Description
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved.No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 36 PROBLEM 11.31 The acceleration due to gravity of a particle falling toward the earth is 2 2 / , a gR r = − where r is the distance from the center of the earth to the particle, R is the radius of the earth, and g is the acceleration due to gravity at the surface of the earth. If 6370 R = km, calculate the escape velocity, that is, the minimum velocity with which a particle must be projected upward from the surface of the earth if it is not to return to earth. (Hint: 0 v = for . r = ∞ ) SOLUTION The acceleration is given by 2 2 dv gR v a dr r = = − Then, 2 2 gR dr v dv r = − Integrating, using the conditions esc 0 at , and v r v v = = ∞ = at r R = esc 0 2 2 v R dr v dv gR r ∞ = − ∫ ∫ esc 0 2 2 1 1 2 v R v gR r ∞ = 2 2 esc 1 1 0 0 2 v gR R − = − esc 2 v gR = 3 2 Now, 6370 km 6370 10 m and 9.81 m/s . R g = = × = Then, ( )( )( ) esc 2 v = 2 esc 111.8 10 m/s v = × � 9.81 3 6370 10 × PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 37 PROBLEM 11.32 The acceleration due to gravity at an altitude y above the surface of the earth can be expressed as ( ) 2 6 32.2 1 / 20.9 10 a y − ] + × ] where a and y are expressed in 2 m/s and feet, respectively. Using this expression, compute the height reached by a projectile fired vertically upward from the surface of the earth if its initial velocity is (a) 720 1200 12,000 v m/s, (b) v m/s, (c) v m/s. SOLUTION The acceleration is given by 6 2 20.9 10 32.2 1 y a × − ] | ` + ' J ] ( J ] 6 2 20.9 10 32.2 1 y dy vdv ady × − ] | ` + ' J ] ( J ] 2 0 max Integrate, using the conditions at 0 and 0 at . Also, use 9.81 m/s and v v y v y y g 3 6370 10 m. R × ( ) ( ) 0 0 2 2 2 0 0 1 v y R dy dy v dv g gR R y ∞ ∞ − − + + ∫ ∫ ∫ max 0 0 2 2 0 1 1 2 y v v gR R y | ` ' J + ( J ( ) 2 2 2 max 0 0 max max max max 1 1 1 0 2 2 gRy v gR v R y gRy R y R R y ] − − − + ] + + ] max Solving for , y 2 0 max 2 0 2 Rv y gR v − Using the given numerical data, ( )( )( ) 2 2 0 0 max 2 2 0 0 2 v v y v v − − 0 ( ) 720 m/s, a v ( )( ) ( ) ( ) 2 max 2 720 y − max y � 0 ( ) 1200 m/s, b v ( )( ) ( ) ( ) 2 max 2 1200 y − max 74250 m y � 0 ( ) 12,000 m/s, c v ( )( ) ( ) ( ) 2 max 2 negative y − Negative value indicates that 0 v is greater than the escape velocity. max y ∞ � 4 637 10 × 3 6370 10 × 3 6370 10 × 4 12498 10 × 9.81 4 12498 10 × 4 637 10 × 720 26532 m 4 637 10 × 1200 4 12498 10 × 4 637 10 × 4 12498 10 × 12,000 12,000 where a and y are expressed in m/s 2 and m, respectively. Using this , PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 38 PROBLEM 11.33 The velocity of a slider is defined by the relation ' sin( ). n v v t ω ϕ = + Denoting the velocity and the position of the slider at 0 t = by 0 v and 0 , x respectively, and knowing that the maximum displacement of the slider is 0 2 , x show that (a) ( ) 2 2 2 0 0 0 ' / 2 n n v v x x ω ω = + , (b) the maximum value of the velocity occurs when 2 0 0 0 3 ( / ) / 2. n x x v x w = − SOLUTION ( ) ( ) Given: sin n a v v t ω ϕ ′ = + At 0, t = 0 0 sin or sin v v v v v ϕ ϕ ′ = = = ′ (1) Let x be maximum at 1 t t = when 0. v = Then, ( ) ( ) 1 1 sin 0 and cos 1 n n t t ω ϕ ω ϕ + = + = ± (2) Using or dx v dx v dt dt = = Integrating, ( ) cos n n v x C t ω ϕ ω ′ = − + At 0, t = 0 0 cos or cos n n v v x x C C x ϕ ϕ ω ω ′ ′ = = − = + Then, ( ) 0 cos cos n n n v v x x t ϕ ω ϕ ω ω ′ ′ = + − + (3) max 0 1 cos using cos 1 n n v v x x t ϕ ω ϕ ω ω ′ ′ = + + + = − Solving for cos , ϕ ( ) max 0 cos 1 n x x v ω ϕ − = − ′ max 0 With 2 , x x = 0 cos 1 n x v ω ϕ = − ′ (4) Using 2 2 2 2 0 0 sin cos 1, or 1 1 n v x v v ω ϕ ϕ + = + − = ′ ′ Solving for gives v′ ( ) 2 2 2 0 0 0 (5) 2 n n v x v x ω ω + ′ = t PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 39 PROBLEM 11.33 CONTINUED ( ) Acceleration: b ( ) cos n n dv a v t dt ω ω ϕ ′ + 2 Let be maximum at when 0. v t t a Then, ( ) 2 cos 0 n t ω ϕ + From equation (3), the corresponding value of x is ( ) 0 0 0 0 2 2 2 2 0 0 0 0 0 2 0 0 cos 1 2 3 1 2 2 2 2 n n n n n n n n v v x v x x x x v v x v x x x x ω ϕ ω ω ω ω ω ω ω ′ ′ ′ | ` + + − − ' J ′ ( J + − − ( ) 0 0 2 0 3 2 n v x x ω ] − ] ] t From Equation (3), the corresponding value of x is PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 40 PROBLEM 11.34 The velocity of a particle is ( ) 0 1 sin / . v v t T π = − Knowing that the particle starts from the origin with an initial velocity 0 v , determine (a) its position and its acceleration at 3 , t T = (b) its average velocity during the interval 0 t = to . t T = SOLUTION 0 ( ) 1 sin dx t a v v dt T π = = − 0 Integrating, using 0 when 0, x x t = = = 0 0 0 0 1 sin x t t t dx v dt v dt T π = = − ∫ ∫ ∫ 0 0 0 0 cos t x v T t x v t T π π ] + ] ] 0 0 0 cos v T t v T x v t T π π π + − (1) When 3 , t T ( ) 0 0 0 0 2 3 cos 3 3 v T v T x v T v T T π π π | · + − − ' J ( J 0 2.36 x v T t 0 cos dv v t a dt T T π π − When 3 , t T 0 cos3 v a T π π − 0 v a T π t ( ) Using equation (1) with , b t T 0 0 1 0 0 2 cos 1 v T v T x v T v T π π π π | · + − − ' J ( J Average velocity is 1 0 ave 0 2 1 x x x v v t T π ∆ − | · − ' J ∆ ( J ave 0 0.363 v v t Using Equation (1) with t = T, PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 41 PROBLEM 11.35 A minivan is tested for acceleration and braking. In the street-start acceleration test, elapsed time is 8.2 s for a velocity increase from 10 km/h to 100 km/h. In the braking test, the distance traveled is 44 m during braking to a stop from 100 km/h. Assuming constant values of acceleration and deceleration, determine (a) the acceleration during the street-start test, (b) the deceleration during the braking test. SOLUTION 10 km/h 2.7778 m/s = 100 km/h 27.7778 m/s = (a) Acceleration during start test. dv a dt = 8.2 27.7778 0 2.7778 adt v dt = ∫ ∫ 8.2 27.7778 2.7778 a = − 2 3.05 m/s a = t (b) Deceleration during braking. dv a v dx = = 44 0 0 27.7778 a dx v dv = = ∫ ∫ ( ) ( ) 0 44 2 0 27.7778 1 2 a x v = ( ) 2 1 44 27.7778 2 a = − 2 8.77 m/s a = − deceleration 2 8.77 m/s a = − = t m/s, , braking: PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 42 PROBLEM 11.36 In Prob. 11.35, determine (a) the distance traveled during the street-start acceleration test, (b) the elapsed time for the braking test. SOLUTION 10 km/h 2.7778 m/s = 100 km/h 27.7778 m/s = (a) Distance traveled during start test. dv a dt = 0 0 t v v a dt dv = ∫ ∫ 0 at v v = − 0 v v a t − = 2 27.7778 2.7778 3.04878 m/s 8.2 a − = = 0 2.7778 3.04878 v v at t = + = + ( ) 8.2 0 0 2.7778 3.04878 t x v dv t dt = = + ∫ ∫ ( )( ) ( )( ) 2 2.7778 8.2 1.52439 8.2 = + 125.3 m x = � (b) Elapsed time for braking test. dv a v dx = 0 0 x v v adx v dv = ∫ ∫ 2 2 0 2 2 v v ax = − ( ) ( )( ) ( ) 2 2 2 0 1 1 0 27.7778 2 2 44 a v v x = − = − 2 8.7682 m/s = − dv a dt = 0 0 t v v a dt dv = ∫ ∫ 0 at v v = − 0 0 27.7778 8.7682 v v t a − − = = − 3.17 s t = � m/s, test: , , test: , , PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 43 PROBLEM 11.37 An airplane begins its take-off run at A with zero velocity and a constant acceleration a. Knowing that it becomes airborne 30 s later at B and that the distance AB is 823 m, determine (a) the acceleration a, (b) the take- off velocity . B v SOLUTION Constant acceleration. 0 0 0, 0 A A v v x x = = = = 0 v v at at = + = (1) 2 2 0 0 1 1 2 2 x x v t at at = + + = (2) At point , B 823 m and 30 s B x x t = = = (a) Solving (2) for a, ( )( ) ( ) 2 2 2 823 2 30 x a t = = 2 m/s a = t (b) Then, ( )( ) 1.8 30 B v at = = 54 m/s B v = t 1.8 Constant acceleration: PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 44 PROBLEM 11.38 Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop safely. A truck enters a 229 m ramp at a high speed 0 v and travels 165 m in 6 s at constant deceleration before its speed is reduced to 0 / 2. v Assuming the same constant deceleration, determine (a) the additional time required for the truck to stop, (b) the additional distance traveled by the truck. SOLUTION Constant acceleration. 0 0 x = 0 v v at = + (1) 2 0 0 1 2 x x v t at = + + (2) Solving (1) for a, 0 v v a t − = (3) Then, ( ) ( ) 2 0 0 0 0 0 0 1 1 1 2 2 2 v v x x v t t x v v t v v t t − = + + = + + = + At 6 s, t = 0 6 1 and 165 m 2 v v x = = ( ) 0 0 0 0 0 1 1 165 165 6 4.5 or 36.7 m/s 2 2 4.5 1 18.3 m/s 2 v v v v v v = + = = = = = Then, from (3), 2 2 18.3 36.7 m/s 3.05 m/s 6 6 a − = = − = − Substituting into (1) and (2), 36.7 3.05 v t = − ( ) 2 1 0 2 x t t = + − At stopping, 0 or 0 12 s s s v t t = − = = ( )( ) ( )( ) 2 1 0 12 12 220.8 m 2 x = + − = ( ) Additional time for stopping 12 s 6 s a = − 6 s t ∆ = t ( ) Additional distance for stopping 220.8 m 165 m b = − 55.8 m d ∆ = t 18.3 36.7 3.05 36.7 3.05 36.7 3.05 229-m Constant acceleration: 0, PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 45 PROBLEM 11.39 A sprinter in a 400-m race accelerates uniformly for the first 130 m and then runs with constant velocity. If the sprinter’s time for the first 130 m is 25 s, determine (a) his acceleration, (b) his final velocity, (c) his time for the race. SOLUTION 2 0 0 1 ( ) During the acceleration phase 2 a x x v t at = + + 0 0 Using 0, and 0, and solving for gives x v a = = 2 2x a t = Noting that 130 m when 25 s, x t = = ( )( ) ( ) 2 2 130 25 a = 0.416 m/s a = t (b) Final velocity is reached at 25 s. t = ( )( ) 0 0 0.416 25 f v v at = + = + 10.40 m/s f v = t (c) The remaining distance for the constant speed phase is 400 130 270 m x ∆ = − = For constant velocity, 270 25.96 s 10.40 x t v ∆ ∆ = = = Total time for run: 25 25.96 t = + 51.0 s t = t PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 46 PROBLEM 11.40 A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude of the rocket was 27.5 m at the end of the powered portion of the flight and that the rocket landed 16 s later. Knowing that the descent parachute failed to deploy so that the rocket fell freely to the ground after reaching its maximum altitude and assuming that 2 9.81m/s , g = determine (a) the speed 1 v of the rocket at the end of powered flight, (b) the maximum altitude reached by the rocket. SOLUTION Constant acceleration. Choose 0 t = at end of powered flight. Then, 2 1 27.5 m 9.81 m/s y a g = = − = − (a) When y reaches the ground, 0 and 16 s. f y t = = 2 2 1 1 1 1 1 1 2 2 f y y v t at y v t gt = + + = + − ( )( ) 2 2 1 1 1 2 2 1 0 27.5 9.81 16 76.76 m/s 16 f y y gt v t − + − + = = = 1 76.8 m/s v = t (b) When the rocket reaches its maximum altitude max , y 0 v = ( ) ( ) 2 2 2 1 1 1 1 2 2 v v a y y v g y y = + − = − − 2 2 1 1 2 v v y y g − = − ( ) ( )( ) 2 max 0 76.76 27.5 2 9.81 y − = − max 328 m y = t Constant acceleration: PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 47 PROBLEM 11.41 Automobile A starts from O and accelerates at the constant rate of 0.75 m/s 2 . A short time later it is passed by bus B which is traveling in the opposite direction at a constant speed of 6 m/s. Knowing that bus B passes point O 20 s after automobile A started from there, determine when and where the vehicles passed each other. SOLUTION Place origin at 0. Motion of auto. ( ) ( ) 2 0 0 0, 0, 0.75 m/s A A A x v a = = = ( ) ( ) ( ) 2 2 0 0 1 1 0 0 0.75 2 2 A A A A x x v t a t t = + + = + + 2 0.375 m A x t = Motion of bus. ( ) ( ) 0 0 ?, 6 m/s, 0 B B B x v a = = − = ( ) ( ) ( ) 0 0 0 6 m B B B B x x v t x t = − = − At 20 , 0. B t s x = = ( ) ( )( ) 0 0 6 20 B x = − ( ) 0 120 m B x = Hence, 120 6 B x t = − When the vehicles pass each other, . B A x x = 2 120 6 0.375 t t − = 2 0.375 6 120 0 t t + − = ( )( )( ) ( )( ) 2 6 (6) 4 0.375 120 2 0.375 t − ± − − = 6 14.697 11.596 s and 27.6 s 0.75 t − ± = = − Reject the negative root. 11.60 s t = t Corresponding values of x A and x B . ( )( ) 2 0.375 11.596 50.4 m A x = = ( )( ) 120 6 11.596 50.4 m B x = − = 50.4 m x = t at 0: , 0.75 m/s 2 . A short time later, it is passed by bus B which is traveling in the PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 48 PROBLEM 11.42 Automobiles A and B are traveling in adjacent highway lanes and at 0 t = have the positions and speeds shown. Knowing that automobile A has a constant acceleration of 0.6 m/s 2 and that B has a constant deceleration of 0.4 m/s 2 , determine (a) when and where A will overtake B, (b) the speed of each automobile at that time. SOLUTION Place the origin at A when t = 0. Motion of A: ( ) ( ) 2 0 0 0, 15 km/h = 4.1667 m/s, 0.6 m/s A A A x v a = = = ( ) 0 4.1667 0.6 A A A v v a t t = + = + ( ) ( ) 2 2 0 0 1 4.1667 0.3 2 A A A A x x v t a t t t = + + = + Motion of B: ( ) ( ) 2 0 0 25 m, 23 km/h = 6.3889 m/s, 0.4 m/s B B B x v a = = = − ( ) 0 6.3889 0.4 B B B v v a t t = + = − ( ) ( ) 2 2 0 0 1 25 6.3889 0.2 2 B B B B x x v t a t t t = + + = + − (a) When and where A overtakes B. A B x x = 2 2 4.1667 0.3 25 6.3889 0.2 t t t t + = + − 2 0.5 2.2222 25 0 t t − − = ( )( )( ) ( )( ) 2 2.2222 2.2222 4 0.5 25 2 0.5 t ± − − = 2.2222 7.4120 9.6343 s and 5.19 s t = ± = − Reject the negative root. . 9.63 s t = t ( )( ) ( )( ) 2 4.1667 9.6343 0.3 9.6343 68.0 m A x = + = ( )( ) ( )( ) 2 25 6.3889 9.6343 0.2 9.6343 68.0 m B x = + − = moves 68.0 m A t moves 43.0 m B t (b) Corresponding speeds. ( )( ) 4.1667 0.6 9.6343 9.947 m/s A v = + = 35.8 km/h A v = t ( )( ) 6.3889 0.4 9.6343 2.535 m/s B v = − = 9.13 km/h B v = t B: speeds: PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 49 PROBLEM 11.43 In a close harness race, horse 2 passes horse 1 at point A, where the two velocities are 2 v = 6.4 m/s and 1 v = 6.2 m/s. Horse 1 later passes horse 2 at point B and goes on to win the race at point C, 366 m from A. The elapsed times from A to C for horse 1 and horse 2 are 1 t = 61.5 s and 2 t = 62.0 s, respectively. Assuming uniform accelerations for both horses between A and C, determine (a) the distance from A to B, (b) the position of horse 1 relative to horse 2 when horse 1 reaches the finish line C. SOLUTION Constant acceleration ( ) 1 2 and a a for horses 1 and 2. Let 0 x = and 0 t = when the horses are at point A. Then, 2 0 1 2 x v t at = + Solving for , a ( ) 0 2 2 x v t a t − = Using 366 m x = and the initial velocities and elapsed times for each horse, ( )( ) ( ) –3 1 1 1 2 2 1 2 366 6.2 61.5 8 ×10 m/s 61.5 x v t a t − − = = = − ( )( ) ( ) 2 2 2 2 2 2 2 2 6.4 62.0 1.6 10 m/s 62.0 x v t a t − − = = = − 1 2 Calculating , x x − ( ) ( ) 2 1 2 1 2 1 2 1 2 x x v v t a a t − = − + − ( ) ( ) ( ) 2 1 2 2 1 2 .2 .004 x x t t t t − = − + − − − = − + At point B, 2 1 2 0 0 B B x x t t − = − + = (a) 0.2 50 s .004 B t = = Calculating B x using data for either horse, Horse 1: ( )( ) ( )( ) 2 1 2 B x = + − 340 m B x = � Horse 2: ( )( ) ( )( ) 2 1 340 m 2 B x = + − = When horse 1 crosses the finish line at 61.5 s, t = (b) ( )( ) ( )( ) 2 1 2 61.5 61.5 x x − = − + 2.8 m x ∆ = � − 2 366 − 2 × 6.2 6.4 –3 8 ×10 − 2 1.6 10 − × .004 50 6.2 –3 8 ×10 − 50 50 6.4 2 1.6 10 − × 50 0.2 .004 0.2 2: 366 m and the initial velocities and elapsed times for each horse, PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 50 PROBLEM 11.44 Two rockets are launched at a fireworks performance. Rocket A is launched with an initial velocity 0 v and rocket B is launched 4 s later with the same initial velocity. The two rockets are timed to explode simultaneously at a height of 73 m, as A is falling and B is rising. Assuming a constant acceleration g = 9.81 m/s 2 determine (a) the initial velocity 0 , v (b) the velocity of B relative to A at the time of the explosion. SOLUTION Choose x positive upward. Constant acceleration a g = − Rocket launch data: Rocket : A 0 0, , 0 x v v t = = = Rocket : B 0 0, , 4 s B x v v t t = = = = Velocities: Rocket : A 0 A v v gt = − Rocket : B ( ) 0 B B v v g t t = − − Positions: 2 0 1 Rocket : 2 A A x v t gt = − ( ) ( ) 2 0 1 Rocket : , 2 B B B B B x v t t g t t t t = − − − ≥ For simultaneous explosions at 73 m when , A B E x x t t = = = ( ) ( ) 2 2 2 2 0 0 0 0 1 1 1 1 2 2 2 2 E E E B E B E B E E B B v t gt v t t g t t v t v t gt gt t gt − = − − − = − − + − 0 Solving for , v 0 2 B E gt v gt = − (1) Then, when , E t t = 2 1 , 2 2 B A E E E gt x gt t gt = − − or 2 2 0 A E B E x t t t g − − = Solving for , E t ( )( ) ( ) ( ) ( )( )( )( ) 2 2 4 1 2 73 2 9.81 4 1 4 4 6.35 s 2 2 A x B B g E t t t ± + ± + = = = Assuming a constant acceleration g = 9.81 m/s 2 , determine (a) the initial t = t E , PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 51 PROBLEM 11.44 CONTINUED (a) From equation (1), ( )( ) ( )( ) 0 9.81 4 9.81 6.35 2 v = − 0 42.67 m/s v = � At time , E t 0 A E v v gt = − ( ) 0 B E B v v g t t = − − (b) ( )( ) 9.81 4 B A B v v gt − = = / 39.24 m/s B A v = � From Equation (1), , PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 52 PROBLEM 11.45 In a boat race, boat A is leading boat B by 38 m and both boats are traveling at a constant speed of 168 km/h. At 0 t = , the boats accelerate at constant rates. Knowing that when B passes A, 8 t = s and 228 A v = km/h, determine (a) the acceleration of A, (b) the acceleration of B. SOLUTION (a) Acceleration of A. ( ) ( ) 0 0 , 168 km/h 46.67 m/s A A A A v v a t v = + = = At 8 s, t = 228 km/h 63.33 m/s A v = = ( ) 0 63.33 46.67 8 A A A v v a t − − = = 2 2.08 m/s A a = � (b) ( ) ( ) 2 0 0 1 2 A A A A x x v t a t = + + ( ) ( ) 2 0 0 1 2 B B B B x x v t a t = + + ( ) ( ) ( ) ( ) ( ) 2 0 0 0 0 1 2 A B A B A B A B x x x x v v t a a t − = − + − + − When 0, t = ( ) ( ) 0 0 38 m A B x x − = and ( ) ( ) 0 0 0 B A v v − = When 8 s, t = 0 A B x x − = Hence, ( )( ) 2 1 0 38 8 , or 1.1875 2 A B A B a a a a = + − − = − 1.1875 2.08 1.1875 B A a a = + = + 2 3.27 m/s B a = � of A: at constant rates. Knowing that when B passes A, t = 8 s and Using this y are expressed in m/s and m.2dy y 1 + 20. 37 . ymax = ∞ � PROPRIETARY MATERIAL.2 ) 2 2 where a and y are expressed in m/s 2and feet.81 m/s 2 and R = 6370 × 103 m. If you are a student using this Manual. compute the height reached by a projectile fired vertically upward from the surface of the earth if its initial velocity is (a) v = 720 m/s. you are using it without permission. using the conditions v = v0 at y = 0 and v = 0 at y = ymax .2 y 1 + 20. without the prior written permission of the publisher. ymax = (c) v0 = 12. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. No part of this Manual may be displayed. Inc. use g = 9. 2 ( 2 )(9.32 The acceleration due to gravity at an altitude y above the surface of the earth can be expressed as a= 1 + y / 20. SOLUTION The acceleration is given by a = − 32. ymax = ( 637 × 10 )( 720) (12498 × 10 ) − ( 720 ) ( 637 × 10 )(1200 ) (12498 × 10 ) − (1200 ) ( 637 × 10 )(12.81 ) (6370 × 103) − v0 4 2 2 6370 × 103 v0 = 2 637 × 104 v0 2 12498 × 104 − v0 (a) v0 = 720 m/s. Using the given numerical data. (c) v = 12.9 × 106 2 Integrate.000 m/s. vdv = ady = − 32. ymax = (b) v0 = 1200 m/s.9 × 106 2 .9 × 106 ( − 32. Also. respectively. 0 v0 ∫ 0 v v0 dv = − g ∫ ∞ 0 ( dy y R 1+ ) 2 = ∞ − gR 2 0 ∫ dy ( R + y )2 1 2 v 2 1 = gR R + y 2 ymax 0 0− 1 2 1 1 gRymax − =− v0 = gR 2 2 R + ymax R R + ymax ymax = ymax = 2 Rv0 2 2 gR − v0 2 v0 ( R + ymax ) = 2 gRymax Solving for ymax . (b) v = 1200 m/s. © 2007 The McGraw-Hill Companies.000 m/s. All rights reserved.000) 4 2 4 2 4 2 4 2 4 ymax = 26532 m � ymax = 74250 m � 2 = negative Negative value indicates that v0 is greater than the escape velocity. respectively. reproduced or distributed in any form or by any means.PROBLEM 11. Using this expression.000 ) (12498 × 10 ) − (12. ( xmax − x0 )ω n −1 v′ (4) v0 x0ω n ′ + ′ − 1 = 1 v v v′ = 2 2 x0ω n −1 v′ or Solving for v′ gives (v 2 0 2 2 + x0ω n 2 x0ω n ) (5) t PROPRIETARY MATERIAL.33 The velocity of a slider is defined by the relation v = v 'sin(ω nt + ϕ ). v = v0 = v′ sin ϕ or sin ϕ = v0 v′ (1) Let x be maximum at t = t1 when v = 0. With xmax = 2 x0 . you are using it without permission. without the prior written permission of the publisher. © 2007 The McGraw-Hill Companies. SOLUTION (a) Given: v = v′ sin (ω nt + ϕ ) At t = 0. Then. x = x0 = C − x = x0 + xmax = x0 + Then. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. v′ cos (ω nt + ϕ ) ωn v′ cos ϕ ωn or C = x0 + v′ cos ϕ ωn (3) At t = 0.PROBLEM 11. and knowing that the maximum displacement of the 2 2 2 slider is 2 x0 . Inc. No part of this Manual may be displayed. Using sin (ω nt1 + ϕ ) = 0 dx =v dt x=C− and or cos (ω nt1 + ϕ ) = ± 1 dx = v dt (2) Integrating. All rights reserved. show that (a) v ' = v0 + x0ω n / 2 x0ω n . v′ v′ cos ϕ − cos (ω nt + ϕ ) ωn ωn v′ v′ cos ϕ + ωn ω using cos ω nt1 + ϕ = −1 Solving for cos ϕ . (b) the maximum ( ) value of the velocity occurs when x = x0 3 − (v0 / x0 wn ) 2 / 2. Using cos ϕ = cos ϕ = sin 2 ϕ + cos 2ϕ = 1. If you are a student using this Manual. respectively. 38 . reproduced or distributed in any form or by any means. Denoting the velocity and the position of the slider at t = 0 by v0 and x0 . 33 CONTINUED (b) Acceleration: a= dv = v′ω n cos (ω nt + ϕ ) dt Let v be maximum at t = t2 when a = 0. Then. All rights reserved. you are using it without permission. 39 . Inc. reproduced or distributed in any form or by any means. No part of this Manual may be displayed. cos (ω nt2 + ϕ ) = 0 From equation (3).PROBLEM 11. the corresponding value ofof is is Equation (3). or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. the corresponding value x x x = x0 + v′ v′ x0ω n v′ cos ϕ = x0 + ′ − 1 = 2 x0 − ωn ωn v ωn 2 2 2 2 v0 + x0ω n 3 1 v0 = x0 − 2 2 x0ω n (2 x0ω n )ω n 2 = 2 x0 − 3 − x0 ( ) v0 x0ωn 2 2 t PROPRIETARY MATERIAL. If you are a student using this Manual. without the prior written permission of the publisher. © 2007 The McGraw-Hill Companies. reproduced or distributed in any form or by any means. Equation (1) with = T . © 2007 The McGraw-Hill Companies. determine (a) its position and its acceleration at t = 3T .36v0T t a= When t = 3T .363v0 t v0T vT 2 cos π − 0 = v0T 1 − π π π dv πv πt = − 0 cos dt T T a=− π v0 cos 3π T a= π v0 t T PROPRIETARY MATERIAL.PROBLEM 11. All rights reserved. Inc. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. you are using it without permission. v0T π t v0T cos − π π T v0T vT 2 cos (3π ) − 0 = 3 − v0T T π π (1) x = 3v0T + x = 2. x1 = v0T + Average velocity is vave = ∆x x1 − x0 2 = = 1 − v0 ∆t T π vave = 0. (b) Using equation (1) with t t = T. (b) its average velocity during the interval t = 0 to t = T . without the prior written permission of the publisher. x t t ∫ 0 dx = ∫ 0 v dt = ∫ 0 v0 1 − sin T dt πt x x 0 vT πt = v0t + 0 cos π T t 0 x = v0t + When t = 3T . 40 .34 The velocity of a particle is v = v0 1 − sin (π t / T ) . SOLUTION (a ) dx πt = v = v0 1 − sin T dt Integrating. If you are a student using this Manual. Knowing that the particle starts from the origin with an initial velocity v0 . using x = x0 = 0 when t = 0. No part of this Manual may be displayed. reproduced or distributed in any form or by any means.PROBLEM 11. © 2007 The McGraw-Hill Companies. you are using it without permission.2 27.77 m/s 2 PROPRIETARY MATERIAL.2 a = 27. SOLUTION 10 km/h = 2.7778 m/s. All rights reserved.7778 − 2. a= dv . or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation.7778 44 a = − 1 ( 27. (b) the deceleration during the braking test.7778 v dv = a (x) 44 0 = 1 2 v 2 ( ) 0 27.35 A minivan is tested for acceleration and braking. No part of this Manual may be displayed.05 m/s 2 t 44 0 ∫ 0 a dx = ∫ 27. braking: a=v dv = dx a = 3.77 m/s 2 t a = − 8. 41 . Inc.7778)2 2 deceleration = − a = 8. If you are a student using this Manual. the distance traveled is 44 m during braking to a stop from 100 km/h. In the braking test.2 s for a velocity increase from 10 km/h to 100 km/h. dt 8. Assuming constant values of acceleration and deceleration.7778 m/s 8. determine (a) the acceleration during the street-start test. elapsed time is 8. In the street-start acceleration test. m/s (a) Acceleration during start test.7778 (b) Deceleration during braking.7778 v dt 100 km/h = 27. without the prior written permission of the publisher.7778 ∫ 0 a dt = ∫ 2. dx x = 125. determine (a) the distance traveled during the street-start acceleration test.2 ) + (1.3 m � ∫ 0 a dx = ∫ v0 v dv x v ax = a= v 2 v0 2 − 2 2 1 2 1 2 v − v0 = 0 − 27. dt 100 km/h = 27. (b) the elapsed time for the braking test.2 v = v0 + at = 2.04878 m/s 2 8.7682 t = 3.7778)(8.7778 m/s ∫ 0 a dt = ∫ v0 dv a= v − v0 t t v at = v − v0 . m/s (a) Distance traveled during start test: test. No part of this Manual may be displayed. you are using it without permission.7778 + 3. a= 27.04878 t x= ∫ 0 v dv = ∫ 0 (2. 11.7778 + 3. dt ∫ 0 a dt = ∫ v0 dv t v at = v − v0 t= v − v0 0 − 27.52439 )(8. If you are a student using this Manual.PROBLEM 11. without the prior written permission of the publisher. © 2007 The McGraw-Hill Companies.7778 = a − 8. All rights reserved. 42 .77782 2x ( 2 )( 44 ) ( ) ( ) = − 8.04878 t ) dt t 8. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation.7778 − 2. SOLUTION 10 km/h = 2.17 s � PROPRIETARY MATERIAL. Inc. reproduced or distributed in any form or by any means. a=v dv .7778 = 3.2 2 = ( 2.7682 m/s 2 a= dv . a= dv .36 In Prob.35.2 ) (b) Elapsed time for braking test: test.7778 m/s. you are using it without permission. v = v0 + at = at x = x0 + v0t + At point B. (a) Solving (2) for a. without the prior written permission of the publisher. All rights reserved. If you are a student using this Manual. vB = at = (1. 1 2 1 2 at = at 2 2 and t = 30 s a = 1.PROBLEM 11. Knowing that it becomes airborne 30 s later at B and that the distance AB is 823 m. No part of this Manual may be displayed. Inc. (b) the takeoff velocity vB .37 An airplane begins its take-off run at A with zero velocity and a constant acceleration a. 43 .8 m/s2 t vB = 54 m/s t x0 = x A = 0 (1) (2) x = xB = 823 m a= 2 x ( 2 )( 823 ) = t2 (30 )2 (b) Then. reproduced or distributed in any form or by any means. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. determine (a) the acceleration a. v0 = v A = 0.8)(30 ) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies. SOLUTION Constant acceleration: Constant acceleration. 3 m/s 2 or 18. from (3). (3) 1 v − v0 2 1 1 t = x0 + (v0 + v ) t = ( v0 + v ) t 2 t 2 2 x6 = 165 m v0 = 165 = 36.7 18.05) t 2 2 At stopping. No part of this Manual may be displayed. Inc. (b) the additional distance traveled by the truck. At t = 6 s. 44 . 1 v0 2 v − v0 t 1 2 at 2 (1) (2) Solving (1) for a.7 m/s 4.3 − 36. x0 = 0 v = v0 + at x = x0 + v0t + a= x = x0 + v0t + v= 165 = v= Then.8 m t ∆t = 6 s t PROPRIETARY MATERIAL.8 m − 165 m ∆d = 55.7 )(12 ) − (a) Additional time for stopping = 12 s − 6 s (b) Additional distance for stopping = 220.8 m 2 x = 0 + ( 36.38 Steep safety ramps are built beside mountain highways to enable vehicles 229-m with defective brakes to stop safely. A truck enters a 229 m ramp at a high speed v0 and travels 165 m in 6 s at constant deceleration before its speed is reduced to v0 / 2.7t − 1 (3.PROBLEM 11. Substituting into (1) and (2). without the prior written permission of the publisher. v = 0 or 0 36.5 and 1 1 v0 + v0 (6 ) = 4. determine (a) the additional time required for the truck to stop. If you are a student using this Manual.05 m/s2 6 6 v = 36.3 2 =− a= m/s = − 3. you are using it without permission. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. ts = 12 s 1 ( 3. reproduced or distributed in any form or by any means.7 − 3.05 ts = 0.05t x = 0 + 36. © 2007 The McGraw-Hill Companies. All rights reserved. Then.05 )(12 )2 = 220.7 − 3. SOLUTION Constant acceleration: Constant acceleration. Assuming the same constant deceleration.5v0 2 2 1 v0 = 18. 40 v t = 51. (b) his final velocity. a= ( 2 )(130 ) ( 25)2 a = 0.0 s t v f = 10.416 )( 25 ) (c) The remaining distance for the constant speed phase is ∆x = 400 − 130 = 270 m For constant velocity. you are using it without permission.40 m/s t t = 25 + 25. If the sprinter’s time for the first 130 m is 25 s. If you are a student using this Manual. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. © 2007 The McGraw-Hill Companies. reproduced or distributed in any form or by any means. v f = v0 + at = 0 + ( 0.96 PROPRIETARY MATERIAL.PROBLEM 11. and solving for a gives a= 2x t2 Noting that x = 130 m when t = 25 s. All rights reserved. 45 . (c) his time for the race. SOLUTION (a) During the acceleration phase x = x0 + v0t + 1 2 at 2 Using x0 = 0. and v0 = 0.96 s 10. Total time for run: ∆t = 270 ∆x = = 25. determine (a) his acceleration.416 m/s t (b) Final velocity is reached at t = 25 s. No part of this Manual may be displayed. Inc. without the prior written permission of the publisher.39 A sprinter in a 400-m race accelerates uniformly for the first 130 m and then runs with constant velocity. 40 A group of students launches a model rocket in the vertical direction.76 ) = 27.5 m and a = − g = − 9.81) ymax = 328 m t PROPRIETARY MATERIAL. they determine that the altitude of the rocket was 27. Choose t = 0 at end of powered flight. All rights reserved. determine (a) the speed v1 of the rocket at the end of powered flight. 46 .5 m at the end of the powered portion of the flight and that the rocket landed 16 s later.81)(16 )2 16 = 76. y f = 0 y f = y1 + v1t + y f − y1 + t v1 = gt 2 = 0 − 27. you are using it without permission. SOLUTION Constant acceleration. Knowing that the descent parachute failed to deploy so that the rocket fell freely to the ground after reaching its maximum altitude and assuming that g = 9.76 m/s v1 = 76. 1 2 1 at = y1 + v1t − gt 2 2 2 1 2 (a) When y reaches the ground. If you are a student using this Manual. reproduced or distributed in any form or by any means. © 2007 The McGraw-Hill Companies. y1 = 27.81 m/s 2 . without the prior written permission of the publisher. Based on tracking data. (b) the maximum altitude reached by the rocket. Inc.PROBLEM 11.5 − ( 2 )(9. No part of this Manual may be displayed. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation.81 m/s 2 t = 16 s.5 + 1 2 (9. v=0 2 2 v 2 = v1 + 2a ( y − y1 ) = v1 − 2 g ( y − y1 ) 2 v 2 − v1 2g 2 y = y1 − ymax 0 − (76. acceleration: Then.8 m/s t (b) When the rocket reaches its maximum altitude ymax . 41 Automobile A starts from O and accelerates at the constant rate of 2 0.375 t 2 0. reproduced or distributed in any form or by any means.it is passed by bus B which is traveling in the later it is passed by bus opposite direction at a constant speed of 6 m/s. 0 = ( xB )0 − ( 6 )( 20 ) Hence.375t 2 m Motion of bus. x A = (0. xB = 0. (vB )0 = − 6 m/s. xB = 120 − 6 t ( xB )0 = 120 m When the vehicles pass each other.75 ) t 2 2 2 x A = ( x A )0 + ( v A )0 t + x A = 0. If you are a student using this Manual. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation.75 and − 27. 0: Motion of auto.6 s t = 11.596 ) = 50. you are using it without permission. Inc. ( x A )0 = 0. All rights reserved.60 s t xB = 120 − (6 )(11. (v A )0 = 0. ( xB )0 = ?.75 m/s 2 1 1 a At 2 = 0 + 0 + ( 0.4 m t PROPRIETARY MATERIAL.375 t 2 + 6 t − 120 = 0 t= t= Reject the negative root.596 s 0. determine when and where the vehicles passed each other. Knowing that bus B passes point O 20 s after automobile A started from there. 47 . . Corresponding values of xA and xB.375 )(11. No part of this Manual may be displayed.697 = 11.PROBLEM 11.4 m x = 50. a A = 0. aB = 0 xB = ( xB )0 − (vB )0 t = ( xB )0 − 6t m At t = 20 s.596 ) = 50. 120 − 6t = 0.375 ) − 6 ± 14.375)( −120 ) ( 2 )(0.75 m/s2. without the prior written permission of the publisher. © 2007 The McGraw-Hill Companies.4 m 2 − 6 ± (6) 2 − ( 4 )(0.. SOLUTION Place origin at 0. xB = x A. A short time later. 19 s Reject the negative root.2 t 2 2 x A = xB (a) When and where A overtakes B: B.3 t 2 = 25 + 6.0 m 2 (b) Corresponding speeds: speeds. SOLUTION Place the origin at A when t = 0. you are using it without permission.6t x A = ( x A )0 + ( v A )0 t + a A = 0. 4. 48 .3889t − 0.6343) = 2.3889 m/s. v A = ( v A )0 + a At = 4. Knowing that automobile A has a constant acceleration of 0.13 km/h t A moves 68.22222 − ( 4 )( 0.6343 s and − 5. t = 9.1667 t + 0.535 m/s B moves 43.2222t − 25 = 0 t= 2.2 t 2 0.4 m/s2. Inc. aB = − 0. x A = ( 4.947 m/s vB = 6.1667 t + 0.6343) − (0.3889 − 0.8 km/h t vB = 9.1667 + 0.5) t = 2.6343) = 68.3889 − (0.2 )(9.3889 )(9. If you are a student using this Manual.0 m t PROPRIETARY MATERIAL.1667 m/s.3t 2 2 = 23 km/h = 6.5 )( − 25 ) ( 2 )(0.42 Automobiles A and B are traveling in adjacent highway lanes and at t = 0 have the positions and speeds shown.6343) = 68.6 m/s 2 Motion of B: ( xB )0 = 25 m.4120 = 9. determine (a) when and where A will overtake B.3889 t − 0. (b) the speed of each automobile at that time. © 2007 The McGraw-Hill Companies.1667 )(9. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation.3)(9.1667 + ( 0.6 m/s2 and that B has a constant deceleration of 0. (v A )0 = 15 km/h = 4.4 )(9. All rights reserved.63 s t xB = 25 + (6.6 )(9.4 m/s 2 1 aBt 2 = 25 + 6. reproduced or distributed in any form or by any means.PROBLEM 11.2222 ± 7.5t 2 − 2.2222 ± 2. without the prior written permission of the publisher.6343) + (0.0 m t v A = 35.0 m 2 . Motion of A: ( x A )0 = 0. (vB )0 vB = ( vB )0 + aBt = 6. v A = 4.6343) = 9. No part of this Manual may be displayed.4t xB = ( xB )0 + ( vB )0 t + 1 a At 2 = 4. 6 × 10− 2 ) t 2 2 x1 − x2 = ( 6.6 × 10− 2 )(50)2 = 340 m 2 2 xB = 340 m� xB = (6.4) t + = − .0 s.43 In a close harness race.0 ) = = − 1. without the prior written permission of the publisher.5) + (.4)(50) + When horse 1 crosses the finish line at t = 61.4 m/s and v1 = 6. Horse 1 later passes horse 2 at point B and goes on to win the race at point C. No part of this Manual may be displayed. The elapsed times from A to C for horse 1 and horse 2 are t1 = 61. x − v2t2 2 366 − ( 6. Let x = 0 and t = 0 when the horses are at point A.5 s.2 − 6.2 )( 61. where the two velocities are v2 = 6.5) ∆x = 2.2)(50) + 1 –3 ( − 8 ×10− )( 50)2 2 1 ( − 1.5 ) = 2 2 t1 (61. you are using it without permission.004 Calculating xB using data for either horse. Then. Solving for a. 366 m from A. and the initial velocities and elapsed times for each horse. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. Horse 1: Horse 2: xB = (6.2 t B + .8 m � PROPRIETARY MATERIAL.6 × 10− 2 m/s 2 2 2 t2 62.004)( 61. (b) the position of horse 1 relative to horse 2 when horse 1 reaches the finish line C.5 s and t2 = 62.5) –3 = − 8 ×10− m/s 2 a2 = Calculating x1 − x2 .PROBLEM 11. horse 2 passes horse 1 at point A.4 )( 62.2)(61. © 2007 The McGraw-Hill Companies. a1 = x − v1t1 2 366 − (6.0 ) ( x1 − x2 = (v1 − v2 ) t + 1 ( a1 − a2 ) t 2 2 1 –3 ( − 8 ×10− ) − ( − 1.2 t + . All rights reserved. 49 . (a) x1 − x2 = 0 tB = 2 − 0.004 tB = 0 0. SOLUTION Constant acceleration ( a1 and a2 ) for horses 1 and 2: 2.2 m/s. reproduced or distributed in any form or by any means. determine (a) the distance from A to B. (b) x1 − x2 = − ( 0.004 t 2 At point B. respectively.2 = 50 s . x = v0t + a= 1 2 at 2 2 ( x − v0t ) t2 Using x = 366 m and the initial velocities and elapsed times for each horse. If you are a student using this Manual. Inc. Assuming uniform accelerations for both horses between A and C. when t = t E . as A is falling and B is rising. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation.81 m/s22.35 s PROPRIETARY MATERIAL. reproduced or distributed in any form or by any means. without the prior written permission of the publisher. All rights reserved. Inc. v = v0 . Rocket launch data: Constant acceleration a = − g Rocket A: Rocket B: Velocities: x = 0. 50 . No part of this Manual may be displayed. SOLUTION Choose x positive upward. 1 2 1 1 2 1 2 2 gt E = v0 (t E − t B ) − g (t E − t B ) = v0t E − v0t B − gt E + gt E t B − gt B 2 2 2 2 v0 = gt E − gt B 2 or 2 t E − t Bt E − (1) 2xA =0 g gt 1 2 x A = gt E − B t E − gt E .. 2 2 2 t B ± t B + ( 4 )(1) 2 xA g Solving for t E . t = 0 x = 0.81 2 = 6. Assuming a constant acceleration g = 9. t = t B = 4 s Rocket A: v A = v0 − gt Rocket B: vB = v0 − g (t − t B ) Positions: Rocket A: x A = v0t − 1 2 gt 2 1 2 g (t − t B ) . you are using it without permission. tE = 2 ( ) = 4± ( 4 )2 + (4)(1)(2)(73) 9. The two rockets are timed to explode simultaneously at a height of 73 m. (b) the velocity of B relative to A at the time of the explosion. © 2007 The McGraw-Hill Companies. 2 t ≥ tB Rocket B: xB = v0 (t − t B ) − For simultaneous explosions at x A = xB = 73 m when t = tE. E v0t E − Solving for v0 . Then. determine (a) the initial Assuming a constant acceleration 9.PROBLEM 11. Rocket A is launched with an initial velocity v0 and rocket B is launched 4 s later with the same initial velocity.44 Two rockets are launched at a fireworks performance. If you are a student using this Manual. v = v0 .81 m/s determine (a) the initial velocity v0 . 67 m/s � vB = v0 − g (t E − t B ) vB/ A = 39. 51 . reproduced or distributed in any form or by any means.81 )(4 ) 2 v0 = 42. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation.81)( 6. without the prior written permission of the publisher. At time t E . All rights reserved.35 ) − v A = v0 − gt E .PROBLEM 11. From Equation (1). If you are a student using this Manual. © 2007 The McGraw-Hill Companies.24 m/s � vB − v A = gt B = (9.44 CONTINUED (a) From equation (1). Inc.81)(4 ) PROPRIETARY MATERIAL. you are using it without permission. No part of this Manual may be displayed. (9. (b) v0 = (9. or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation.PROBLEM 11.33 m/s 63. No part of this Manual may be displayed. © 2007 The McGraw-Hill Companies.08 m/s 2 � 1 aB t 2 2 (b) x A = ( x A )0 + ( v A )0 t + 1 x A − xB = ( x A )0 − ( xB )0 + ( v A )0 − ( vB )0 t + ( a A − aB ) t 2 2 When t = 0. At t = 0 .45 In a boat race. t t = 8 s and at constant rates.1875 PROPRIETARY MATERIAL. Hence.33 − 46. Knowing that when B passes A.08 + 1. A: v A = ( v A )0 + a At .27 m/s 2 � x A − xB = 0 1 ( a A − aB )(8)2 . If you are a student using this Manual. without the prior written permission of the publisher.1875 aB = 3. aA = v A − ( v A )0 t = (v A )0 = 168 km/h = 46. All rights reserved. = 8 s and v A = 228 km/h. When t = 8 s. Knowing that when B passes A. Inc. determine (a) the acceleration of A. you are using it without permission. 52 .67 m/s v A = 228 km/h = 63. (b) the acceleration of B. the boats accelerate at constant rates. reproduced or distributed in any form or by any means. SOLUTION (a) Acceleration of A.67 8 1 a At 2 2 xB = ( xB )0 + ( vB )0 t + a A = 2.1875 = 2. At t = 8 s. 2 or aB = a A + 1. 0 = 38 + ( x A )0 − ( xB )0 = 38 m and (vB )0 − (vA )0 = 0 a A − aB = − 1. boat A is leading boat B by 38 m and both boats are traveling at a constant speed of 168 km/h.
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