Problem 13.1 In Example 13.2, suppose that the vehi-cle is dropped from a height h = 6m. (a) What is thedownward velocity 1 s after it is released? (b) What isits downward velocity just before it reaches the ground?hSolution: The equations that govern the motion are:a = g = 9.81 m/s2v = gts = 12gt2+h(a) v = gt = (9.81 m/s2)(1 s) = 9.81 m/s.The downward velocity is 9.81 m/s.(b) We need to rst determine the time at which the vehicle hits thegrounds = 0 = 12gt2+h t =_2hg=_ 2(6 m)9.81 m/s2 = 1.106 sNow we can solve for the velocityv = gt = (9.81 m/s2)(1.106 s) = 10.8 m/s.The downward velocity is 10.8 m/s.Problem 13.2 The milling machine is programmed sothat during the interval of time from t = 0 to t = 2 s,the position of its head (in inches) is given as a functionof time by s = 4t 2t3. What are the velocity (in in/s)and acceleration (in in/s2) of the head at t = 1 s?sSolution: The motion is governed by the equationss = (4 in/s)t (2 in/s2)t2,v = (4 in/s) 2(2 in/s2)t ,a = 2(2 in/s2).At t = 1 s, we have v = 0, a = 4 in/s2.9c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.3 In an experiment to estimate the accel-eration due to gravity, a student drops a ball at a distanceof 1 m above the oor. His lab partner measures the timeit takes to fall and obtains an estimate of 0.46 s.(a) What do they estimate the acceleration due to grav-ity to be?(b) Let s be the balls position relative to the oor.Using the value of the acceleration due to gravitythat they obtained, and assuming that the ball isreleased at t = 0, determine s (in m) as a functionof time.ss 0Solution: The governing equations area = gv = gts = 12gt2+h(a) When the ball hits the oor we have0 = 12gt2+h g = 2ht2 = 2(1 m)(0.46 s)2 = 9.45 m/s2g = 9.45 m/s2(b) The distance s is then given bys = 12(9.45 m/s2) +1 m. s = (4.73 m/s2)t2+1.0 m.Problem 13.4 The boats position during the intervalof time from t = 2 s to t = 10 s is given by s = 4t +1.6t20.08t3m.(a) Determine the boats velocity and acceleration att = 4 s.(b) What is the boats maximum velocity during thisinterval of time, and when does it occur?Solution:s = 4t +1.6t20.08t3v = dsdt= 4 +3.2t 0.24t2a) v(4s) = 12.96 m/s2a(4s) = 1.28 m/s2b) a = 3.2 0.48t = 0 t = 6.67sv(6.67s) = 14.67 m/sa = dvdt= 3.2 0.48t10c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.5 The rocket starts from rest at t = 0 andtravels straight up. Its height above the ground as afunction of time can be approximated by s = bt2+ct3,where b and c are constants. At t = 10 s, the rocketsvelocity and acceleration are v = 229 m/s and a = 28.2m/s2. Determine the time at which the rocket reachessupersonic speed (325 m/s). What is its altitude whenthat occurs?sSolution: The governing equations ares = bt2+ct3,v = 2bt +3ct2,a = 2b +6ct .Using the information that we have allows us to solve for the constantsb and c.(229 m/s) = 2b(10 s) +3c(10 s)2,(28.2 m/s2) = 2b +6c(10 s).Solving these two equations, we nd b = 8.80 m/s2, c = 0.177 m/s3.When the rocket hits supersonic speed we have(325 m/s) = 2(8.80 m/s2)t +3(0.177 m/s3)t2 t = 13.2 s.The altitude at this time iss = (8.80 m/s2)(13.2 s)2+(0.177 m/s3)(13.2 s)3s = 1940 m.Problem 13.6 The position of a point during the inter-val of time from t = 0 to t = 6 s is given by s = 12t3+6t2+4t m.(a) What is the maximum velocity during this intervalof time, and at what time does it occur?(b) What is the acceleration when the velocity is amaximum?Solution:s = 12t3+6t2+4t mv = 32t2+12t +4 m/sa = 3t +12 m/s2Maximum velocity occurs where a = dvdt= 0 (it could be a minimum)This occurs at t = 4 s. At this point dadt= 3 so we have a maximum.(a) Max velocity is at t = 4 s. where v = 28 m/s and(b) a = 0 m/s211c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.7 The position of a point during the inter-val of time from t = 0 to t = 3 seconds is s = 12 +5t2t3(a) What is the maximum velocity during this intervalof time, and at what time does it occur?(b) What is the acceleration when the velocity is amaximum?Solution:(a) The velocity is dsdt= 10t 3t2. The maximum occurs whendvdt= 10 6t = 0, from whicht = 106 = 1.667 seconds.This is indeed a maximum, since d2vdt2 = 6 < 0. The maximumvelocity isv =_10t 3t2_t =1.667 = 8.33 /s(b) The acceleration is dvdt= 0 when the velocity is a maximum.Problem 13.8 The rotating crank causes the positionof point P as a function of time to be s = 0.4 sin(2t ) m.(a) Determine the velocity and acceleration of P att = 0.375 s.(b) What is the maximum magnitude of the velocityof P?(c) When the magnitude of the velocity of P is amaximum, what is the acceleration of P?PsSolution:s = 0.4 sin(2t )v = dsdt= 0.8 cos(2t ) a) v(0.375s) = 1.777 m/sa(0.375) = 11.2 m/s2b) vmax = 0.8 = 2.513 m/s2c) vmax t = 0, n a = 0a = dvdt= 1.62sin(2t )Problem 13.9 For the mechanism in Problem 13.8,draw graphs of the position s, velocity v, and acce-leration a of point P as functions of time for 0 t 2 s.Using your graphs, conrm that the slope of the graphof s is zero at times for which v is zero, and the slopeof the graph of v is zero at times for which a is zero.Solution:12m.mc 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.10 A seismograph measures the horizon-tal motion of the ground during an earthquake. An engi-neer analyzing the data determines that for a 10-s intervalof time beginning at t = 0, the position is approximatedby s = 100 cos(2t ) mm. What are (a) the maximumvelocity and (b) maximum acceleration of the groundduring the 10-s interval?Solution:(a) The velocity isdsdt= (2)100 sin(2t ) mm/s = 0.2 sin(2t ) m/s.The velocity maxima occur atdvdt= 0.42cos(2t ) = 0,from which2t = (2n 1)2 , or t = (2n 1)4 ,n = 1, 2, 3, . . . M, where (2M 1)4 10 seconds.These velocity maxima have the absolute valuedsdtt =(2n1)4= [0.2] = 0.628 m/s.(b) The acceleration isd2sdt2 = 0.42cos(2t ).The acceleration maxima occur atd3sdt3 = d2vdt2 = 0.83sin(2t ) = 0,from which 2t = n, or t = n2, n = 0, 1, 2, . . . K, whereK2 10 seconds.These acceleration maxima have the absolute valuedvdtt =n2= 0.42= 3.95 m/s2.Problem 13.11 In an assembly operation, the robotsarm moves along a straight horizontal line. During aninterval of time from t = 0 to t = 1 s, the position of thearm is given by s = 30t220t3mm. (a) Determine themaximum velocity during this interval of time. (b) Whatare the position and acceleration when the velocity is amaximum?sSolution:s = 30t220t3mmv = 60t 60t2mm/sa = 60 120t mm/s2dadt= 120 mm/s3(a) Maximum velocity occurs when dvdt= a = 0. This occurs at0 = 60 120t or t = 1/2 second. (since da/dt < 0, we havea maximum). The velocity at this time isv = (60)_12_60_14_ mm/sv = 15 mm/s(b) The position and acceleration at this time ares = 7.5 2.5 mms = 5 mma = 0 mm/s213c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.12 In Active Example 13.1, the accelera-tion (in m/s2) of point P relative to point O is given asa function of time by a = 3t2. Suppose that at t = 0 theposition and velocity of P are s = 5 m and v = 2 m/s.Determine the position and velocity of P at t = 4 s.O Pss Solution: The governing equations area = (3 m/s4)t2v = 13(3 m/s4)t3+(2 m/s)s = 112(3 m/s4)t4+(2 m/s)t +(5 m)At t = 4 s, we have s = 77 m, v = 66 m/s.Problem 13.13 The Porsche starts from rest at timet = 0. During the rst 10 seconds of its motion, itsvelocity in km/h is given as a function of time by v =22.8t 0.88t2, where t is in seconds. (a) What is thecars maximum acceleration in m/s2, and when does itoccur? (b) What distance in km does the car travel dur-ing the 10 seconds?Solution: First convert the numbers into meters and seconds22.8kmhr_1000 m1 km__ 1hr3600 s_= 6.33 m/s0.88kmhr_1000 m1 km__ 1 hr3600 s_= 0.244 m/sThe governing equations are thens = 12(6.33 m/s)(t2/s) 13(0.88 m/s)(t3/s2),v = (6.33 m/s)(t /s) (0.88 m/s)(t2/s2),a = (6.33 m/s2) 2(0.88 m/s)(t /s2),ddta = 2(0.88 m/s)(1/s2) = 1.76 m/s3.The maximum acceleration occurs at t = 0 (and decreases linearlyfrom its initial value).amax = 6.33 m/s2@ t = 0In the rst 10 seconds the car travels a distances =_12_22.8kmhr_ (10 s)2s 13_0.88kmhr_ (10 s)3s2_ _ 1 hr3600 s_s = 0.235 km.14c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.14 The acceleration of a point is a =20t m/s2. When t = 0, s = 40 m and v = 10 m/s.What are the position and velocity at t = 3 s?Solution: The velocity isv =_ a dt +C1,where C1 is the constant of integration. Thusv =_ 20t dt +C1 = 10t2+C1.At t = 0, v = 10 m/s, hence C1 = 10 and the velocity is v =10t210 m/s. The position iss =_ v dt +C2,where C2 is the constant of integration.s =_ (10t210) dt +C2 =_103_t310t +C2.At t = 0, s = 40 m, thus C2 = 40. The position iss =_103_t310t +40 m.At t = 3 seconds,s =_103 t310t +40_t =3= 100 m.The velocity at t = 3 seconds isv =_10t210_t =3 = 80 m/s .Problem 13.15 The acceleration of a point is a =60t 36t2 2t = 0, s = 0 and v = 20 m/s.What are position and velocity as a function of time?Solution: The velocity isv =_ a dt +C1 =_ (60t 36t2) +C1 = 30t212t3+C1.At t = 0, v = 20 m/s, hence C1 = 20, and the velocity as a functionof time isv = 30t212t3+20 m/s .The position iss =_ v dt +C2 =_ (30t212t3+20) +C2= 10t33t4+20t +C2.At t = 0, s = 0, hence C2 = 0, and the position iss = 10t33t4+20t15m/s . Whenmc 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.16 As a rst approximation, a bioengineerstudying the mechanics of bird ight assumes that thesnow petrel takes off with constant acceleration. Videomeasurements indicate that a bird requires a distance of4.3 m to take off and is moving at 6.1 m/s when it does.What is its acceleration?Solution: The governing equations area = constant, v = at, s = 12at2.Using the information given, we have6.1 m/s = at, 4.3 m = 12at2.Solving these two equations, we nd t = 1.41 s and a = 4.33 m/s2.Problem 13.17 Progressively developing a more real-istic model, the bioengineer next models the accelerationof the snow petrel by an equation of the form a =C(1 +sin t ), where C and are constants. From videomeasurements of a bird taking off, he estimates that = 18/s and determines that the bird requires 1.42 sto take off and is moving at 6.1 m/s when it does. Whatis the constant C?Solution: We nd an expression for the velocity by integrating theaccelerationa = C(1 +sin t ),v = Ct + C(1 cos t ) = C_t + 1 1cos t_.Using the information given, we have6.1 m/s = C_1.42 s + s18 s18 cos[18(1.42)]_Solving this equation, we nd C = 4.28 m/s2.16c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.18 Missiles designed for defense againstballistic missiles have attained accelerations in excessof 100 gs, or 100 times the acceleration due to gravity.Suppose that the missile shown lifts off from the groundand has a constant acceleration of 100 gs. How longdoes it take to reach an altitude of 3000 m? How fast isit going when it ranches that altitude?Solution: The governing equations area = constant, v = at, s = 12at2Using the given information we have3000 m = 12100(9.81 m/s2)t2 t = 2.47 sThe velocity at that time isv = 100(9.81 m/s2)(2.47 s) = 243 m/s. v = 2430 m/s.Problem 13.19 Suppose that the missile shown liftsoff from the ground and, because it becomes lighter asits fuel is expended, its acceleration (in gs) is given asa function of time in seconds bya = 1001 0.2t.What is the missiles velocity in kilometres per hour 1safter liftoff?Solution: We nd an expression for the velocity by integrating theacceleration (valid only for 0 < t < 5s).v =_ t0adt =_ t0100g1 0.2(t /s)dt = 100gs0.2 ln_ 11 0.2[t /s]_At time t = 1 s, we havev = 100( . 2)(1 s)0.2 ln_ 11 0.2_ = .v = 0 m h.179 81 m/s1095 m/s394 k /c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.20 The airplane releases its drag para-chute at time t = 0. Its velocity is given as a functionof time byv = 801 +0.32tm/s.What is the airplanes acceleration at t = 3 s?Solution:v = 801 +0.32t; a = dvdt= 25.6(1 +0.32t )2 a(3 s) = 6.66 m/s2Problem 13.21 How far does the airplane in Problem13.20 travel during the interval of time from t = 0 tot = 10 s?Solution:v = 801 +0.32t; s =_ 10 s0801 +0.32tdt = 250 ln_1 +3.21_= 359 mProblem 13.22 The velocity of a bobsled is v =10t t = 2 s, the position is s = 25 m. Whatis its position at t = 10 s?Solution: The equation for straight line displacement under con-stant acceleration iss = a(t t0)22 +v(t0)(t t0) +s(t0).Choose t0 = 0. At t = 2, the acceleration isa =_dv(t )dt_t =2= 10 m/s2,the velocity is v(t0) = 10(2) = 20 m/s, and the initial displacement iss(t0) = 25 m. At t = 10 seconds, the displacement iss = 102 (10 2)2+20(10 2) +25 = 505 m18m/s. Whenc 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.23 In September, 2003, Tony Schumacherstarted from rest and drove 402 km in 4.498 seconds ina National Hot Rod Association race. His speed as hecrossed the nish line wa 5 8 /h. Assume that thecars acceleration can be expressed by a linear functionof time a = b +ct .(a) Determine the constants b and c.(b) What was the cars speed 2 s after the start ofthe race?Solution:a = b +ct , v = bt + ct22 , s = bt22 + ct36Both constants of integration are zero.(a) 528 km/h = b(4.498 s) + c2(4.498 s)2= b2(4.498 s)2+ c6(4.498 s)3 b = 2c = 3(b) v = b(2 s) + c2(2 s)2= 9 /sProblem 13.24 The velocity of an object is v = 200 2t2m/s. When t = 3 seconds, its position is s = 600 m.What are the position and acceleration of the object att = 6 s?Solution: The acceleration isdv(t )dt= 4t m/s2.At t = 6 seconds, the acceleration is a = 24 m/s2. Choose the ini-tial conditions at t0 = 3 seconds. The position is obtained from thevelocity:s(t t0) =_ 63v(t ) dt +s(t0) =_200t 23t3_63+600 = 1070 m .Problem 13.25 An inertial navigation system mea-sures the acceleration of a vehicle from t = 0 to t = 6 sand determines it to be a = 2 +0.1t m/s2. At t = 0,the vehicles position and velocity are s = 240 m, v =42 m/s, respectively. What are the vehicles position andvelocity at t = 6 s?Solution:a = 2 +0.1t m/s2v0 = 42 m/s s0 = 240 mIntegratingv = v0+2t +0.1t2/2s = v0t +t2+0.1t3/6 +s0Substituting the known values at t = 6 s, we getv = 55.8 m/ss = 531.6 m1940254 m/s9.5 m/s8 ms 2 kmkmc 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.26 In Example 13.3, suppose that the chee-tahs acceleration is constant and it reaches its top speedof 120 km/h in 5 s. What distance can it cover in 10 s?Solution: The governing equations while accelerating area = constant, v = at, s = 12at2.Using the information supplied, we have_ _ = a(5 s) a = 2The distance that he travels in the rst 10 s (5 seconds acceleratingand then the last 5 seconds traveling at top speed) iss = 12( 2) (5 s)2+_ _ (10 s 5 s) = .s = .Problem 13.27 The graph shows the airplanes accel-eration during its takeoff. What is the airplanes velocitywhen it rotates (lifts off) at t = 30 s?at9 m/ s23 m/s25 s 0030 sSolution: Velocity = Area under the curvev =12(3 m/s2+9 m/s2)(5 s) +(9 m/s2)(25 s) =255 m/sProblem 13.28 Determine the distance traveled duringits takeoff by the airplane in Problem 13.27.Solution: for 0 t 5 sa =_6 m/s25 s_t +(3 m/s2), v =_6 m/s25 s_ t22 +(3 m/s2)ts =_6 m/s25 s_ t36 +(3 m/s2)t22v(5 s) = 30 m/s, s(5 s) = 62.5 mfor 5 s t 30 sa = 9 m/s2, v = (9 m/s2)(t 5 s) +30 m/s,s =(9 m/s2)(t 5 s)22 +(30 m/s)(t 5 s) +62.5 m s(30 s) = 362520120 1000 m3600 s 6. 67 m/s6.67 m/s 120 1000 m3600 s m 250m 250mc 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.29 The car is traveling at 48 km/h whenthe trafc light 90 m ahead turns yellow. The drivertakes one second to react before he applies the brakes.(a) After he applies the brakes, what constant rate ofdeceleration will cause the car to come to a stopjust as it reaches the light?(b) How long does it take the car to travel the 90 m?90 m /h 48 kmSolution: for 0 t 1 sa = 0, v = = , s = ( )ts(1 s) =for t > 1 sa = c (constant), v = ct + , s = ct22 +( )t +At the stop we have90 m = ct22 +( )t + a) c = . 7 /s2b) t = 11.41 s0 = ct +Problem 13.30 The car is traveling at 48 km/h whenthe trafc light 90 m ahead turns yellow. The drivertakes 1 s to react before he applies the accelerator. Ifthe car has a constant acceleration of 2 m/s2and thelight remains yellow for 5 s, will the car reach the lightbefore it turns red? How fast is the car moving when itreaches the light?Solution: First, convert the initial speed into m/s.= .At the end of the 5 s, the car will have traveled a distanced = ( )(1 s) +

12 ( 2)(5 s 1 s)2+( )(5 s 1 s)

= .When the light turns red, the driver will still be 7.35 m from the light.No.To nd the time at which the car does reach the light, we solve90 m=( )(1 s) +

12 ( 2)(t 1 s)2+ ( )(t 1 s)

t = 5. .The speed at this time isv = +( 2) (5. 1 s) =v = . m h.2148 km/h 13.33 m/s 13.33 m/s13. 33 m13.33 m/s 13.33 m13. 33 m/s1 1 m48 km/h 13.33 m/s13. 33 m/s 2 m/s 13.33 m/s m 82.6513.33 m/s 2 m/s 13.33 m/s34 s13.33 m/s 2 m/s 34 s 22.01 m/s.79 2 k /13.33 m/s 13.33 m/s13.33 mc 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.31 A high-speed rail transportation sys-tem has a top speed of 100 m/s. For the comfort of thepassengers, the magnitude of the acceleration and decel-eration is limited to 2 m/s2. Determine the time requiredfor a trip of 100 km.Strategy: A graphical approach can help you solve thisproblem. Recall that the change in the position from aninitial time t0 to a time t is equal to the area denedby the graph of the velocity as a function of time fromt0 to t .Solution: Divide the time of travel into three intervals: The timerequired to reach a top speed of 100 m/s, the time traveling at topspeed, and the time required to decelerate from top speed to zero. Fromsymmetry, the rst and last time intervals are equal, and the distancestraveled during these intervals are equal. The initial time is obtainedfrom v(t1) = at1, from which t1 = 100/2 = 50 s. The distance trav-eled during this time is s(t1) = at21/2 from which s(t1) = 2(50)2/2 =2500 m. The third time interval is given by v(t3) = at3+100 = 0,from which t3 = 100/2 = 50 s. Check. The distance traveled is s(t3) =a2t23 +100t3, from which s(t3) = 2500 m. Check. The distance trav-eled at top speed is s(t2) = 100000 2500 2500 = 95000 m= 95 km. The time of travel is obtained from the distance traveled atzero acceleration: s(t2) = 95000 = 100t2, from which t2 = 950. Thetotal time of travel is ttotal = t1+t2+t3 = 50 +950 +50 = 1050 s= 17.5 minutes .A plot of velocity versus time can be made and the area under thecurve will be the distance traveled. The length of the constant speedsection of the trip can be adjusted to force the length of the trip to bethe required 100 km.Problem 13.32 The nearest star, Proxima Centauri, is4.22 light years from the Earth. Ignoring relative motionbetween the solar system and Proxima Centauri, sup-pose that a spacecraft accelerates from the vicinity ofthe Earth at 0.01 g (0.01 times the acceleration due togravity at sea level) until it reaches one-tenth the speedof light, coasts until it is time to decelerate, then decel-erates at 0.01 g until it comes to rest in the vicinity ofProxima Centauri. How long does the trip take? (Lighttravels at 3 108m/s.)Solution: The distance to Proxima Centauri isd = (4.22 light - year)(3 108m/s)(365.2422 day)_86400 s1 day_= 3.995 1016m.Divide the time of ight into the three intervals. The time required toreach 0.1 times the speed of light ist1 = va= 3 107m/s0.0981 m/s2 = 3.0581 108seconds.The distance traveled iss(t1) = a2t21 +v(0)t +s(0),where v(0) = 0 and s(0) = 0 (from the conditions in the problem),from which s(t1) = 4.587 1015m. From symmetry, t3 = t1, ands(t1) = s(t3). The length of the middle interval is s(t2) = d s(t1) s(t3) = 3.0777 1016m. The time of ight at constant velocity ist2 = 3.0777 1016m3 107 = 1.026 109seconds.The total time of ight is ttotal = t1+t2+t3 = 1.63751 109sec-onds. In solar years:ttotal =_1.63751 109sec__ 1 solar years365.2422 days__ 1 days86400 sec_= 51.9 solar years22c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.33 A race car starts from rest and accel-erates at a = 5 +2t 2for 10 seconds. The brakesare then applied, and the car has a constant acceler-ation a = 30 m/s2until it comes to rest. Determine(a) the maximum velocity, (b) the total distance trav-eled; (c) the total time of travel.Solution:(a) For the rst interval, the velocity isv(t ) =_ (5 +2t ) dt +v(0) = 5t +t2since v(0) = 0. The velocity is an increasing monotone function;hence the maximum occurs at the end of the interval, t = 10 s,from whichvmax = 150 m/s .(b) The distance traveled in the rst interval iss(10) =_100 (5t +t2) dt =_52t2+ 13t3_100= 583.33 .The time of travel in the second interval isv(t210) = 0 = a(t210) +v(10), t2 10 s,from which(t210) = 15030 = 5, and(c) the total time of travel is t2 = 15. The total distance traveled iss(t210) = a2(t210)2+v(10)(t210) +s(10),from which (b)s(5) = 302 52+150(5) +583.33 = 958.33 mProblem 13.34 When t = 0, the position of a point iss = 6 m and its velocity is v = 2 m/s. From t = 0 to t =6 s, the acceleration of the point is a = 2 +2t2m/s2.From t = 6 s until it comes to rest, its acceleration isa = 4 m/s2.(a) What is the total time of travel?(b) What total distance does the point move?Solution: For the rst interval the velocity isv(t ) =_ (2 +2t2) dt +v(0) =_2t + 23t3_+2 m/s.The velocity at the end of the interval is v(6) = 158 m/s. The dis-placement in the rst interval iss(t ) =_ _2t + 23t3+2_ dt +6 =_t2+ 16t4+2t_+6.The displacement at the end of the interval is s(6) = 270 m. For thesecond interval, the velocity is v(t 6) = a(t 6) +v(6) = 0, t 6,from which(t 6) = v(6)a= 1584 = 39.5.The total time of travel is(a) ttotal = 39.5 +6 = 45.5 seconds.(b) The distance traveled iss(t 6) = 42 (t 6)2+v(6)(t 6) +s(6)= 2(39.5)2+158(39.5) +270,from which the total distance is stotal = 3390 m23m/smc 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.35 Zoologists studying the ecology of theSerengeti Plain estimate that the average adult cheetahcan run 100 km/h and that the average springbuck canrun 65 km/h. If the animals run along the same straightline, start at the same time, and are each assumed tohave constant acceleration and reach top speed in 4 s,how close must the a cheetah be when the chase beginsto catch a springbuck in 15 s?Solution: The top speeds are Vc = 100 km/h = 27.78 m/s for thecheetah, and Vs = 65 km/h = 18.06 m/s. The acceleration is ac =Vc4 = 6.94 m/s2for the cheetah, and as = Vs4 = 4.513 m/s2for thespringbuck. Divide the intervals into the acceleration phase and thechase phase. For the cheetah, the distance traveled in the rst issc(t ) = 6.942 (4)2= 55.56 m. The total distance traveled at the endof the second phase is stotal = Vc(11) +55.56 = 361.1 m. For thespringbuck, the distance traveled during the acceleration phase isss(t ) = 4.5132 (4)2= 36.11 m. The distance traveled at the end of thesecond phase is ss(t ) = 18.06(11) +36.1 = 234.7 m. The permissibleseparation between the two at the beginning for a successful chase isd = sc(15) ss(15) = 361.1 234.7 = 126.4 mProblem 13.36 Suppose that a person unwisely drivescers beginconstant acceleration at the instant they are passed andincrease their speed to 129 km/h in 4 s , how long does ittake them to be even with the pursued car?Solution: The conversion from mi/h to m/s isThe acceleration of the police car isa = ( )( . )4 s = 2.The distance traveled during acceleration iss(t1) = .2 (4)2+ ( . )(4) = .The distance traveled by the pursued car during this acceleration issc(t1) = ( . ) t1= (4) =The separation between the two cars at 4 seconds isd = = .This distance is traversed in the timet2 = ( )( . )= .The total time is ttotal = + 4 = seconds.Problem 13.37 If = 1 rad and ddt= 1 rad/s, whatis the velocity of P relative to O?Strategy: You can write the position of P relative toO as s = (2 m) cos +(2 m) cos and then take thederivative of this expression with respect to time todetermine the velocity.s2 mOP2 mSolution: The distance s from point O iss = (2 m) cos +(2 m) cos .The derivative isdsdt= 4 sin ddt.For = 1 radian and ddt= 1 radian/second,dsdt= v(t ) = 4(sin(1 rad)) = 4(0.841) = 3.37 m/s24120 km/h in a 88 km/h zone and passes a police car going88 km/h in the same direction. If the police ofkmh = 1000 m3600 s = 0.278 m/s129 88 0 278 m/s2.85 m/s2 8588 0 278 m 121120 0 278 33. 36 . 133. 4 m133.4 121 12.4 m12.4129 88 0 278 1. 091. 09 5. 09c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.38 In Problem 13.37, if = 1 rad, d/dt= 2 rad/s and d2/dt2= 0, what are the velocity andacceleration of P relative to O?Solution: The velocity isdsdt= 4 sin ddt= 4(sin(1 rad))(2) = 6.73 m/s .The acceleration isd2sdt2 = 4 cos _ddt_24 sin _d2dt2_,from whichd2sdt2 = a = 4 cos(1 rad)(4) = 8.64 m/s2Problem 13.39 If = 1 rad and ddt= 1 rad/s, whatis the velocity of P relative to O?sOP200 mm400 mmSolution: The acute angle formed by the 400 mm arm with thehorizontal is given by the sine law:200sin = 400sin ,from whichsin =_200400_sin .For = 1 radian, = 0.4343 radians. The position relative to O is.s = 200 cos +400 cos .The velocity isdsdt= v(t ) = 200 sin _ddt_400 sin _ddt_.From the expression for the angle, cos _ddt_= 0.5 cos _ddt_,from which the velocity isv(t ) = (200 sin 200 tan cos )_ddt_.Substitute: v(t ) = 218.4 mm/s .25c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.40 In Active Example 13.4, determine thetime required for the planes velocity to decrease from50 m/s to 10 m/s.Solution: From Active Example 13.4 we know that the accelera-tion is given bya = (0.004/m)v2.We can nd an expression for the velocity as a function of time byintegratinga = dvdt= (0.004/m)v2 dvv2 = (0.004/m)t_ 10 m/s50 m/sdvv2 =_1v_10 m/s50 m/s=_ 1 s10 m + 1 s50 m_=_ 4 s50 m_= (0.004/m)tt =_ 4 s50 m__1 m0.004_= 20 s.t = 20 s.Problem 13.41 An engineer designing a system to con-trol a router for a machining process models the sys-tem so that the routers acceleration (in cm/s ) during an2interval of time is given by a = 0.4v, where v is thevelocity of the router in cm/s. When t = 0, the position iss =0 and the velocity is v =2 cm/s. What is the positionat t = 3 s?sSolution: We will rst nd the velocity at t = 3.a = dvdt= _0.4s_v,_ v22 cm/sdvv=_ 3 s0_0.4s_dt,ln_ v22 cm/s_=_0.4s_(3 s) = 1.2v2 = (2 cm/s)e1 . 2= 0.602 cm/s.Now we can nd the positiona = vdvds=_0.4s_v,_ 0.602 cm/s2 cm/svdvv=_0.4s__ s20ds(0.602 cm/s) (2 cm/s)=_0.4s_s2s2 = 1.398 cm/s0.4/ s = 3.49 cms2 = 3.49 cm26c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.42 The boat is moving at 10 m/s whenits engine is shut down. Due to hydrodynamic drag, itssubsequent acceleration is a = 0.05v2m/s2, where vis the velocity of the boat in m/s. What is the boatsvelocity 4 s after the engine is shut down?Solution:a = dvdt= (0.05 m1)v2_ v10 m/sdvv2 = (0.05 m1)_ t0dt 1vv10 m/s= (0.05 m1)tv = 10 m/s1 +(0.5 s1)tv(4 s) = 3.33 m/sProblem 13.43 In Problem 13.42, what distance doesthe boat move in the 4 s following the shutdown ofits engine?Solution: From Problem 13.42 we knowv = dsdt= 10 m/s1 +(0.5 s1)t s(4 s) =_ 4 s010 m/s1 +(0.5 s1)tdts(4 s) = (20 m) ln_2 +(1 s1)(4 s)2_= 21.97 mProblem 13.44 A steel ball is released from rest in acontainer of oil. Its downward acceleration is a = 2. 4 0.6v 2v is the balls velocity in cm/s. What isthe balls downward velocity 2 s after it is released?Solution:a = dvdt= (2.4 cm/s) (0.6 s1)v_ v0dv(2.4 cm /s) (0.6 s1)v=_ t0dt 53 ln_v +4 cm/s4 cm/s_= t v = (4 cm/s)_1 e(0.6 s1)t_v(2 s) = 2.795 cm/sProblem 13.45 In Problem 13.44, what distance doesthe ball fall in the rst 2 s after its release?Solution: From 13.44 we knowv = dsdt= (4 cm/s)_1 e(0.6 s1)t_s(2 s) =_ t0(4 cm/s)_1 e(0.6 s1)t_dt= 20 cm3_e(0.6 s1)t1_+(4 cm/s )ts(2 s) = 3.34 cm.27cm/s , wherec 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.46 The greatest ocean depth yet discov-ered is the Marianas Trench in the western Pacic Ocean.A steel ball released at the surface requires 64 minutesto reach the bottom. The balls downward accelerationis a = 0.9g cv, where g = 9.81 m/s2and the constantc = 3.02 s1. What is the depth of the Marianas Trenchin kilometers?Solution:a = dvdt= 0.9g cv.Separating variables and integrating,_ v0dv0.9g cv=_ t0dt = t .Integrating and solving for v,v = dsdt= 0.9gc(1 ect).Integrating,_ s0ds =_ t00.9gc(1 ect) dt.We obtains = 0.9gc_t + ectc 1c_.At t = (64)(60) = 3840 s, we obtains = 11,225 m.Problem 13.47 The acceleration of a regional airlinerduring its takeoff run is a = 14 0.0003v2 2v is its velocity in /s. How long does it take the airlinerto reach its takeoff speed of 200 m/s?Solution:a = dvdt= (14 /s2) (0.0003 m1)v2_ 200 m/s0dv(14 m/s2) (0.0003 m1)v2 =_ t0dtt = 25.1 sProblem 13.48 In Problem 13.47, what distance doesthe airliner require to take off?Solution:a = vdvds= (14 m/s2) (0.0003 m1)v2_ 200 m/s0vdv(14 m/s2) (0.0003 m1)v2 =_ s0dss = 324328m/s , wherem mmc 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.49 A sky diver jumps from a helicopterand is falling straight down at 30 m/s when her parachuteopens. From then on, her downward acceleration isapproximately a = g cv2, where g = 9.81 m/s2andc is a constant. After an initial transient period shedescends at a nearly constant velocity of 5 m/s.(a) What is the value of c, and what are its SI units?(b) What maximum deceleration is the sky diversubjected to?(c) What is her downward velocity when she has fallen2 meters from the point at which her parachuteopens?Solution: Assume c > 0.(a) After the initial transient, she falls at a constant velocity, so thatthe acceleration is zero and cv2= g, from whichc = gv2 = 9.81 m/s2(5)2m2/s2 = 0.3924 m1(b) The maximum acceleration (in absolute value) occurs when theparachute rst opens, when the velocity is highest:amax = |g cv2| = |g c(30)2| = 343.4 m/s2(c) Choose coordinates such that distance is measured positive down-ward. The velocity is related to position by the chain rule:dvdt= dvdsdsdt= vdvds= a,from whichv dvg cv2 = ds.Integrate:_ 12c_ln |g cv2| = s +C.When the parachute opens s = 0 and v = 30 m/s, from whichC = _ 12c_ln |g 900c| = 7.4398.The velocity as a function of distance is ln |g cv2| =2c(s +C). For s = 2 m,v = 14.4 m/sProblem 13.50 The rocket sled starts from rest andaccelerates at a = 30 +2t m/s2until its velocity is400 m/s. It then hits a water brake and its accelerationis a = 0.003v2m/s2until its velocity decreases to100 m/s. What total distance does the sled travel?Solution: Acceleration Phasea = 30 +2t m/s2v = 30t +t2m/ss = 15t2+t3/3 mWhen v = 400 m/s, acceleration ends. At this point, t = 10 s and s =1833 m. Deceleration Phase starts at s1 = 1833 m, v1 = 400 m/s. Letus start a new clock for the deceleration phase. vf = 100 m/sa = vdvds= 0.003v2_ sfs1ds = 1(0.003)_ vfv1v dvv2sf 1833 m = 1(0.003)[ln(100) ln(400)]sf = 2300 m29c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.51 In Problem 13.50, what is the sledstotal time of travel?Solution: From the solution to Problem 13.50, the accelerationtakes 10 s. At t = 10 s, the velocity is 400 m/s. We need to nd outhow long it takes to decelerate from 400 m/s to 100 m/s and add thisto the 10 s required for acceleration. The deceleration is given asa = dvdt= 0.003v2m/s20.003_ td0dt =_ 100400dvv20.003td = 1v100400= _ 1100 1400_0.003td = 3400td = 2.5 st = 10 +td = 12.5 sProblem 13.52 A cars acceleration is related to itsposition by a = 0.01s m/s2. When s = 100 m, the caris moving at 12 m/s. How fast is the car moving whens = 420 m?Solution:a = vdvds= 0.01s m/s2_ vf12v dv = 0.01_ 420100s ds_v22_vf12 m/s= 0.01_s22_420 m100 mv2f2 = 1222 +0.01(42021002)2vf = 42.5 m/sProblem 13.53 Engineers analyzing the motion of alinkage determine that the velocity of an attachmentpoint is given by v = A +4s2A is a con-stant. When s = 2 m, its acceleration is measured anddetermined to be a = 320 m/s . What is its velocity of2the point when s = 2 m?Solution: The velocity as a function of the distance isvdvds= a.Solve for a and carry out the differentiation.a = vdvds= (A+4s2)(8s).When s = 2 m, a = 320 m/s , from which2A = 4.The velocity at s = 2 m isv = 4 +4(22) = 20 /s30m/s, wheremc 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.54 The acceleration of an object is givenas a function of its position in feet by a = 2 s2( 2).When s = 0, its velocity is v = 1 m/s. What is the veloc-ity of the object when s = 2 m?Solution: We are givena = vdvds=_ 22_s2,_ v1 m/svdv =_ 22__ 2m0s2dsv22 (1 m/s)22 =_ 22_ (2 m)33v = 3.42 /s.Problem 13.55 Gas guns are used to investigatethe properties of materials subjected to high-velocityimpacts. A projectile is accelerated through the barrelof the gun by gas at high pressure. Assume that theacceleration of the projectile is given by a = c/s, wheres is the position of the projectile in the barrel in metersand c is a constant that depends on the initial gas pressurebehind the projectile. The projectile starts from rest ats = 1.5 m and accelerates until it reaches the end of thebarrel at s = 3 m. Determine the value of the constantc necessary for the projectile to leave the barrel with avelocity of 200 m/s.sSolution:a = vdvds= cs, _ 200 m/s0vdv =_ 3 m1.5 mcsds (200 m/s)22 = c ln_ 3 m1.5 m_c = 28.85 103m2/s2Problem 13.56 If the propelling gas in the gas gundescribed in Problem 13.55 is air, a more accuratemodeling of the acceleration of the projectile is obtainedby assuming that the acceleration of the projectile isgiven by a = c/s, where = 1.4 is the ratio of specicheats for air. (This means that an isentropic expansionprocess is assumed instead of the isothermal processassumed in Problem 13.55.) Determine the value of theconstant c necessary for the projectile to leave the barrelwith a velocity of 200 m/s.Solution:a = vdvds= cs1.4, _ 200 m/s0vdv =_ 3 m1.5 mcs1.4ds(200 m/s)22 = 2.5c_(3m)0.4(1.5m)0.4_c = 38.86 103m2.4/s231m/sm-sm s -m-smc 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.57 A spring-mass oscillator consists of amass and a spring connected as shown. The coordinates measures the displacement of the mass relative to itsposition when the spring is unstretched. If the springis linear, the mass is subjected to a deceleration pro-portional to s. Suppose that a = 4s m/s2, and that yougive the mass a velocity v = 1 m/s in the position s = 0.(a) How far will the mass move to the right before thespring brings it to a stop?(b) What will be the velocity of the mass when it hasreturned to the position s = 0?sSolution: The velocity of the mass as a function of its posi-tion is given by v dv/ds = a. Substitute the given acceleration, sepa-rate variables, and integrate: v dv = 4s ds, from which v2/2 =2s2+C. The initial velocity v(0) = 1 m/s at s = 0, from whichC = 1/2. The velocity is v2/2 = 2s2+1/2.(a) The velocity is zero at the position given by0 = 2(s1)2+ 12,from which s1 = _14 = 12 m.Since the displacement has the same sign as the velocity,s1 = +1/2 mis the distance traveled before the spring brings it to a stop.(b) At the return to s = 0, the velocity is v = _22 = 1 m/s. Fromthe physical situation, the velocity on the rst return is negative(opposite the sign of the initial displacement),v = 1 m/s .Problem 13.58 In Problem 13.57, suppose that att = 0 you release the mass from rest in the positions = 1 m. Determine the velocity of the mass as a func-tion of s as it moves from the initial position to s = 0.Solution: From the solution to Problem 13.57, the velocity as afunction of position is given byv22 = 2s2+C.At t = 0, v = 0 and s = 1 m, from which C = 2(1)2= 2. The velocityis given byv(s) = (4s2+4)12 = 21 s2m/s.From the physical situation, the velocity is negative (opposite the signof the initial displacement):v = 21 s2m/s .[Note: From the initial conditions, s2 1 always.]32c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.59 A spring-mass oscillator consists of amass and a spring connected as shown. The coordinates measures the displacement of the mass relative to itsposition when the spring is unstretched. Suppose thatthe nonlinear spring subjects the mass to an accelerationa = 4s2s3m/s2and that you give the mass a velocityv = 1 m/s in the position s = 0.(a) How far will the mass move to the right before thesprings brings it to a stop?(b) What will be the velocity of the mass when it hasreturned to the position s = 0?sSolution:(a) Find the distance when the velocity is zero.a = vdvds=_4s2_s +_ 2m2s2_s3_ 01 m/svdv =_ d0__4s2_s +_ 2m2s2_s3_ds (1 m/s)22 = _ 4s2_ d22 _ 2m2s2_ d44Solving for d we nd d = 0.486 m.(b) When the cart returns to the position s = 0, we havea = vdvds=_4s2_s +_ 2m2s2_s3_ v1 m/svdv =_ 00__4s2_s +_ 2m2s2_s3_ds = 0v22 (1 m/s)22 = 0, v = 1 m/s.The cart will be moving to the left, so we choose v = 1 m/s.Problem 13.60 The mass is released from rest withthe springs unstretched. Its downward acceleration isa = 32.2 50s 2s is the position of the massmeasured from the position in which it is released.(a) How far does the mass fall? (b) What is themaximum velocity of the mass as it falls?sSolution: The acceleration is given bya = dvdt= dvdsdsdt= vdvds= 32.2 50s 2.Integrating, we get_ v0v dv =_ s0(32.2 50s) ds or v22 = 32.2s 25s2.(a) The mass falls until v = 0. Setting v = 0, we get 0 = (32.2 25s)s. We nd v = 0 at s = 0 and at s = 1.288 m. Thus, themass falls 1.288 m before coming to rest.(b) From the integration of the equation of motion, we have v2=2(32.2s 25s2). The maximum velocity occurs where dvds= 0.From the original equation for acceleration, we have a = vdvds=(32.2 50s) 2assume that v = 0 at this point. Thus, 0 = (32.2 50s), or s =(32.2/50) v =vMAX. Substituting this value for s intothe equation for v, we getv2MAX = 2_(32.2)250 (25)(32.2)2502_,or vMAX = 4.55 /s33m/s , wherem/sm/s . Since we want maximum velocity, we canm whenmc 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.61 Suppose that the mass in Problem13.60 is in the position s = 0 and is given a downwardvelocity of 10 m/s.(a) How far does the mass fall?(b) What is the maximum velocity of the mass asit falls?Solution:a = vdvds= (32.2 m/s2) (50 s2)s_ v10 m/svdv =_ s0[(32.2 m/s2) (50 s2)s]dsv22 (10 m/s)22 = (32.2 m/s2) s (50 s2)s22v2= (10 m/s)2+(64.4 m/s2)s (50 s2)s2(a) The mass falls until v = 00 = (10 m/s)2+(64.4 m/s2)s (50 s2)s2s = 2.20 m(b) The maximum velocity occurs when a = 00 = (32.2 m/s2) (50 s2)s s = 0.644 mv2= (10 m/s)2+(64.4 m/s2)(0.644 m) (50 s2)(0.644 m)2v = 10.99 /sProblem 13.62 If a spacecraft is 161 km above the sur-face of the earth, what initial velocity v0 straight awayfrom the earth would be required for the vehicle toreach the moons orbit , m from the center ofthe earth? The radius of the earth is 6372 km. Neglectthe effect of the moons gravity. (See Example 13.5.)0161 km382,942 kmSolution: For computational convenience, convert the accelerationdue to Earths gravity into the units given in the problem, namely milesand hours:g =_ .1 s2__ 1 km __36002s21 h2_= . 2.The velocity as a function of position is given byvdvds= a = gR2Es2 .Separate variables,v dv = gR2Edss2 .Integrate:v2= 2gR2E_12_+C.Suppose that the velocity at the distance of the Moons orbit is zero.Then0 = 2( . )_ 2 _+C,from which C = 2/h . At the 161 km altitude, the2equation for the velocity isv20 = 2 g_ R2ERE+161_+C.From whichv0 = =Converting:v0 =_ 1 h__ 1 km__ 1 h3600 s_= ,Check: Use the result of Example 13.5v0 =_2gR2E_ 1s0 1H_,(where H > s0 always), and H = , ,from which v0 = ,4 m/h. check.34m382 942 k9 81 m1000 m 127137 6 km/h127137 6 637238294226960164 km1553351991 39,413 km/h39413 km 1000 m10 948 m/s .382 94239 13 kc 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.63 The moons radius is 1738 km. Themagnitude of the acceleration due to gravity of the moonat a distance s from the center of the moon is4.89 1012s2 m/s2.Suppose that a spacecraft is launched straight up fromthe moons surface with a velocity of 2000 m/s.(a) What will the magnitude of its velocity be when itis 1000 km above the surface of the moon?(b) What maximum height above the moons surfacewill it reach?Solution: Set G = 4.89 1012m3/s2, r0 = 1.738 106m,v0 = 2000 m/sa = vdvds= Gs2 _ vv0vdv = _ rr0Gs2ds v22 v022 = G_1r 1r0_v2= v02+2G_r0rrr0_(a) v(r0+1.0 106m) = 1395 m/s(b) The maximum velocity occurs when v = 0r = 2G2Gr0v02 = 6010 km h = r r0 = 4272 kmProblem 13.64* The velocity of an object subjectedonly to the earths gravitational eld isv =_v20 +2gR2E_1s 1s0__1/2,where s is the objects position relative to the center ofthe earth, v0 is the velocity at position s0, and RE is theearths radius. Using this equation, show that the objectsacceleration is given as a function of s by a = gR2E/s2.Solution:v = v20 +2gR2E_1s 1s0_1/2a = dvdt= vdvdsRewrite the equation given as v2= v20 + 2gR2Es 2gR2Es0Take the derivative with respect to s.2vdvds= 2gR2Es2Thusa = vdvds gR2Es2Problem 13.65 Suppose that a tunnel could be drilledstraight through the earth from the North Pole to theSouth Pole and the air was evacuated. An object droppedfrom the surface would fall with the acceleration a =gs/RE, where g is the acceleration of gravity at sealevel, RE is radius of the earth, and s is the distance of theobject from the center of the earth. (The acceleration dueto gravitation is equal to zero at the center of the earthand increases linearly with the distance from the center.)What is the magnitude of the velocity of the droppedobject when it reaches the center of the earth?SNTunnelREsSolution: The velocity as a function of position is given byvdvds= gsRE.Separate variables and integrate:v2= _ gRE_s2+C.At s = RE, v = 0, from which C = gRE. Combine and reduce:v2= gRE_1 s2R2E_At the center of the earth s = 0, and the velocity is v =gRE35c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.66 Determine the time in seconds requiredfor the object in Problem 13.65 to fall from the surfaceof the earth to the center. The earths radius is 6370 km.Solution: From Problem 13.65, the acceleration isa = vdvds= gREs_ v0v du = _ sRE_ gRE_s dsv2=_ gRE_(R2Es2)Recall that v = ds/dtv = dsdt= _ gRE_R2Es2_ 0REds_R2Es2= _ gRE_ tf0dt_ gREtf = sin1_ sRE_0RE= sin1(1)tf = _REg2 = 1266 s = 21.1 minProblem 13.67 In a second test, the coordinates of theposition (in m) of the helicopter in Active Example 13.6are given as functions of time byx = 4 +2t ,y = 4 +4t +t2.(a) What is the magnitude of the helicopters velocityat t = 3 s?(b) What is the magnitude of the helicopters acceler-ation at t = 3 s?yxSolution: We havex = (4 m) +(2 m/s)t , y = (4 m) +(4 m/s)t +(1 m/s2)t2,vx = (2 m/s), vy = (4 m/s) +2(1 m/s2)t ,ax = 0, ay = 2(1 m/s2).At t = 3 s, we havex = 10 m, vx = 2 m/s, ax = 0,y = 25 m, vy = 10 m/s, ay = 2 m/s2.Thusv =_v2x +v2y = 10.2 m/s v = 10.2 m/s.a =_02+(2 m/s2)2= 2 m/s2a = 2 m/s2.36c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.68 In terms of a particular referenceframe, the position of the center of mass of the F-14at the time shown (t = 0) is r = 10i +6j +22k (m).The velocity from t = 0 to t = 4 s is v = (52 +6t )i +(12 +t2)j (4 +2t2)k (m/s). What is the position ofthe center of mass of the plane at t = 4 s?Solution:r0 = 10i +6j +22k mv = (52 +6t )i +(12 +t2)j (4 +2t2)k m/sx4 =_ 40vx dt = 52t +3t2+x0x4 = (52)(4) +3(4)2+10 m = 266.0 my4 =_ 40vy dt = 12t +t3/3 +y0y4 = 12(4) +(4)3/3 +6 m = 75.3 mz4 =_ 40vz dt = (4t +2t3/3) +z0z4 = 4(4) 2(4)3/3 +22 = 36.7 mr|t =4s = 266i +75.3j 36.7k (m)Problem 13.69 In Example 13.7, suppose that the anglebetween the horizontal and the slope on which the skierlands is 30 instead of 45. Determine the distance d tothe point where he lands.453 md20Solution: The skier leaves the 20 surface at 10 m/s.The equations areax = 0, ay = 9.81 m/s2,vx = (10 m/s) cos 20, vy = (9.81 m/s2)t (10 m/s) sin 20sx = (10 m/s) cos 20t , sy = 12(9.81 m/s2)t2(10 m/s) sin 20tWhen he hits the slope, we havesx = d cos 30 = (10 m/s) cos 20t,sy = (3 m) d sin 30 = 12(9.81 m/s2)t2(10 m/s) sin 20tSolving these two equations together, we nd t = 1.01 s, d = 11.0 m.37c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.70 A projectile is launched from groundlevel with initial velocity v0 = 20 m/s. Determine itsrange R if (a) 0 = 30; (b) 0 = 45 (c) 0 = 60.R0u0yxSolution: Set g = 9.81 m/s2, v0 = 20 m/say = g, vy = gt +v0 sin 0, sy = 12gt2+v0 sin 0tax = 0, vx = v0 cos 0, sx = v0 cos 0tWhen it hits the ground, we have0 = 12gt2+v0 sin 0t t = 2v0 sin 0gR = v0 cos 0t R = v02sin 20ga) 0 = 30 R = 35.3 mb) 0 = 45 R = 40.8 mc) 0 = 60 R = 35.3 mProblem 13.71 Immediately after the bouncing golfball leaves the oor, its components of velocity are vx =0.662 m/s and vy = 3.66 m/s.(a) Determine the horizontal distance from the pointwhere the ball left the oor to the point where ithits the oor again.(b) The ball leaves the oor at x = 0, y = 0. Deter-mine the balls y coordinate as a function of x.(The parabolic function you obtain is shown super-imposed on the photograph of the ball.)xySolution: The governing equations areax = 0, ay = g,vx = vx0, vy = gt +vy0,x = vx0t , y = 12gt2+vy0t(a) When it hits the ground again0 = 12gt2+vy0t t = 2vy0gx = vx0t = vx0_2y0g_= 2vx0vy0gx = 2(0.662 m/s)(3.66 m/s)9.81 m/s2 x = 0.494 m.(b) At any point of the ight we havet = xvx0, y = 12g_ xvx0_2+vy0_ xvx0_y = 12_ 9.81 m/s2[0.662 m/s]2_x2+ 3.66 m/s0.662 m/sxy = _11.2m_x2+5.53x.38c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.72 Suppose that you are designing amortar to launch a rescue line from coast guard vesselto ships in distress. The light line is attached to a weightred by the mortar. Neglect aerodynamic drag and theweight of the line for your preliminary analysis. If youwant the line to be able to reach a ship 91 m awaywhen the mortar is red at 45 above the horizontal,what muzzle velocity is required?y45xSolution: From 13.70 we know thatR = v02sin 20gv0 =_ Rgsin 20v0 =_ ( )( 2)sin(90)=Problem 13.73 In Problem 13.72, what maximumheight above the point where it was red is reached bythe weight?Solution: From Problem 13.70 we havevy = gt +v0 sin 0, sy = 12gt2+v0 sin 0tWhen we reach the maximum height,0 = gt +v0 sin 0 t = v0 sin 0gh = 12gt2+v0 sin 0t h = 12g_v0 sin 0g_2+v0 sin 0_v0 sin 0g_h = v02sin202gPutting in the numbers we haveh = ( )2sin2(45)2( 2)=3991 9.81 m/s29.9 m/s29.9 m/s9.81 m/s 22.78 mc 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.74 When the athlete releases the shot, itis 1.82 m above the ground and its initial velocity is v0 =13.6 m/s. Determine the horizontal distance the shot trav-els from the point of release to the point where it hitsthe ground.30v0Solution: The governing equations areax = 0,vx = v0 cos 30,sx = v0 cos 30t ,ay = gvy = gt +v0 sin 30sy = 12gt2+v0 sin 30t +hWhen it hits the ground, we havesx = (13.6 m/s) cos 30t,sy = 0 = 12(9.81 m/s2)t2+(13.6 m/s) sin 30t +1.82 m.Solving these two equations, we nd t = 1.62 s, sx = 19.0 m.Problem 13.75 A pilot wants to drop survey markersat remote locations in the Australian outback. If he iesat a constant velocity v0 = 40 m/s at altitude h = 30 mand the marker is released with zero velocity relative tothe plane, at what horizontal d from the desired impactpoint should the marker be released?dh0Solution: We want to nd the horizontal distance traveled by themarker before it strikes the ground (y goes to zero for t > 0.)ax = 0 ay = gvx = vx0 vy = vy0 gtx = x0+vx0t y = y0+vy0t gt2/2From the problem statement, x0 = 0, vy0 = 0, vx0 = 40 m/s, and y0 =30 m The equation for y becomesy = 30 (9.81)t2/2Solving with y = 0, we get tf = 2.47 s.Substituting this into the equation for x, we getxf = 40tf = 98.9 m40c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.76 If the pitching wedge the golfer isusing gives the ball an initial angle 0 = 50, what rangeof velocities v0 will cause the ball to land within 3 m ofthe hole? (Assume the hole lies in the plane of the ballstrajectory).30 m3 mv00Solution: Set the coordinate origin at the point where the golferstrikes the ball. The motion in the horizontal (x) direction is given byax = 0, Vx = V0 cos 0, x = (V0 cos 0)t . The motion in the vertical(y) direction is given byay = g, Vy = V0 sin 0gt, y = (V0 sin 0)t gt22 .From the x equation, we can nd the time at which the ball reachesthe required value of x (27 or 33 metres). This time istf = xf/(V0 cos 0).We can substitute this information the equation for Y with Yf = 3 mand solve for V0. The results are: For hitting (27,3) , V0 = 31.2 m/s.For hitting (33,3) metre, V0 = 34.2 /s.Problem 13.77 A batter strikes a baseball 0.9 m abovehome plate and pops it up. The second baseman catchesit 1.8 m above second base 3.68 s after it was hit.What was the balls initial velocity, and what wasthe angle between the balls initial velocity vector andthe horizontal? SecondbaseHome plate27 m27mSolution: The equations of motion g = 2ax = 0 ay = gvx = v0 cos 0 vy = gt +v0 sin 0sx = v0 cos 0t sy = 12gt2+v0 sin 0t +When the second baseman catches the ball we have=v0 cos 0(3.68 s)= 12 ( 2) (3.68 s)2+ v0 sin 0(3.68 s)Solving simultaneously we ndv0 = , 0 = 60.441m9.81 m/s0. 9 m38.2 m1.8 m 9.81 m/s21 m/smetrec 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.78 A baseball pitcher releases a fastball withan initial velocity v0= be the initialangle of the balls velocity vector above the horizontal.When it is released, the ball is 1.83 m above the groundand 17.7 m from the batters plate. The batters strikezone extends from 0.56 m above the ground to 1.37 mabove the ground. Neglecting aerodynamic effects,determine whether the ball will hit the strike zone (a) if = 1; (b) if = 2. Solution: The initial velocity is v0 = =velocity equations are(1) dvxdt= 0, from which vx = v0 cos .(2) dvydt= g, from which vy = gt +v0 sin .(3) dxdt= v0 cos , from which x(t ) = v0 cos t , since the initialposition is zero.(4) dydt= gt +v0 sin , from whichy(t ) = g2t2+v0 sin t + ,since the initial position is y(0) = d =the height is h. The time of passage across the home plate isx(tp) = d = v0 cos tp, from whichtp = dv0 cos .Substitute:y(tp) = h = g2_ dv0 cos _2+d tan + .For = 1, h = . , Yes, the pitcher hits the strike zone.For = 2, h = . No, the pitcher misses the strike zone.Problem 13.79 In Problem 13.78, assume that thepitcher releases the ball at an angle = 1 above thehorizontal, and determine the range of velocities v0 (instrike zone.Solution: From the solution to Problem 13.78,h = g2_ dv0 cos _2+d tan + ,where d = h Solve for the initial velocity:v0 =_ gd22 cos2(d tan + h).For h = . v0 = h = . v0 =pitcher will hit the strike zone for velocities of release of. v0 . ,and a release angle of = 1. Check: The range of velocities in milesper hour is 112. v0161.1 km/h, which is within the range ofmajor league pitchers, although the 160.9 km upper value is achievableonly by a talented few (Nolan Ryan, while with the Houston Astros,would occasionally in a game throw a 168.9 km fast ball, as measuredby hand held radar from behind the plate).42144.8 km/h. Let144.8 km/h 40.3 m/s.The1.831.83 m. At a distance 17.7 m,1.831 2 m1 5 mm/s) within which he must release the ball to hit the1.8317.7 m, and 1.37 0.56 m.1.831 37, 44.74 m/s. For 0 56 m, 31.15 m/s. The31 15 44 74 m/s1km/ h1.37 m0.5617.7mmc 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.80 A zoology student is provided with abow and an arrow tipped with a syringe of sedativeand is assigned to measure the temperature of a blackrhinoceros (Diceros bicornis). The range of his bowwhen it is fully drawn and aimed 45 above the horizon-tal is 100 m. A truculent rhino charges straight towardhim at 30 km/h. If he fully draws his bow and aims 20above the horizontal, how far away should the rhino bewhen the student releases the arrow?20Solution: The strategy is (a) to determine the range and ight timeof the arrow when aimed 20 above the horizontal, (b) to determine thedistance traveled by the rhino during this ight time, and then (c) toadd this distance to the range of the arrow. Neglect aerodynamic dragon the arrow. The equations for the trajectory are: Denote the constantsof integration by Vx, Vy, Cx, Cy, and the velocity of the arrow by VA.(1) dvxdt= 0, from which vx = Vx. At t = 0, Vx = VA cos .(2) dvydt= g, from which vy = gt +Vy. At t = 0, Vy =VA sin .(3) dxdt= vx = VA cos , from which x(t ) = VA cos t +Cx. Att = 0, x(0) = 0, from which Cx = 0.(4) dydt= vy = gt +VA sin , from whichy = g2t2+VA sin t +Cy.At t = 0, y = 0, from which Cy = 0. The time of ight isgiven byy(tight) = 0 =_g2tight+VA sin _tight,from whichtight = 2VA sin g.The range is given byx(tight) = R = VA cos tight = 2V2A cos sin g.The maximum range (100 meters) occurs when the arrow isaimed 45 above the horizon. Solve for the arrow velocity: VA =gRmax = 31.3 m/s. The time of ight when the angle is 20 istight = 2VA sin g= 2.18 s,and the range is R = VA cos tight = 64.3 m. The speed ofthe rhino is 30 km/h = 8.33 m/s. The rhino travels a distanced = 8.33(2.18) = 18.2 m. The required range when the arrow isreleased isd +R = 82.5 m .43c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.81 The crossbar of the goalposts inAmerican football is yc=kick a eld goal, the ball must make the ball go betweenthe two uprights supporting the crossbar and be abovethe crossbar when it does so. Suppose that the kickerattempts a 36.58 m eld goal , and kicks the ball withan initial velocity v0 = 0=40. By whatvertical distance does the ball clear the crossbar?ycxc0v0Solution: Set the coordinate origin at the point where the ballis kicked. The x (horizontal) motion of the ball is given by ax = 0,Vx = V0 cos 0, x = (V0 cos 0)t . The y motion is given by ay = g,Vy =V0 sin 0gt , y =(V0 sin 0)t gt 22 . Set x =xc=nd the time tc at which the ball crossed the plane of the goal posts.Substitute this time into the y equation to nd the y coordinate YBof the ball as it passes over the crossbar. Substituting in the numbers(g = 2), we get tc=2.24 s and yB = .clears the crossbar by .0 .Problem 13.82 An American football quarterbackstands at A. At the instant the quarterback throws thefootball, the receiver is at B running at 6.1m/s towardC, where he catches the ball. The ball is thrown at anangle of 45 above the horizontal, and it is thrown andcaught at the same height above the ground. Determinethe magnitude of the balls initial velocity and the lengthof time it is in the air.90CBReceiverspathPath of the ballASolution: Set x as the horizontal motion of the football, y as thevertical motion of the football and z as the horizontal motion of thereceiver. Set g = 2, 0 = 45. We haveaz = 0, vz = , sz = ( )tay = g, vy = gt +v0 sin 0, sy = 12gt2+v0 sin 0tax = 0, vx = v0 cos 0, sx = v0 cos 0tWhen the ball is caught we havesz = ( )t0 = 12gt2+v0 sin 0tsx = v0 cos 0tsx2= sz2+( )2We can solve these four equations for the four unknowns sx, sz, v0, tWe nd t = 1.67 s, v0 =443.05 m above the ground. To21.3 m/s and36.58 m and9. 81 m/s 6 11 m. Thus, the ball3 7 m9.81 m/s6.1 m/s 6.1 m/s6.1 m/s9.1 m11.6 m/s9.1 mc 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.83 The cliff divers of Acapulco, Mexicomust time their dives that they enter the water at thecrest (high point) of a wave. The crests of the wavesare 1 m above the mean water depth h = 4 m. Thehorizontal velocity of the waves is equal to gh. Thedivers aiming point is 2 m out from the base of the cliff.Assume that his velocity is horizontal when he beginsthe dive.(a) What is the magnitude of the drivers velocity whenhe enters the water?(b) How far from his aiming point must a wave crestbe when he dives in order for him to enter the waterat the crest?2 m 6.4 m1 m26 mhSolution:t = 0, vy = 0, y = 27 m, x0 = 0ay = g = 9.81 m/sVy = V0y0 gty = y0gt2/2y = 1 m at tIMPACTfor an ideal dive to hit the crest of the wavet1 = tIMPACT = 2.30 sVy(t1) = 22.59 m/sax = 0Vx = Vx0XI = Vx0t1+X0At impact XI = 8.4 m.For impact to occur as planned, thenVx = 8.4/t1 = 3.65 m/s = constantThe velocity at impact is(a) |V| =_(Vx)2+[Vy(t1)]2= 22.9 m/sThe wave moves at gh = 6.26 m/s.The wave crest travels 2.30 seconds while the diver is in their s =ght1 = 14.4 m.27 m8.4 m45c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.84 A projectile is launched at 10 m/s froma sloping surface. The angle = 80. Determine therange R.a10 m/sR30Solution: Set g = 9.81 m/s2, v0 = 10 m/s.The equations of motion areax = 0, vx = v0 cos(8030), sx = v0 cos 50tay = g, vy = gt +v0 sin(8030)t , sy = 12gt2+v0 sin 50tWhen the projectile hits we haveRcos 30 = v0 cos 50t t = 2.32 s, R = 17.21 mRsin 30 = 12gt2+v0 sin 50tProblem 13.85 A projectile is launched at 100 m/s at60 above the horizontal. The surface on which it landsis described by the equation shown. Determine the pointof impact.yy = 0.001x260100 m/sxSolution: The motion in the x direction is ax = 0, vx = V0 cos 0,x = (V0 cos 0)t , and the motion in the y direction is given byay = g, vy = (V0 sin 0) gt , y = (V0 sin 0)t gt2/2. We knowthat V0 = 100 m/s and 0 = 60. The equation of the surface uponwhich the projectile impacts is y = 0.001x2. Thus, the time ofimpact, tI, can be determined by substituting the values of x andy from the motion equations into the equation for the surface.Hence, we get (V0 sin 0)tI gt2I2 = 0.001(V0 cos 0)2t2I . Evaluatingwith the known values, we get tI = 6.37 s Substituting this valueinto the motion equations reveals that impact occurs at (x, y) =(318.4, 101.4) .46mc 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.86 At t = 0, a steel ball in a tank of oilis given a horizontal velocity v = 2i (m/s). The compo-nents of the balls acceleration in m/s2are ax = 1.2vx,ay = 8 1.2vy, az = 1.2vz. What is the velocity ofthe ball at t = 1 s?xySolution: Assume that the effect of gravity is included in the givenaccelerations. The equations for the path are obtained from:(1) dvxdt= ax = 1.2vx. Separate variables and integrate:dvxvx= 1.2 dt,from which ln(vx) = 1.2t +Vx. At t = 0, vx(0) = 2, fromwhichln_vx2_= 1.2t .Inverting: vx(t ) = 2e1.2t.(2) dvydt= ay = 8 1.2vy. Separate variables and integrate:dvy81.2 +vy= 1.2 dt,from whichln_ 81.2 +vy_= 1.2t +Vy.At t = 0, vy(0) = 0, fromln_1 + 1.28 vy_= 1.2t .Inverting: vy(t ) = 81.2(e1.2t1).(3) dvzdt= az = 1.2vz, from which ln(vz) = 1.2t +Vz. Invert toobtain vz(t ) = Vze1.2t. At t = 0, vz(0) = 0, hence Vz = 0 andvz(t ) = 0. At t = 1 second,vx(1) = 2e1.2= 0.6024 m/s , andvy(1) = _ 812_(1 e1.2) = 4.66 m/s , orv = 0.602i 4.66j (m/s) .47c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.87 In Problem 13.86, what is the positionof the ball at t = 1 s relative to its position at t = 0?Solution: Use the solution for the velocity components from Prob-lem 13.86. The equations for the coordinates:(1) dxdt= vx = 2e1.2t, from whichx(t ) = _ 21.2_e1.2t+Cx.At t = 0, x(0) = 0, from whichx(t ) =_ 21.2_(1 e1.2t).(2) dydt=_ 81.2_(e1.2t1), from whichy(t ) = _ 81.2__e1.2t1.2 +t_+Cy.At t = 0, y(0) = 0, from whichy(t ) = _ 81.2__e1.2t1.2 +t 11.2_.(3) Since vz(0) = 0 and z(0) = 0, then z(t ) = 0. At t = 1,x(1) =_ 21.2_(1 e1.2) = 1.165 m .y(1) = _ 81.2__e1.21.2 +1 11.2_= 2.784 m , orr = 1.165i 2.784j (m) .Problem 13.88 The point P moves along a circularpath with radius R. Show that the magnitude of its veloc-ity is |v| = R|d/dt |.Strategy: Use Eqs. (13.23).PyxSolution:x = Rcos y = Rsin vx = Rsin _ddt_vy = Rcos _ddt_|V| =_V2x +V2y|V| =_R2sin2_ddt_2+R2cos2_ddt_2|V| =_R2_ddt_2(sin2 +cos2)|V| = Rddt48c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.89 If y = 150 mm, dydt= 300 mm/s, andd2ydt2 = 0, what are the magnitudes of the velocity andacceleration of point P?300 mmPySolution: The equation for the location of the point P is R2=x2+y2, from which x = (R2y2)12 = 0.2598 m, anddxdt= _yx__dydt_= 0.1732 m/s,d2xdt2 = 1x_dydt_2+ yx2_dxdt__dydt__yx__d2ydt2_= 0.4619 m/s2.The magnitudes are:|vP| =__dxdt_2+_dydt_2= 0.3464 m/s|ap| =__d2xdt2_2+_d2ydt2_2= 0.4619 m/s2Problem 13.90 A car travels at a constant speed of100 km/h on a straight road of increasing grade whosevertical prole can be approximated by the equationshown. When the cars horizontal coordinate is x =400 m, what is the cars acceleration? y = 0.0003x2yxSolution: Denote C = 0.0003 and V = 100 km/h = 27.78 m/s. Themagnitude of the constant velocity isV =__dydt_2+_dxdt_2The equation for the road is y = Cx2from whichdydt= 2Cx_dxdt_.Substitute and solve:dxdt= V_(2Cx)2+1= 27.01 m/s.dxdtis positive (car is moving to right in sketch). The acceleration isd2xdt2 = ddt_ V_(2Cx)2+1_= 4C2Vx((2Cx)2+1)32_dxdt_= 0.0993 m/s2.d2ydt2 = 2C_dxdt_2+2Cx_d2xdt2_= 0.4139 m/s2, ora = 0.099i +0.414j(m/s2)49c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.91 Suppose that a projectile has the initialconditions shown in Fig. 13.12. Show that in terms ofthe x

y

coordinate system with its origin at the high-est point of the trajectory, the equation describing thetrajectory isy

= g2v20 cos20(x

)2.xyyxSolution: The initial conditions are t = 0, x(0) = 0, y(0) = 0,vx(0) = v0 cos 0, and vy(0) = v0 sin 0. The accelerations are ax(t )= 0, ay(t ) = g. The path of the projectile in the x, y system isobtained by solving the differential equations subject to the initialconditions:x(t ) = (v0 cos 0)t , y(t ) = g2t2+(v0 sin 0)tEliminate t from the equations by substitutingt = xv0 cos 0to obtainy(x) = gx22v20 cos20+x tan 0.At the peak,dydxpeak= 0,from whichxp = v20 cos 0 sin 0g,and yp = v20 sin202g.The primed coordinates: y

= y yp x

= x xp. Substitute andreduce:y

= g(x

+xp)22v20 cos20+(x

+xp) tan 0yp.y

= g2v20 cos20((x

)2+x2p +2x

xp) +(x

+xp) tan 0 v20 sin202g.Substitute xp = v20 cos 0 sin 0g,y

= g(x

)22v20 cos20 v20 sin202gx

tan 0+x

tan 0+ v20 sin20g v20 sin202g.y

= g2v20 cos20(x

)250c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.92 The acceleration components of apoint are ax = 4 cos 2t , ay = 4 sin 2t , az = 0. Att = 0, its position and velocity are r = i, v = 2j. Showthat (a) the magnitude of the velocity is constant; (b) thevelocity and acceleration vectors are perpendicular; (c)the magnitude of the acceleration is constant and pointstoward the origin; (d) the trajectory of a point is a circlewith its center at the origin.Solution: The equations for the path are(1) dvxdt= ax = 4 cos(2t ), from which vx(t ) = 2 sin(2t ) +Vx.At t = 0, vx(0) = 0, from which Vx = 0. dxdt= vx = 2 sin(2t ),from which x(t ) = cos(2t ) +Cx. At t = 0, x(0) = 1, from whichCx = 0.(2) dvydt= ay = 4 sin(2t ), from which vy(t ) = 2 cos(2t ) +Vy. Att = 0, vy(0) = 2, from which Vy = 0. dydt= vy = 2 cos(2t ),from which y(t ) = sin(2t ) +Cy. At t = 0, y(0) = 0, from whichCy = 0.(3) For az = 0 and zero initial conditions, it follows that vz(t ) = 0and z(t ) = 0.(a) The magnitude of the velocity is|v| =_(2 sin(2t ))2+(2 cos(2t ))2= 2 = const.(b) The velocity is v(t ) = i2 sin(2t) +j2 cos(2t ). The accel-eration is a(t ) = i4 cos(2t ) j4 sin(2t ). If the two areperpendicular, the dot product should vanish: a(t ) v(t ) =(2 sin(2t ))(4 cos(2t )) +(2 cos(2t ))(4 sin(2t )) = 0,and it does(c) The magnitude of the acceleration:|a| =_(4 cos(2t ))2+(4 sin(2t ))2= 4 = const .The unit vector parallel to the acceleration ise = a|a| = i cos(2t ) j sin(2t),which always points to the origin.(d) The trajectory path is x(t ) = cos(2t ) and y(t ) = sin(2t ).These satisfy the condition for a circle of radius 1:1 = x2+y2Problem 13.93 When an airplane touches down att = 0, a stationary wheel is subjected to a constant angu-lar acceleration = 110 rad/s2until t = 1 s.(a) What is the wheels angular velocity at t = 1 s?(b) At t = 0, the angle = 0. Determine in radiansand in revolutions at t = 1 s.Solution: = 110 rad/s2 = t +0 = (12t2) +0t +0From the problem statement, 0 = 0 = 0(a) At t = 1 s, = (110)(1) +0 = 110 rad/s(b) At t = 1 s, = 110(1)2/2 = 55 radians (8.75 revolutions)51c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.94 Let L be a line from the center of theearth to a xed point on the equator, and let L0 be axed reference direction. The gure views the earth fromabove the north pole.(a) Is d/dt positive or negative? (Remember that thesun rises in the east.)(b) Determine the approximate value of d/dt in rad/sand use it to calculate the angle through which theearth rotates in one hour.LL0uSolution:(a) ddt> 0.(b) ddt 2 rad24(3600) s= 7.27 105rad/s.In one hour (7.27 105rad/s)(1 hr)_3600 s1 hr_= 0.262 rad = 15ddt 7.27 105rad/s, 15.Problem 13.95 The angular acceleration of the line Lrelative to the line L0 is given as a function of time by = 2.5 1.2t rad/s2. At t = 0, = 0 and the angularvelocity of L relative to L0 is = 5 rad/s. Determine and at t = 3 s.LL0uSolution: = 2.5 1.2t = 2.5t 0.6t2+5 = 1.25t20.2t3+5t(3) = 1.25(3)20.2(3)3+5(3) = 20.85 rad(3) = 2.5(3) 0.6(3)2+5 = 7.1 rad/s52c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.96 In Active Example 13.8, suppose thatthe angular acceleration of the rotor is = 0.000022,where is the angular velocity of the rotor in rad/s. Howlong does it take the rotor to slow from 10,000 rpm to1000 rpm?LL0uSolution: Let = k2, where k = 0.0002.Thena = ddt= k2,_ 21d2 =_ t0kdt , 12+ 11= kt t = 1k_ 11 12_Convert the numbers to rad/s1 = 10000 rpm_2 radrev__1 min60 sec_= 1047 rad/s2 = 1000 rpm_2radrev__1 min60 sec_= 104.7 rad/sUsing the numbers in the problemt = 10.0002_ 11047 rad/s 1104.7 rad/s_t = 430 s.Problem 13.97 The astronaut is not rotating. He hasan orientation control system that can subject him toa constant angular acceleration of 0.1 rad/s2about thevertical axis is either direction. If he wants to rotate180 about the vertical axis (that is, rotate so that heis facing toward the left) and not be rotating in his neworientation, what is the minimum time in which he couldachieve the new orientation?Solution: He could achieve the rotation in minimum time by accel-erating until he has turned 90 and then decelerating for the same timewhile he rotates the nal 90.Thus the time needed to turn 90 (/2 rad) is = 0.1 rad/s2, = t = 12t2t =_2=_ 2(/2)0.1 rad/s2 = 5.605 s.The total maneuver time is 2t . 2t = 11.2 s.53c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.Problem 13.98 The astronaut is not rotating. He hasan orientation control system that can subject him toa constant angular acceleration of 0.1 rad/s2about thevertical axis in either direction. Refer to problem 13.97.For safety, the control system will not allow his angularvelocity to exceed 15 per second. If he wants to rotate180 about the vertical axis (that is, rotate so that heis facing toward the left) and not be rotating in his neworientation, what is the minimum time in which he couldachieve the new orientation?Solution: He could achieve the rotation in minimum time by accel-erating until he has reached the angular velocity limit, then coastinguntil he has turned 90. He would then continue to coast until heneeded to decelerate to a stop at the 180 position. The maneuver issymmetric in the spin up and the spin down phases.We will rst nd the time needed to reach the angular velocity limitand also nd the angle through which he has rotated in this time. = 0.1 rad/s2 = t1 t1 = = (15/180) rad/s0.1 rad/s2 = 2.62 s.1 = 12t2= 12(0.1 rad/s2)(2.62 s)2= 0.343 rad.Now we need to nd the coast time (constant angular velocity)2 rad 0.343 rad =_ 15180_ rad/s t2 t2 = 4.69 sThus the total time to complete a 90 turn is t = t1+t2 = 2.62 s +4.69 s = 7.31 s.The time for the full 180 turn is 2t 2t = 14.6 s.Problem 13.99 The rotor of an electric generator isrotating at 200 rpm when the motor is turned off. Dueto frictional effects, the angular acceleration of the rotorafter the motor is turned off is = 0.01 rad/s2, where is the angular velocity in rad/s.(a) What is the rotors angular velocity one minuteafter the motor is turned off?(b) After the motor is turned off, how many revolutionsdoes the rotor turn before it comes to rest?Strategy: To do part (b), use the chain rule to writethe angular acceleration as = ddt= ddddt= dd.Solution: Let = k, where k = 0.01 s1. Note that 200 rpm =20.9 rad/s.(a) One minute after the motor is turned off = ddt= k _ 0d=_ t0kdt ln_ 0_= kt = 0ekt= (20.9 rad/s)e(0.01/s)(60 s)= 11.5 rad/s. = 11.5 rad/s (110 rpm).(b) When the rotor comes to rest = dd= k _ 00d =_ 0kd 0 = k = 1k(0) = 10.01 s1(20.9 rad/s) = 2094 rad_ 1 rev2 rad_= 333 rev. = 333 rev.54c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the p