Solutions manual fishbane physics for scientists and engineers 3rd

CHAPTER Straight-Line Motion Answers to Understanding the Concepts Questions You should be worried about something that might happen to bring the car in front of you to a stop Your stopping time depends on two factors: Your fixed reaction time, and the time required for your brakes to bring your car to a stop At a higher initial speed, you travel farther during the time it takes you to react and apply the brakes, and you travel farther in the time it takes your brakes to bring you to a stop Both factors, then, argue for increasing spacing with increasing speed The velocity of the chalk is zero at that point, but the acceleration remains g, downward In fact if the acceleration were zero as well then the chalk would maintain zero velocity there –– i.e., it would “freeze” at the top of its path! We have seen that for a fixed acceleration gx, the relation between fall distance d and fall time t is, 1/2 assuming the falling object starts from rest, d = !gxt Thus for fixed d, t = (2d/gx) The variation from – l/2 planet to planet for t, that is, with gx, is then (gx) The larger gx, the smaller the fall time The speed 1/2 1/2 of the object at the end of the fall is v = gxt = (2dgx) The speed increases with gx like (gx) Both you and the bowling balls would be falling at the same rate (g), so there is no reason to worry that any of them would crash onto you The acceleration of a falling object is equal to g only in a true free fall, which is devoid of any air resistance In reality, as the object falls, it encounters an air resistance which increases with its speed Initially, the object is not moving very fast so the air resistance exerted on it is not yet significant, and it falls with an acceleration close to g As it speeds up, however, it picks up more and more air resistance so its acceleration gradually diminishes, until it reaches a certain speed, at which the upward air resistance equals the downward gravitational pull, whereupon its acceleration is zero and the its speed can no longer increase This speed is therefore referred to as the terminal speed So no, the speed of a falling object cannot increase indefinitely Certainly if the (negative) acceleration has a constant magnitude, the velocity cannot remain positive Indeed, if the initial velocity has magnitude v0, and the acceleration has the constant magnitude a, then the velocity varies with time according to v = v0 – at, and v = at a time t = v0/a; for times greater than this the velocity is negative However, the acceleration could steadily decrease in magnitude, while remaining negative, such that the velocity could remain positive A physical example occurs when a rocket is sent away from Earth with enough initial speed to leave the Solar System we say that its initial speed exceeds the “escape speed.” If we say that “up” is the positive direction, then the acceleration is negative while the velocity is positive As the object moves away, the force of gravity on it, and hence its acceleration, decreases in magnitude while remaining negative For a fast enough start, the object never comes to rest or turns around Incidentally, the escape speed from Earth is about 11.2 km/s Treat the jump of the astronaut as a projectile motion Then the height he can reach is h = v0 /2g, which is inversely proportional to g Since the astronaut can jump 1.2 m on the surface of Earth, assuming that his initial jumping speed does not change, then he would be able to jump as much as (0.8 m)[(9.8 m/s )/(1.6 m/s ) ] = m on the surface of the Moon Note that we neglected the height of the Download Full Version: Solution manual Fishbane Physics for Scientists and Engineers 3rd in https://getbooksolutions.com/download/solutions-manual-fishbane-physics-for-scientists-andengineers-3rd © 2005 Pearson Education, Inc., Upper Saddle River, NJ 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 Page 2-1 Chapter 2: Straight-Line Motion astronaut To get a more precise result, we need to find how much the center of mass of the astronaut 2 can rise on the surface of Earth, and multiply that number by [(9.8 m/s )/(1.6 m/s ) ] = 6.1 to obtain the corresponding value on the Moon Common sense tells us that a sudden decrease in speed, i.e., a large deceleration, can cause damage to our body The airbag prolongs the deceleration process as the object’s speed decreases to zero in a collision So the magnitude of the deceleration of the object is reduced, lowering the chance of injury or damage True This description is consistent with the case when the object is undergoing a uniformly accelerated motion with a pointing to the left, i.e., in the negative x-direction (so a < 0) The velocity of the object as a function of time is v = v0 + at Since the object starts out moving to the right v0 > But since a v0/t0 A t0 O Velocity A – (v0/t0) v0 O Time Time 2 28 For constant acceleration we have x = x0 + v0t + !at and v = v0 + 2a(x – x0) From the v-equation: (128 mi/h) = + 2a[# mi/ – 0], which gives a = 3.28 ⋅ 10 mi/h From 2 the x-equation: # mi = + + ! (3.28 ⋅ 10 mi/h )t , which gives t = 0.00391 h = 14.1 s 29 We convert the speeds to ft/s: (25 mi/h)(5.28 ⋅ 10 ft/mi)/(3600 s/h) = 36.7 ft/s; 50 mi/h = 73.3 ft/s For constant2 acceleration: 2 v =v + 2a(x – x ); (73.3 ft/s) = (36.7 ft/s) + 2a(1000 ft – 0), which gives a = 2.02 ft/s 2 (0.61 m/s ) © 2005 Pearson Education, Inc., Upper Saddle River, NJ 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 Page 2-10 Fishbane, Gasiorowicz, and Thornton 30 For constant acceleration the average speed is ! (v0 + v); thus x = vavt = ! (v0+ v)t: 6000 ft = ! [0 + (212 mi/h)(5280 ft/mi)(1 h/3600 s)]t, which gives t = 38.6 s 31 For constant acceleration, the velocity is given by v = v0 + at: v1 = + (0.50 m/s )(1.0 s) = 0.50 m/s; v2 = + (0.50 m/s )(2.0 s) = 1.0 m/s For constant acceleration the average velocity is !(v1+ v2): v av = !(0.50 m/s + 1.0 m/s) = 0.75 m/s 32 For constant acceleration the average speed is !(v0 + v); thus x = vavt = !(v0+ v)t: x = !(0 + 4.2 ⋅ 10 mi/h)(125 s)(1 h/3600 s), which gives x = 73 mi 33 (a) For constant acceleration: v = v0 + at = 8.0 m/s + (– 0.50 m/s )t = 8.0 – 0.50t, easterly with t in s, v in m/s 2 (b) For the position: x – x0 = v0t + !at = (8.0 m/s)(5.0 s) + !(– 0.50 m/s )(5.0) = 34 m to the east 34 (a) For constant acceleration: v = v0 + at = 10 m/s + (– 0.2 m/s )t Plug in t = s and s to obtain v = 9.8 m/s and 9.6 m/s (b) At t = s, v = 9.6 m/s So vav = ! (v0 + v) = ! (10 m/s + 9.6 m/s) = 9.8 m/s 35 The initial speed of the weight just before it touches the surface after falling through a distance h in the 1/2 air is v0 = (2gh) , where h = m The weight then falls through a distance ∆x in the mud, undergoing 2 an acceleration a, reaching final speed v = From v = v0 + 2a∆x = 2gh + 2a∆x = 4 a = – gh/∆x = – (9.8 m/s )(6 m)/(0.004 m) = – ⋅ 10 m/s , so its magnitude is ⋅ 10 m/s The average speed of the weight as it decelerates in the mud is vav = !(v0 + v) = !(2gh) 1/2 = ![(2(9.8 m/s )(6 m)] t = ∆x/ vav = 0.004 m/(5.42 m/s) = ⋅ 10 –4 1/2 = 5.42 m/s, so s 36 (a) x (m) v (m/s) 80 400 32 t (s) (b) We find the time the airplane rolls from x1 = v0t1; 400 m = (80 m/s)t1, which gives t1 = s We find the time until the airplane stops from v = v0 + a(t2 – t1); 32 t (s) = (80 m/s) + (– 3.0 m/s )(t2 – s), so t2 = 32 s For the position, we have x2 = x1 + v0(t2 – t1)+ !a(t2 – t1) 2 = (400 m) + (80 m/s)(27 s) + !(– 3.0 m/s )(27 s) = 1.5 km © 2005 Pearson Education, Inc., Upper Saddle River, NJ 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 Page 2-11 37 (a) (b) We find the distance before the brakes are applied from x1 = v0t1 = (20 m/s)(1.2 s) = 24 m We find the acceleration while the brakes are applied from v = v0 + 2a(x2 – x1); Distance (m) Chapter 2: Straight-Line Motion 80 40 0 Time (s) Speed (m/s) = (20 m/s) + 2a(80 m – 24 m), which gives a = – 3.6 m/s 2 20 10 0 Time (s) 38 Leg I : ∆x1 = v0t + !a1t = + !(3.0 m/s )(4 s) = 24 m; v1 = v0 + at = + (3.0 m/s )(4 s) = 12 m/s Leg II: ∆x2 = v1t + !a2t = (12 m/s)(7 s) + = 84 m 2 Leg III: ∆x = v t + !a t = (12 m/s)(15 s) + !(1.0 m/s )(15 s) = 293 m; v3 = v + a t = 12 m/s + (1.0 m/s )(15 s) = 27 m/s 2 Leg IV: v4 =v3 + 2a(∆x4); = (27 m/s) + 2(– 2.5 m/s ) ∆x4 , which gives ∆x4 = 146 m Total displacement: x = 24 m + 84 m + 293 m + 146 m = 547 m 39 We will choose a coordinate system with the origin at the release point and take down as the positive direction, so for the penny: x0 = 0; v0 = + m/s; a = + 9.8 m/s 2 For the constant acceleration: x – x = v t + !at ; 20 m – = (1 m/s)t + !(9.8 m/s )t 0 Solving the quadratic gives t = 1.92 s, – 2.12 s Because the penny starts at t = 0, the positive answer is the physically possible answer: t = 1.92 s 40 (a) ∆x1 = v0t + !at ; 2 10 m – = + ! (2.8 m/s )t ; the positive answer is t = 2.7 s (b) ∆x2 = v0t + !at ; 2 50 m – = + !(2.8 m/s )t ; the positive answer is t = 6.0 s The velocity at this time is v = v0 + at = + (2.8 m/s )(6.0 s) = 17 m/s (c) Because the initial velocity for the second 50 m is the final velocity from the first 50 m, ∆x3 = v0t + !at 2 100 m – 50 m = (17 m/s)t + !(2.8 m/s )t ; the positive answer is t = 2.5 s (d) ∆x4 = v0t + !at ; 2 100 m – = + !(2.8 m/s )t ; the positive answer is t = 8.5 s (e) The early times are longer, but the later times are shorter, because runners cannot maintain a high constant acceleration 41 From the symmetry of the motion, the initial acceleration lasted for one-half the total time and one-half the total distance traveled 2 (a) Thus v2 = v + 2a(x – x ); (144 m/s) = + 2a(3500 m – 0), which gives a = 2.96 m/s 0 (b) To find the time we use v = v0 + at; 144 m/s = + (2.96 m/s )t, which gives t = 48.6 s and a total time = 2t = 97.2 s © 2005 Pearson Education, Inc., Upper Saddle River, NJ 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 Page 2-12 Fishbane, Gasiorowicz, and Thornton 42 We use a coordinate system with the origin at the corner and t = when you turn the corner The bus position is given by xb = x0 + v0t + !at = 30 m + + ! (0.6 2 m/s )t and its speed is vb = v0 + at = + (0.6 m/s )t Your speed is vp and your position is given by xp = x0 + vpt = + vpt For you to catch the bus, your position must be the same as the bus’s position For you to catch the bus with a minimum speed, your speed must be the same as the bus’s speed when you catch it This can be seen from the x-t graph; the slope is the speed Position v > minimum bus v = minimum O Time Thus, xb = xp , or 2 30 m + !(0.6 m/s )t = vpt and vb = vp , or (0.6 m/s )t = vp These equations have two unknowns: t and vp Solving, we get t = (a) We choose the origin at the initial passing point and change units of the speeds: 75 mi/h = (75 mi/h)(5280 ft/mi)(1 h/3600 s) = 110 ft/s; 85 mi/h = 125 ft/s The speeder’s position is xs = x0 + vst = + (110 ft/s)t For the police car, the distance traveled while accelerating is xp1 = vavt1 = !(125 ft/s + 0)(13 s) = 813 ft Thereafter its position is xp = xp1 + vp(t – t1) = 813 ft + (125 ft/s)(t – 13 s) The police car will overtake the speeder when xs = xp; (110 ft/s)t = 813 ft + (125 ft/s)(t – 13 s) 10 Distance (1000 ft) 43 10 s and vp = 6.0 m/s 0 20 40 60 80 Time (s) The solution gives (c) t = 54 s and (b) xs = xp = 5940 ft (1.13 mi) 44 With the origin at the point where the brakes are applied, v0 = 35 mi/h = 51 ft/s and a = – 3.0 m/s = 9.8 ft/s We can find the distance she travels before stopping from 2 v =v + 2a(x – x ); = (51 ft/s) + 2(– 9.8)(x – 0), which gives x = 133 ft Because this is greater than 90 ft, she does not stop before arriving at the light 45 We use v = v0 + at; = 58.7 ft/s + (– 0.5g)t, which gives t = 3.65 s © 2005 Pearson Education, Inc., Upper Saddle River, NJ 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 Page 2-13 Chapter 2: Straight-Line Motion 46 (a) For the initial acceleration: Car A: aA1 = 5.0 m/s 2 To find the speed: vA1 = v0 + 2aA1(x1 – x0); 2 vA1 = + 2(5.0 m/s )(200 m – 0), which gives vA1 = 44.7 m/s To find the time: v A1 = v + a A1 t A1 ; 44.7 m/s = + (5.0 m/s )t A1 , which gives t Car B: a B1 = 4.5 m/s 2 = v + 2a (x – x ); To find the speed: v B1 B1 A1 = 8.94 s vB1 = + 2(4.5 m/s )(200 m – 0), which gives vB1 = 42.4 m/s To find the time: v B1 = v + a B1 t B1 ; 42.4 m/s = + (4.5 m/s )t B1 , which gives t B1 = 9.43 s (b) For the second acceleration: Car A: aA2 = 2.5 m/s 2 To find the speed: vA2 = vA1 + 2aA2(x2 – x1); 2 vA2 = (44.7 m/s) + 2(2.5 m/s )(400 m – 200 m), which gives vA2 = 54.8 m/s To find the time: vA2 = vA1 + aA2(tA2 – tA1); 54.8 m/s = 44.7 m/s + (2.5 m/s )(tA2 – 8.94 s), which gives tA2 = 13.0 s Car B: aB2 = 3.0 m/s 2 To find the speed: vB2 = vB1 + 2aB2(x2 – x1); 2 vB2 = (42.4 m/s) + 2(3.0 m/s )(400 m – 200 m), which gives vB2 = 54.8 m/s To find the time: vB2 = vB1 + aB2(tB2 – tB1); 54.8 m/s = 42.4 m/s + (3.0 m/s )(tB2 – 9.43 s), which gives tB2 = 13.6 s 47 (a) For uniform acceleration: ∆x = vav∆t; ⋅ 10 –2 m = !(5 ⋅ 10 m/s + ⋅ 10 m/s) ∆t, which gives ∆t = 8.0 ⋅ 10 2 –9 s (b) To find the acceleration: v = v0 + 2a(x – x0); –2 14 (5 ⋅ 10 m/s) = (3 ⋅ 10 m/s) + 2a(2 ⋅ 10 m), which gives a = 6.3 ⋅ 10 m/s 48 To find the acceleration: v = v0 + at; -4 (50 mi/h)(1 h/3600 s) = + a(18 s), which gives a = 7.7 ⋅ 10 mi/s -4 -2 After 36 s: v = v0 + at = + (6.94 ⋅ 10 mi/s )(36 s) = 2.8 ⋅ 10 mi/s = 100 mi/h = 45 m/s During the first 20 s: x – x0 = v0t + !at = + !(7.7 ⋅ 10 km During min: x – x0 = + !(7.7 ⋅ 10 -4 2 -4 2 mi/s )(20 s) = 0.15 mi = 0.25 mi/s )(36 s) = 0.50 mi = 0.80 km 49 We use a coordinate system with the origin at the ground and up positive At the instant the object is dropped: 2 y elev = y0elev + v0elevt1 + !at1 = + + !at1 = !a(3 s) = (4.5 s )a; velev = v0elev + at1 = + at1 = a(3 s); Thus for the object’s motion: y0 = (4.5 s )a, v0 = (3 s)a; ao = – 9.8 2 m/s : y – y0 = v0t2 + !a0 t2 ; 2 – (4.5 s )a = (3 s)a(3.5 s) + !(– 9.8 m/s )(3.5 s) , which gives a = 4.0 2 2 m/s y0 = (4.5 s )a = (4.5 s )(4.0 m/s ) = 18 m © 2005 Pearson Education, Inc., Upper Saddle River, NJ 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 Page 2-14 Fishbane, Gasiorowicz, and Thornton 50 We use a coordinate system with the origin at the initial position of B and up positive The acceleration of A is negative, with the magnitude of the acceleration of B, which is positive For the motion of the two objects, we have y = y0 + v0t + !at1 ; 2 y A = (1.2 m) + + !(– 0.3 m/s )t , 2 y B = + + !(+ 0.3 m/s )t The two objects bump into each other when yA = yB: 2 2 1.2m + !(– 0.3 m/s )t = !(+ 0.3 m/s )t , which gives t = s 51 For uniform acceleration: x = vavt = !(0 + 60 mi/h)(4 s)(1 h/3600 s)(1.6 ⋅ 10 m/mi) = 54 m 52 We use a coordinate system with the origin where she slips and down positive During the 2 free fall: v1 = v0 + 2a(x1 – x0) = + 2(9.8 m/s )(8 m – 0), or v1 = 12.5 m/s 2 While the rope is stretching: v = v + 2a (x – x ); 2 = (12.5 m/s) + 2(– 5)(9.8 m/s )(x 2 53 We find the acceleration from v = v = (600 m/s) + 2a(20 ⋅ 10 We find the time from v = v0 + at; –2 2 – x ), which gives x – x = 1.6 m + 2a(x – x ); 5 m), which gives a = – 9.0 ⋅ 10 m/s ; |a| = 9.0 ⋅ 10 m/s –4 = 600 m/s – (– 9.0 ⋅ 10 m/s )t, we get t = 6.7 ⋅ 10 s 54 We use a coordinate system with the origin at the top of the tower and down positive For the motion of the object, we have y = y0 + v0t + !at1 ; 2 54.5 m = + + ! (+ 9.8 m/s )t , which gives t = 3.34 s 55 We take the free fall for 97 stories ≈ 970 ft The speed at that point is found from v 2 = v0 + 2a(x – x0) = + 2(32 ft/s)(970 ft), which gives v ≈ 250 ft/s If we take this to be the average speed for the rest of the fall, the time to fall the last stories is ∆t ≈ 30 ft/(250 ft/s) ≈ 0.1 s This is not much time to say anything! 56 We use a coordinate system with the origin at the drum and up positive For all the sinkers, v0 = We take t = when the string is released The time for the nth sinker to hit the drum is found from: 1/2 y = y + v t + !(– g)t ; = y + + !(– g)t , which gives t = (2y /g) 0n n n 0n n n With t1 = 0, equal time intervals requires tn = (n – 1)t2 , n = 3, 4, 1/2 1/2 (2y /g) = (n – 1)(2y /g) , or y = (n – 1) y Thus 0n y 03 = (2) y 02 02 = 0n In terms of the initial positions: 02 0n 4(10 cm) = 40 cm; y04 = (3) y02 = 9(10 cm) = 90 cm; y 05 = (4) y02 = 16(10 cm) = 160 cm 57 We use a coordinate system with the origin at the release point and down positive 2 From y = y + v0y t + !at ; m = + + ![(9.8 m/s )/6]t , we get t = 1.1 s 58 We use a coordinate system with the origin at the release point and down positive To find the time for the object to hit the ground, we have y = y0 + v0t + !at ; 2 10 m = + + !(+ 25.9 m/s )t , which gives a positive answer of t = 0.88 s © 2005 Pearson Education, Inc., Upper Saddle River, NJ 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 Page 2-15 Chapter 2: Straight-Line Motion 59 We use a coordinate system with the origin at the ground and up positive and label the first rock A and the second rock B To find the time for rock A to hit the ground: yA = y0A + v0AtA + !at ; 2 = 10 m + (22 m/s) tA + !(– 9.8 m/s ) tA , which gives positive answer of tA = 4.9 s To find the time for rock B to hit the ground: yB = y0B + v0BtB + !at ; 2 = 10 m + + !(– 9.8 m/s )tB , which gives positive answer of tB = 1.4 s Thus rock B must be released ∆t = tA – tB = 3.5 s after rock A 60 We use a coordinate system with the origin at the release point and up positive, label the first rock A and the second rock B and call the height of the cliff H For rock A: y = y + v 0A tA + !at 2; A A – H = + + !(– g)(4.15 s) , or H = (8.61 s )g For rock B to reach the highest point: v 2=v2 + 2(– g)(y top – y ); top 0B 1/2 = v + 2(– g)(2 m – 0), which gives v 0B = [(4 m)g] 0B Thus y = y + v t + !at ; B 0B – H = + [(4 m)g] 1/2 (6.30 s) + !(– g)(6.30 s) Combining the equations from A and B gives g = 1.27 m/s and H = 10.9 m 61 We use a coordinate system with the origin at the ground and up positive 2 For the fall: v1 = v0 + 2a(y1 – y0); 2 v1 = + 2(– 9.8 m/s )(0 – 25 m), which gives v1 = – 22 2 m/s For the rebound: v2 = v1 + 2a(y2 – y0); 2 = (22 m/s) + 2(– 9.8 m/s )(y2 – 0); which gives y2 = 25 2 m For a bounce of 20 m: v4 = v3 + 2a(y3 – y0); 2 = v3 + 2(– 9.8 m/s )(20 m – 0), which gives v3 = 20 m/s 62 We use a coordinate system with the origin at the ground and up positive From the symmetry of the up and down motion, we know that the ball takes !(3.6 s) = 1.8 s to reach the highest point from the ground and !(2.2 s) = 1.1 s to reach the highest point after it passes the window Thus, after being thrown, the ball takes 0.7 s to reach the window From y = y0 + v0yt + !at we can write: to reach the window: 10 m = + v0(0.7 s) + !(– g)(0.7 s) , and to return to the ground: = + v0(3.6 s) + !(– g)(3.6 s) Solving these two equations for the two unknowns gives us v0 = 18 m/s and g = 9.85 m/s 63 The velocity is found from v v dv = t – t dt , which gives v – v0 = – 8 3/2 t , or v = (15 m/s) – 3/2 t , with v in m/s and t in s 0 If we set v = 0, we find the time to come to a stop: t = 3.16 s The displacement is found from x x dx = t t vdt = 3/2 v0 –3 t dt, which gives x – x0 = v0t – 16 5/2 15 t , with x inm and t in s 16 The displacement when it stops is x – x0 = 15 m/s 3.16 s – 15 3.16s 5/2 = 28.4m © 2005 Pearson Education, Inc., Upper Saddle River, NJ 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 Page 2-16 Fishbane, Gasiorowicz, and Thornton 64 (a) a(t) = dv/dt = 10 – 2t, so t v= (b) t ∫0a(t)dt =∫0(10 − 2t)dt =[10t− t2 ]t0 = 10t− t2 t x(t) = t ∫0v(t)dt =∫0 (10t − t t 5t )dt = − t = 5t − t At t = s, x(t) = 5(5) – @(5) = 83 m 65 For simplicity, temporarily suppress the units in the calculation below a(t) = dv/dt = d(4.0 + 8.0 e –0.5t )/dt = 8.0(– 0.5) e –0.5t =–4e –0.5t Plug this into the expression for v(t): –0.5t –0.5 t v(t) = 4.0 + 8.0 e = 4.0 – (– e )= 4.0 – a(t), or a(t) = – !v, where t is in s, v in m/s, and a in m/s 66 The displacement is found from x t dx = 0 vdt = t v0 at v0 e dt, which gives x = At t = s, we have x = (1 m/s)/(0.5 /s)[e a e at t (0.5 /s)(2 s) v at = e –1 a – 1] = 3.4 m 67 (a) y (m) 0.1 0.1 0.2 0.4 0.6 0.8 1.2 t (s) (c) (d) The instantaneous velocity is found from the slope The line at t = 0.15 s gives v0.15 = 0.15 m/s r ˆ 0.94 m/s From v = A cos(3πt) j : – 0.147 m/s = A cos[3π(0.15 s)], which gives A = 1.0 Velocity (m/s) (b) 0.5 0.0 t (s) 0.0 –0.5 –1.0 0.2 0.4 0.6 0.8 © 2005 Pearson Education, Inc., Upper Saddle River, NJ 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 Page 2-17 Chapter 2: Straight-Line Motion 68 (a) Take up as positive and put the origin at the ground level Then the acceleration of the falling body is a(t) = dv/dt = – g(t) = – (g – hg’) To eliminate t, write v = dh/dt, or dt = dh/v, so dv dv dv dv v = = = = −g + g'h, dt dh/v dh dh dv = 2(− g0 + g'h)dh Integrate both sides: v ∫0 dv = y (m) v (m/s) 60 50 h (−g + g'h)dh, ∫ h0 v = − g0h+ g'h 2 h = 2g0 (h0 − h) + g'(h − h0 ), v 10 h0 y 40 30 20 ) − h0 v = 2g0 (h0 − h) + g'( h 2 (b) a(t) = dv/dt = d h/dt = – g(t) = – (g0 – hg’), or Time (s) 2 d h/dt – hg’ + g0 = To avoid solving this the falling distance h0 – h ≈ !g0 t Thus differential equation for h(t), note that g’ is small so dv/dt = – g0 + hg’ ≈ – g0 + h0 g’ – !g0 g’ t , which we integrate over t to obtain v(t) ≈ (– g0+ h0 g’) t – g0 g’ t /6 69 Because the acceleration is a function of time, a = αt , we obtain the speed by integrating: v f t dv = t adt = αt dt, f 0 α3 which gives v = t Thus t = 3v f 1/3 α 70 (a) The velocity is found from v t ge dv = – bt dt , which gives v=– g – bt e b t = g 1–e b – bt 9.8m/s – (0.5 /s)t = 0.5 /s – e = 19.6 m/s – e The displacement is found from y t t dy = vdt = 0 y =g t + 1e– bt b t – (0.5 /s)t – bt g – e dt, which gives b – bt – (0.5 /s)t – (0.5 /s)t = g bt – + e = 9.8 m/s (0.5 /s)t– + e = 39.2 m (0.5 /s)t – + e b (0.5 /s) b (c) We find the time to fall 50 m from – (0.5 /s)t ; or 2.276 = 0.5t + e 50 m = 39.2 m (0.5 /s)t – + e A numerical solution of this equation gives t = 4.32 s (d) When b = 0, we have 2 y = y + v t + !gt ; 50 m = + + !(9.8 m/s )t , which gives – 0.5t t = 3.19 s, less than when air resistance is present © 2005 Pearson Education, Inc., Upper Saddle River, NJ 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 Page 2-18 Fishbane, Gasiorowicz, and Thornton 71 If 20 transfers take 15 s, the time for one transfer is τ = 0.75 s To juggle two objects, one will be in the air for τ, which means !τ to reach the highest point and !τ to fall back to the hand We use a coordinate system with the origin at the highest point and down positive 2 From y = y0 + v0t + !at ; h2 = + + !g(!τ) , we get h = !g[!(0.75 s)] = 0.7 m To juggle three objects, two must be in the air during a transfer so each one must be in the air for a time of 2τ or fall for a time of τ Thus 2 h = !g(τ) = !g(0.75 s) = m 72 We use a coordinate system with the origin at the fingers with up positive The point on the ruler that will be grabbed is L from the fingers Its motion can be represented by 2 y = y0 + v0t + !at ; = L + + !(– g)t , which gives t = (2L/g) 1/2 1/2 = [2(5 in)(1 ft/12 in)/(32 ft/s )] = 0.16 s 73 The height reached is determined by the initial velocity We assume the same initial velocity of the 2 jumper on the Moon and Earth With a vertical velocity of at the highest point, from v = v0 + 2ah we get v0 = 2gEh E = 2gMh M, or h M = (gE/gM)h E = (6)(2 m) = 12 m 74 The speed attained in a fall through a height h can be found from 2 v1 = v0 + 2a(y – y0); v1 = + 2gh If we assume you stop by flexing your knees a distance 2 ∆: v2 = v1 + 2a(y – y0); = 2gh + 2a∆, which gives a = – gh/∆ = – 9.8(8 ft/∆ ft) m/s 75 We use a coordinate system with the origin at the throw and up positive Then v0 = m/s From v = v0 + at = v0 – gt; – m/s = m/s – g(7 s), we get g = 1.6 m/s To find the highest point, we have 2 v = v0 + 2a(y – y0); 2 = (7 m/s) + 2(– 1.6 m/s )h, which gives h = 15 m 76 We use a coordinate system with the origin at June’s initial position and the positive direction toward Bill 2 Then aJ = 1.0 m/s and aB = – 0.9 m/s : 2 x =x + v t + !a t = + + !(1.0 m/s )t J 0J J 2 x =x + v t + !a t = 20 m + + !(– 0.9 m/s )t B 0B B 2 When they meet: x = x , or 0.50t = 20 – 0.45t , which gives J B t = 4.6 s and xJ = 11 m 77 Let da/dt = J = constant Integrate over time: a(t) = a0 + Jt, where a0 is the initial acceleration at t = Integrate over t again: v(t) = v0 + a0t + Jt /2 Integrate over t once again: x(t) = x0 + v0t + a0t /2 + Jt /6 © 2005 Pearson Education, Inc., Upper Saddle River, NJ 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 Page 2-19 Chapter 2: Straight-Line Motion 78 v(t) = dx/dt = 4t – t, so the displacement from t1 to t2 is t t2 t2 2 4 ∆x ∫ v(t)dt =∫t = t2 − t2 − t1 (4t − t)dt = t − t − t1 t 2 t1 1 Plug in t1 = 0.5 s and t2 = 1.5 s to obtain ∆x = m The average speed is vav = ∆x/∆t = m/(1.5 s – 0.5 s) = m/s a(t) = dv/dt = d(4t – t)/dt = 12t – 1, which is negative before t1 = √(1/12) s = 0.29 s and positive thereafter; while v(t) = 4t – t, which is negative before t2 = 0.5 s and positive thereafter So between 0.5 s and 1.5 s both v and a have the same direction, indicating that the speed increases monotonically during this time interval So vmin occurs at t = 0.5 s while vmax at t = 1.5 s Plug these values of t into the expression v(t) = 4t – t to obtain vmin = and vmax = 12 m/s Note that vav = m/s is indeed within this range [Since the acceleration is not a constant here, however, we generally cannot write vav = !(vmin + vmax).] = 79 (a) Take up as the y-axis The velocity of the ball just before it hits the switch after it falls through a distance h is ˆ ˆ r 1/2 ˆ 1/2 v = (2gh) (– j )= [2(9.8 m/s )(10 m)] (– j ) = (−14 m/s )j (b) The ball rebounds to a height h’, so its the velocity as it leaves the switch is 1/2 ˆ r 1/2 ˆ ˆ v = (2gh’) j = [2(9.8 m/s )(9 m)] j = (13 m/s )j ˆ ˆ r rr ˆ (c) a av = (v f – v i)/∆t = [(13.3 m/s) j – (–14 m/s) j ]/0.002 s = (1.4⋅ 10 m/s )j i f 80 Our coordinate system has the origin at the archer, up positive and t = when the balloon is dropped For the balloon: yB = y0B + v0B t + !at = 200 m + + ! (– 9.8 2 m/s )t For the arrow: 2 yA = y0A + v0A t + !gt = + (40 m/s)(t – s) + !(– 9.8 m/s )(t – s) The arrow intercepts the balloon when yB = yA: 2 2 (200 m) – !(9.8 m/s )t = (40 m/s)(t – 5) – !(9.8 m/s )(t – s) _ 2 The time of intercept is t = 5.87 s, which means y = 200 m – !(9.8 m/s )(5.87 s) = 31 m 81 We use a coordinate system with the origin at the top of the snowbank with down positive To 2 find the distance: v = v0 + 2a(y – y0); 2 = (40 m/s) + 2(– 50)(9.8 m/s )(y – y0), which gives y – y0 = 1.6 m To find the time: v = v0 + at; = 40 m/s + (– 50)(9.8 m/s )t, which gives t = 0.08 s 82 We use a coordinate system with the origin at the ground and up positive For all the weights, v0 = and a = – g Take t = when the string is released, and the lowest weight is at the ground The time for the jth sinker to hit the drum is found from: 2 y = y0j + v0tj + !(– g)tj ; = y0j + + !(– g)tj , which gives tj = (2y0j/g) 1/2 1/2 , with j = 0, 1, 2, …, n If we call t1 = (2L0/g) the time for the second lowest weight to hit the ground, for equal time intervals we have tj – tj–1 = t1 , j = 2, 3, …, n This gives the sequence t2 = 2t1 , t3 = 3t1 , t4 = 4t1 , … In terms of the1/2 initial positions: 1/2 (2y /g) = j (2L /g) , or y = j L 0j 0j Because each y0 is the sum of the corresponding L’s, we have y02 = L0 + L1 = 4L0 , L0 + L1 + L2 = 9L0 , L0 + L1 + L2 + L3 = 16L0 , … Successively combining these, we get L1 = 3L0 , L2 = 5L0 , L3 = 7L0 , …, or Lj = (2j + 1)L0 , j = 1, 2,…, n © 2005 Pearson Education, Inc., Upper Saddle River, NJ 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 Page 2-20 ... (x – x0), and we see that the motion is uniformly accelerated, with zero initial speed (v0 = 0) 14 No Velocity and acceleration can have different signs For example, if a car is moving forward,... be reproduced, in any form or by any means, without permission in writing from the publisher Page 2-8 Fishbane, Gasiorowicz, and Thornton + a∆t and ∆x = v ∆t + !a∆t (c) For the separate legs... point and up positive, label the first rock A and the second rock B and call the height of the cliff H For rock A: y = y + v 0A tA + !at 2; A A – H = + + !(– g)(4.15 s) , or H = (8.61 s )g For