   g , v2  0, and y  y2  y1  h. v12  2a1h  2( g )h Cancel the negative signs and set the two expressions for v12 equal to each other. 2a0d  2 gh Solve for a0 . a0  h g d Assess: The answer is independent of the mass of the ball. The units check out. 2.74. Solve: (a) With vx  2P 1/2 t , we have m ax  dvx 2P 1 1/2 P   t  dt m 2 2mt © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension (b) The quantity 2-41 2 P 2(3.6 104 W)   60 m2 /s3. Thus vx  (60 m2 /s3 )t m 1200 kg At t  2 s, vx  (60 m2 /s3 )(2 s)  11 m/s ( 25 mph). At t  10 s, vx  (60 m2 /s3 )(10 s)  24 m/s (  50 mph). (c) At t  2 s, ax   P (3.6 104 W)   2.7 m/s 2. Similarly, at t  10 s, ax  1.2 m/s2. 2mt 2(1200 kg)(2 s) 2.75. Model: The masses are particles. Visualize: Solve: The rigid rod forms the hypotenuse of a right triangle, which defines a relationship between x2 and y1: x22  y12  L2 . Taking the time derivative of both sides yields 2 x2 We can now use v2x  dx2 dy  2 y1 1  0 dt dt dx2 dy and v1y  1 to write x2v2 x  y1v1y  0. dt dt y  y Thus v2 x    1  v1 y . But from the figure, 1  tan   v2 x  v1y tan  . x x 2  2 Assess: As x2 decreases (v2 x  0), y1 increases (v1y  0), and vice versa. 2.76. Solve: A comparison of the given equation with the constant-acceleration kinematic equation 1 x1  x0  v0 (t1  t0 )  ax (t1  t0 )2 2 yields the following information: x0  0 m, x1  64 m, t0  0,t 1  4 s, and v0  32 m/s (a) After landing on the deck of a ship at sea with a velocity of 32 m/s, a fighter plane is observed to come to a complete stop in 4.0 seconds over a distance of 64 m. Find the plane’s deceleration. (b) © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-42 Chapter 2 1 (c) 64 m  0 m  (32 m/s)(4 s  0 s)  ax (4 s  0 s) 2 2 64 m  128 m  (8 s2 )ax  ax  8 m/s2 The deceleration is the absolute value of the acceleration, or 8 m/s2 . 2.77. Solve: (a) A comparison of this equation with the constant-acceleration kinematic equation (v1y )2  v02y  2(a y )( y1  y0 ) yields the following information: y0  0 m, y1  10 m, a y  98 m/s 2 , and v1y  10 m/s It is clearly a problem of free fall. On a romantic Valentine’s Day, John decided to surprise his girlfriend, Judy, in a special way. As he reached her apartment building, he found her sitting in the balcony of her second floor apartment 10 m above the first floor. John quietly armed his spring-loaded gun with a rose, and launched it straight up to catch her attention. Judy noticed that the flower flew past her at a speed of 10 m/s. Judy is refusing to kiss John until he tells her the initial speed of the rose as it was released by the spring-loaded gun. Can you help John on this Valentine’s Day? (b) (c) (10 m/s)2  v02y  2(9.8 m/s2 )(10 m  0 m)  v0 y  17.2 m/s Assess: The initial velocity of 17.2 m/s, compared to a velocity of 10 m/s at a height of 10 m, is very reasonable. 2.78. Solve: A comparison with the constant-acceleration kinematic equation (v1x )2  (v0x )2  2ax ( x1  x0 ) yields the following quantities: x0  0 m, v0 x  5 m/s, v1x  0 m/s, and ax  (9.8 m/s2 )sin10. (a) A wagon at the bottom of a frictionless 10° incline is moving up at 5 m/s. How far up the incline does it move before reversing direction and then rolling back down? © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-43 (b) (c) (0 m/s)2  (5 m/s) 2  2(9.8 m/s 2 )sin10 ( x1  0 m)  25(m/s)2  2(9.8 m/s 2 )(0.174) x1  x1  7.3 m 2.79. Solve: (a) From the first equation, the particle starts from rest and accelerates for 5 s. The second equation gives a position consistent with the first equation. The third equation gives a subsequent position following the second equation with zero acceleration. A rocket sled accelerates from rest at 20 m/s 2 for 5 s and then coasts at constant speed for an additional 5 s. Draw a graph showing the velocity of the sled as a function of time up to t = 10 s. Also, how far does the sled move in 10 s? (b) 1 (c) x1  (20 m/s2 )(5 s)2  250 m 2 v1  20 m/s 2 (5 s)  100 m/s x2  250 m  (100 m/s)(5 s)  750 m © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-44 Chapter 2 Challenge Problems 2.80. Model: The rocket and the bolt will be represented as particles to investigate their motion. Visualize: The initial velocity of the bolt as it falls off the side of the rocket is the same as that of the rocket, that is, vB0  vR1 and it is positive since the rocket is moving upward. The bolt continues to move upward with a deceleration equal to g  9.8 m/s2 before it comes to rest and begins its downward journey. Solve: To find aR we look first at the motion of the rocket: yR1  yR0  vR0 (tR1  tR0 )  1 aR (tR1  tR0 ) 2 2 1 aR (4.0 s  0 s) 2  8aR 2 To find aR we must determine the magnitude of yR1 or yB0  Let us now look at the bolt’s motion:  0 m  0 m/s  1 yB1  yB0  vB0 (tB1  tB0 )  aB (tB1  tB0 )2 2 1 0  yR1  vR1(6.0 s  0 s)  (9.8 m/s 2 )(6.0 s  0 s) 2 2  yR1  176.4 m  (6.0 s) vR1 Since vR1  vR0  aR (tR1  tR0 )  0 m/s  4 aR  4 aR the above equation for yR1 yields yR1  176.4  6.0(4aR ). We know from the first part of the solution that yR1  8aR . Therefore, 8aR  176.4  24.0aR and hence aR  5.5 m/s2. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-45 2.81. Model: Use the particle model. Visualize: Solve: (a) Substituting into the constant-acceleration kinematic equation 1 10   x2  x1  v1 (t2  t1)  a1(t2  t1) 2  100 m  x1  v1  t2    0 m 2 3   100  x1 10 t2   v1 3 Let us now find v1 and x1 as follows:  10  v1  v0  a0 (t1  t0 )  0 m/s  (3.6 m/s 2 )  s  0 s   12 m/s 3   2 x1  x0  v0 (t1  t0 )  1 1  10  a0 (t1  t0 )2  0 m  0 m  (3.6 m/s 2 )  s  0 s   20 m 2 2  3  The expression for t2 can now be solved as t2  100 m  20 m 10 s   10 s 12 m/s 3 (b) The top speed = 12 m/s which means v1  12 m/s. To find the acceleration so that the sprinter can run the 100-meter dash in 9.9 s, we use v1  v0  a0 (t1  t0 )  12 m/s  0 m/s  a0t1  t1  12 m/s a0 1 1 1 x1  x0  v0 (t1  t0 )  a0 (t1  t0 )2  0 m  0 m  a0t12  a0t12 2 2 2 Since x2  x1  v1(t2  t1)  12 a1(t2  t1)2 , we get 1 2 a0t1  (12 m/s)(9.9 s  t1)  0 m 2 Substituting the above equation for t1 in this equation, 100 m  2  12 m/s   1   12 m/s  2 100 m    a0    (12 m/s)  9.9 s    a0  3.8 m/s a  2   a0  0   (c) We see from parts (a) and (b) that the acceleration has to be increased from 3.6 m/s 2 to 3.8 m/s 2 for the sprint time to be reduced from 10 s to 9.9 s, that is, by 1%. This decrease of time by 1% corresponds to an increase of acceleration by © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-46 Chapter 2 3 8  3 6  100,  56, 3 6 2.82. Solve: (a) The acceleration is the time derivative of the velocity. ax  dvx d  [a(1  ebt )]  abebt dt dt With a  11.81 m/s and b  0.6887 s1, ax  8.134e0.6887t m/s2. At the times t  0 s, 2 s, and 4 s, this has the values 8.134 m/s2 , 2.052 m/s2 , and 0.5175 m/s2 . dx dx , the position x is the integral of the velocity. With vx   a  aebt and the initial condition that dt dt xi  0 m at ti  0 s, (b) Since vx  x t t o dx  o dt  o ae 2 bt dt Thus t t a a a x  at  e2 bt  at  e2 bt  b b o b o This can be written a little more neatly as x a (bt  ebt  1) b  17.15(0.6887t  e06887t  1) m (c) By trial and error, t = 9.92 s yields x = 100.0 m. Assess: Lewis’s actual time was 9.93 s. 2.83. Model: We will use the particle-model to represent the sprinter and the equations of kinematics. Visualize: Solve: Substituting into the constant-acceleration kinematic equations, 1 1 1 1 x1  x0  v0 (t1  t0 )  a0 (t1  t0 ) 2  0 m  0 m  a0 (4 s  0 s) 2  a0t12  a0 (4.0 s) 2 2 2 2 2  x1  (8 s 2 )a0 v1  v0  a0 (t1  t0 )  0 m/s  a0 (4.0 s  0 s)  v1  (4.0 s) a0 From these two results, we find that x1  (2 s)v1. Now, 1 x2  x1  v1(t2  t1)  a1(t2  t1)2 2  100 m  (2 s)v1  v1(10 s  4 s)  0 m  v1  12.5 m/s © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-47 Assess: Using the conversion 2.24 mph = 1 m/s, v1  12.5 m/s  28 mph. This speed as the sprinter reaches the finish line is physically reasonable. 2.84. Model: The balls are particles undergoing constant acceleration. Visualize: Solve: (a) The positions of each of the balls at t1 is found from kinematics. 1 2 1 gt1  v0t1  gt12 2 2 1 2 1 2 ( y1)B  ( y0 )B  (v0 y )B t1  gt1  h  gt1 2 2 In the particle model the balls have no physical extent, so they meet when ( y1)A  ( y1)B. This means ( y1)A  ( y0 )A  (v0 y )A t1  v0t1  Thus the collision height is ycoll  h  1 2 1 h gt1  h  gt12  t1  2 2 v0 1 2 gh2 gt1  h  . 2 2v02 (b) We need the collision to occur while ycoll  0. Thus h So hmax  gh2 gh 2v0 2v0  0  1 2  h 2 2v02 g 2v02 . g (c) Ball A is at its highest point when its velocity (v1y )A  0. (v1y )A  (v0 y )A  gt1  0  v0  gt1  t1  In (a) we found that the collision occurs at t1  v0 g h h v0 v2 . Equating these,  h 0 . v0 v0 g g Assess: Interestingly, the height at which a collision occurs while Ball A is at its highest point is exactly half of hmax . © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-48 Chapter 2 2.85. Model: The space ships are represented as particles. Visualize: Solve: The difficulty with this problem is how to describe ―barely avoid.‖ The Klingon ship is moving with constant speed, so its position-versus-time graph is a straight line from xK0  100 km. The Enterprise will be decelerating, so its graph is a parabola with decreasing slope. The Enterprise doesn’t have to stop; it merely has to slow quickly enough to match the Klingon ship speed at the point where it has caught up with the Klingon ship. (You do the same thing in your car when you are coming up on a slower car; you decelerate to match its speed just as you come up on its rear bumper.) Thus the parabola of the Enterprise will be tangent to the straight line of the Klingon ship, showing that the two ships have the same speed (same slopes) when they are at the same position. Mathematically, we can say that at time t1 the two ships will have the same position ( xE1  xK1) and the same velocity (vE1  vK1). Note that we are using the particle model, so the ships have zero length. At time t1, vK1  vK0 xK1  xK0  vK0t1 1 xE1  vE0t1  at12 2 vE1  vE0  at1 Equating positions and velocities at t1: 1 xK0  vK0t1  vE0t1  at12 vK0  vE0  at1 2 We have two simultaneous equations in the two unknowns a and t1. From the velocity equation, t1  (vK0  vE0 )/a © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Kinematics in One Dimension 2-49 Substituting into the position equation gives 2 xK0  (vK0  vE0 )   a 2 (vK0  vE0 ) 1  (vK0  vE0 )  (vK0  vE0 ) 2  a  2  a 2  a 2a  (vK0  vE0 )2 (20,000 m/s  50,000 m/s) 2   4500 m/s 2 2 xK0 2(100,000 m) The magnitude of the acceleration is 4500 m/s2. Assess: The deceleration is 4500 m/s2 , which is a rather extreme  460 g. Fortunately, the Enterprise has other methods to keep the crew from being killed. © Copyright 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.