Análise do Movimento de Partículas: Posição, Velocidade e Aceleração

CHAPTER 11 PROBLEM 111 The motion of a particle is defined by the relation x 15t4 30t2 5t 10 where x and t are expressed in meters and seconds respectively Determine the position the velocity and the acceleration of the particle when t 4 s SOLUTION Given x 15t4 30t2 5t 10 v dxdt 6t3 60t 5 a dvdt 18t2 60 Evaluate expressions at t 4 s x 1544 3042 54 10 66 m v 643 604 5 149 ms a 1842 60 228 ms2 PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission PROBLEM 113 The motion of a particle is defined by the relation x 53 t3 52 t2 30 t 8 where x and t are expressed in feet and seconds respectively Determine the time the position and the acceleration when v 0 PROBLEM 114 The motion of a particle is defined by the relation x 6 t2 8 40 cos πt where x and t are expressed in inches and seconds respectively Determine the position the velocity and the acceleration when t 6 s The motion of a particle is defined by the relation x 6t4 2t3 12t2 3t 3 where x and t are expressed in meters and seconds respectively Determine the time the position and the velocity when a 0 The motion of a particle is defined by the relation x 2t3 15t2 24t 4 where x is expressed in meters and t in seconds Determine a when the velocity is zero b the position and the total distance traveled when the acceleration is zero The motion of a particle is defined by the relation x t3 6t2 36t 40 where x and t are expressed in feet and seconds respectively Determine a when the velocity is zero b the velocity the acceleration and the total distance traveled when x 0 PROBLEM 118 The motion of a particle is defined by the relation x t³ 9t² 24t 8 where x and t are expressed in inches and seconds respectively Determine a when the velocity is zero b the position and the total distance traveled when the acceleration is zero SOLUTION We have x t³ 9t² 24t 8 Then v dxdt 3t² 18t 24 and a dvdt 6t 18 a When v 0 3t² 18t 24 3t² 6t 8 0 or t 2t 4 0 or t 200 s and t 400 s b When a 0 6t 18 0 or t 3 s At t 3 s x₃ 3³ 93² 243 8 or x₃ 1000 in First observe that 0 t 2 s v 0 2 s t 3 s v 0 Now At t 0 x₀ 8 in At t 2 s x₂ 2³ 92² 242 8 12 in Then x₂ x₀ 12 8 20 in x₃ x₂ 10 12 2 in Total distance traveled 20 2 in 220 in PROBLEM 119 The acceleration of a particle is defined by the relation a 8 ms² Knowing that x 20 m when t 4 s and that v 16 ms determine a the time when the velocity is zero b the velocity and the total distance traveled when t 11 s SOLUTION We have dvdt a 8 ms² Then dv 8 dt C constant v 8t C ms Also dxdt v 8t C At t 4 s x 20 m x₂⁴ dx 8t C dt or x 20 4t² Ct₄ or x 4t² Ct 4 84 m When v 16 ms x 4 m 16 8t C C 16 8t When v 16 ms x 4 m 4 4t² Ct 4 84 Combining 0 4t² 16 8tt 4 80 Simplifying t² 4t 4 0 or t 2 s and C 32 ms v 8t 32 ms x 4t² 32t 44 m a When v 0 8t 32 0 or t 400 s b Velocity and distance at 11 s At t 0 x₀ 44 m At t 4 s x₄ 20 m At t 11 s x₁₁ 411² 3211 44 176 m PROBLEM 119 Continued Now observe that 0 t 4 s v 0 4 s t 11 s v 0 Then x₄ x₀ 20 44 64 m x₁₁ x₄ 176 20 196 m Total distance traveled 64 196 m 260 m The acceleration of a particle is directly proportional to the time t At t 0 the velocity of the particle is v 16 ins Knowing that v 15 ins and that x 20 in when t 1 s determine the velocity the position and the total distance traveled when t 7 s At t 0 x₀ 133 At t 4 s x₄ 134³ 164 133 47 in The total distance traveled 4267 45 in 877 in PROBLEM 1112 The acceleration of a particle is defined by the relation a k² a Knowing that v 32 fts when t 0 and that v 32 fts when t 4 s determine the constant k b Write the equations of motion knowing also that x 0 when t 4 s SOLUTION a k² 1 dvdt a k² t 0 v 32 fts and t 4 s v 32 fts a ₃₂² dv ₀⁴ k² dt 32 32 13 k4³ k 300 fts⁴ b Substituting k 3 fts⁴ into 1 dvdt a 3t² a 3t² t 0 v 32 fts ₃₂² dv ₀³ t² dt v 32 1330³ v t³ 32 t 4 s x 0 ₀⁴ dx ₀⁴ t³ 32dt x 14 t⁴ 32t₄ x 144⁴ 324 x 1464 128 x 16 128 x 112 PROPRIETARY MATERIAL 2009 The McGrawHill Companies Inc All rights reserved PROBLEM 1113 The acceleration of a particle is defined by the relation a A 6t² where A is a constant At t 0 the particle starts at x 8 m with v 0 Knowing that at t 1 s v 30 ms determine a the times at which the velocity is zero b the total distance traveled by the particle when t 5 s SOLUTION We have a A 6t² A constant Now dvdt a A 6t² At t 0 v 0 ₀ʹ dv ₀ʹ A 6t²dt or v A 2t³ ms At t 1 s v 30 ms 30 4A 21³ or A 32 ms² and v 32t 2t³ Also dxdt v 32t 2t³ At t 0 x 8 m ₈ʹ dx ₈ 32t 2t²dt or x 8 16t² 12 t⁴ m a When v 0 32t 2t³ 2t16 t² 0 t 0 and t 400 s b At t 4 s x₄ 8 164² 124⁴ 136 m t 5 s x₅ 8 165² 125⁴ 955 m PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 19 PROBLEM 1114 It is known that from t 2 s to t 10 s the acceleration of a particle is inversely proportional to the cube of the time t When t 2 s v 15 ms and when 10 s 036 ms t v Knowing that the particle is twice as far from the origin when t 2 s as it is when t 10 s determine a the position of the particle when t 2 s and when t 10 s b the total distance traveled by the particle from t 2 s to t 10 s SOLUTION We have 3 k a t k constant Now 3 dv k a dt t At t 2 s v 15 ms 3 15 2 v t k dv dt t or 2 2 2 1 1 15 2 2 k v t or 2 1 1 15 ms 2 4 k v t At 10 s 036 ms t v 2 1 1 036 15 2 4 10 k or k 128 m s and 2 64 1 ms v t a We have 2 64 1 dx v dt t Then 2 64 1 dx dt C t C constant or 64 m x t C t Now 2 2 10 x x 64 64 2 2 10 2 10 C C or C 12 m and 64 12 m x t t Now observe that 0 t 4 s v 0 4 s t 5 s v 0 Then x₄ x₀ 136 8 128 m x₅ x₄ 955 136 405 m Total distance traveled 128 405 m 1685 m PROPRIETARY MATERIAL 2009 The McGrawHill Companies Inc All rights reserved At t 2 s x2 2 642 12 or x2 352 m t 10 s x10 10 6410 12 or x10 1760 m Note A second solution exists for the case x2 0 x10 0 For this case C 22 415 m and x2 1115 m x10 5 1315 m b When v 0 1 64t2 0 or t 8 s At t 8 s x8 8 648 12 172 m Now observe that 2 t 8 s v 0 8 t 10 s v 0 Then x8 x2 172 352 18 m x10 x8 176 172 04 m Total distance traveled 18 04 m 1840 m Note The total distance traveled is the same for both cases a k xx When x 06 ft v 15 fts Separate and integrate using x 06 ft v 15 fts v dv k dxx 12 v215 k lnx06 12 v2 12 152 k lnx06 When v 9 fts x 12 ft 12 92 12 152 k ln1206 Solve for k k 103874 ft2s2 a Substitute k 103874 ft2s2 and x 15 ft into 1 12 v2 12 152 103874 ln1506 v 589 fts b For v 0 0 12 152 103874 lnx06 lnx06 1083 x 1772 ft We have v dydx a k x Ax When x 1 ft v 0 01 v dydx k x Ax dx or 12 v2k k12 x2 4 ln x01 k12 12 0 At x 2 ft 12 v2k k12 22 2 12 k32 4 ln 2 At x 8 ft 12 v2k k12 82 1 k315 4 ln 8 Now v2 2 12 v2 k12 315 4 ln 8 255 When x 16 ft v 29 fts 12 292 k12 162 255 ln12 Noting that ln16 4 ln 2 and ln12 ln2 we have 841 k236 ln 255ln2 4 ln2 1 k 1832 s2 PROBLEM 1117 A particle oscillates between the points x 40 mm and x 160 mm with an acceleration a k100 x where a and x are expressed in mms² and mm respectively and k is a constant The velocity of the particle is 18 mms when x 100 mm and is zero at both x 40 mm and x 160 mm Determine a the value of k b the velocity when x 120 mm PROBLEM 1118 A particle starts from rest at the origin and is given an acceleration a kx 4² where a and x are expressed in mms² and m respectively and k is a constant Knowing that the velocity of the particle is 4 ms when x 8 m determine a the value of k b the position of the particle when v 45 ms c the maximum velocity of the particle PROBLEM 1119 A piece of electronic equipment that is surrounded by packing material is dropped so that it hits the ground with a speed of 4 ms After impact the equipment experiences an acceleration of a kx where k is a constant and x is the compression of the packing material If the packing material experiences a maximum compression of 20 mm determine the maximum acceleration of the equipment PROBLEM 1120 Based on experimental observations the acceleration of a particle is defined by the relation a 01 sin x08 where a and x are expressed in ms² and meters respectively Knowing that b 08 m and that v 1 ms when x 0 determine a the velocity of the particle when x 1 m b the position where the velocity is maximum c the maximum velocity SOLUTION We have v dvdx a 01 sin x08 When x 0 v 1 ms or 12 v² 1 01x 08 cos x08₀¹ or 12 v² 01x 08 cos x08 03 a When x 1 m 12 v² 011 08 cos 108 03 or 12 v² 011 08 cos 125 03 or v 0323 ms b When v vmax a 0 01 sin x08 0 or x 0080134 m c When x 0080134 m 12 v²max 010080134 08 cos 008013408 03 0504 m²s² or vmax 1004 ms PROBLEM 1121 Starting from x 0 with no initial velocity a particle is given an acceleration a 08v² 49 where a and v are expressed in ms² and ms respectively Determine a the position of the particle when v 24 ms b the speed of the particle when x 40 m SOLUTION We have v dvdx a 08v² 49 When x 0 v 0 or ₀ᵛ dvv² 49 ₀ˣ 08dx or v² 49₀ 08x or v² 49 7 08x a When v 24 ms 24² 49 7 08x or x 225 m b When x 40 m v² 49 7 0840 or v² 49 7 32 or v 384 ms PROBLEM 1122 The acceleration of a particle is defined by the relation a kv where k is a constant Knowing that x 0 and v 81 ms at t 0 and that v 36 ms when x 18 m determine a the velocity of the particle when x 20 m b the time required for the particle to come to rest SOLUTION a We have v dvdx a kv so that When x 0 v 81 ms or ₈₁ᵛ v dv ₀ˣ kdx or 23 v32 729 kx or 23 36322 729 k18 or k 19 ms² Finally When x 20 m 23 v32 729 1920 or v32 159 or v 293 ms b We have dvdt a 19v At t 0 v 81 ms ₈₁ᵛ dvv 19₀ᵗ dt or 2v₈₁ 19t or 2v 9 19t or 29 19t or t 0947 s PROBLEM 1123 The acceleration of a particle is defined by the relation a 08v where a is expressed in ins² and v in ins Knowing that at t 0 the velocity is 40 ins determine a the distance the particle will travel before coming to rest b the time required for the particle to come to rest c the time required for the particle to be reduced by 50 percent of its initial value SOLUTION a a v dvdx 08v dv 08dx Separate and integrate with v 40 ins when x 0 v40 dv 08x40 dx Distance traveled For v 0 x 4008 x 500 in b a dvdt 08v Separate v40 dvv 08dt ln v ln 40 08t For v 0 we get t c For v 0540 ins 20 ins t 125 ln4020 0866 s PROBLEM 1124 A bowling ball is dropped from a boat so that it strikes the surface of a lake with a speed of 25 fts Assuming the ball experiences a downward acceleration of a 10 09y² when in the water determine the velocity of the ball when it strikes the bottom of the lake SOLUTION v₀ 25 fts x x₀ 30 ft a 10 09y² kc² v² where k 09 ft¹ and c² 1009 11111 ft²s² c 33333 fts Since v₀ c write a v dvdx kv² c² Integrating 12 lnv² c² v₀ kx x₀ v² c² v₀² c²ekx x₀ 11111 25² 11111e20930 11111 389x10¹⁹ 11111 ft²s² PROBLEM 1125 The acceleration of a particle is defined by the relation a 041 kv where k is a constant Knowing that at t 0 the particle starts from rest at x 4 m and that when t 15 s v 4 ms determine a the constant k b the position of the particle when v 6 ms c the maximum velocity of the particle SOLUTION a We have dydt a 041 kv At t 0 v 0 v0 dy1 kv 0 dt or 1k ln1 kv 0 04t or ln1 kv 04kt 1 At t 15 s v 4 ms ln1 4k 04k15 6k Solving yields k 0145703 sm b We have vdx a 041 kv When x 4 m v 0 0 dy1 kv 0 04dx When v 6 ms 60145703 10145703² ln1 0145703 6 04x 4 x 1452 m The maximum velocity occurs when a 0 a 0 041k vmax 0 or vmax 1 0145 703 or vmax 686 ms A particle is projected to the right from the position x 0 with an initial velocity of 9 ms If the acceleration of the particle is defined by the relation a 06v32 where a and v are expressed in ms² and ms respectively determine a the distance the particle will have traveled when its velocity is 4 ms b the time when v 1 ms c the time required for the particle to travel 6 m At t 0 x 0 x 0 dx t 0 9 1 09t² dt or x 91 0 1 09 09t dt 101 1 1 09t 10 9t 1 09t When x 6 m 6 9t 1 09t or t 1667 s PROBLEM 1127 Based on observations the speed of a jogger can be approximated by the relation v 751 004x03 where v and x are expressed in mih and miles respectively Knowing that x 0 at t 0 determine a the distance the jogger has run when t 1 h b the joggers acceleration in fts2 at t 0 c the time required for the jogger to run 6 mi PROBLEM 1128 Experimental data indicate that in a region downstream of a given louvered supply vent the velocity of the emitted air is defined by v 018v0x where v and x are expressed in ms and meters respectively and v0 is the initial discharge velocity of the air For v0 36 ms determine a the acceleration of the air at x 2 m b the time required for the air to flow from x 1 to x 3 m PROBLEM 1129 The acceleration due to gravity at an altitude y above the surface of the earth can be expressed as a 322 1 v209x1062 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 1800 fts b 3000 fts c 36700 fts PROBLEM 1129 Continued c v0 36700 fts ymax 36700² 644 36700² 20910⁶ or The velocity 36700 fts is approximately the escape velocity vR from the earth For vR ymax PROBLEM 1130 The acceleration due to gravity of a particle falling toward the earth is a gR²r² 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 R 3960 mi calculate the escape velocity that is the minimum velocity with which a particle must be projected vertically upward from the surface of the earth if it is not to return to the earth Hint v 0 for r SOLUTION We have v dvdr a gR²r² When r R v ve r ve 0 Then 0 ve dv R gR²r² dr or 12 ve² gR² 1R 1 or ve 2gR 2322 fts²3960 mi5280 ft1 mi¹² or ve 36710³ fts PROBLEM 1131 The velocity of a particle is v v01 sinπtT Knowing that the particle starts from the origin with an initial velocity v0 determine a its position and its acceleration at t 3T b its average velocity during the interval t 0 to t T SOLUTION a We have dxdt v v0 1 sinπtT At t 0 x 0 0 x dx 0 t v0 1 sinπtT dt or x v0 t Tπ cosπtT0 v0 t Tπ cos0 Tπ At t 3T x3T v0 3T Tπ cos3π Tπ v0 3T 2Tπ or Also a dvdt ddtv0 1 sinπtT v0 πT cosπtT At t 3T a3T v0 πT cos3π πv0T b Using Eq 1 At t 0 x0 v0 0 Tπ cos0 Tπ 0 At t T xT v0 T π cos πT T T π v0 T 2T π v0 T 2T π 0363v0T Now vave xT x0 Δt 0363v0T 0 T 0 or vave 0363v0 PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 42 PROBLEM 1132 The velocity of a slider is defined by the relation sin n v v ω t φ Denoting the velocity and the position of the slider at t 0 by 0v and x0 respectively and knowing that the maximum displacement of the slider is 2 0 x show that a 2 2 2 0 0 2 0 n n v v x x ω ω b the maximum value of the velocity occurs when x 2 0 0 0 3 n 2 x v x ω SOLUTION a At 0 0 t v v 0 sin 0 sin v v v φ φ Then 2 2 0 cos v v v φ Now sin n dx v v t dt ω φ At 0 0 t x x 0 0 sin x t n x dx v t dt ω φ or 0 0 1 cos t n n x x v ω t φ ω or 0 cos cos n n v x x t φ ω φ ω Now observe that xmax occurs when cos 1 ωnt φ Then max 0 0 2 cos 1 n v x x x φ ω Substituting for cosφ 2 2 0 0 1 1 n v v v x v ω or 1 2 2 0 0 n x v v v ω Squaring both sides of this equation 2 2 2 2 2 0 0 0 2 n n x x v v v ω ω or 2 2 2 0 0 0 2 n n v x v x ω ω Q E D PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 43 PROBLEM 1132 Continued b First observe that vmax occurs when 2 nt π ω φ The corresponding value of x is max 0 0 cos cos 2 cos v n n v x x v x π φ ω φ ω Substituting first for cosφ and then for v max 2 2 0 0 12 2 2 2 2 2 0 0 0 0 0 12 4 2 2 2 4 4 2 2 2 0 0 0 0 0 0 0 2 0 12 2 2 2 2 0 0 0 2 0 2 2 2 0 0 0 2 0 1 2 1 2 4 2 1 2 2 v n n n n n n n n n n n n v v v x x v v x x v x x v v x x x v x x x v x x v x x ω ω ω ω ω ω ω ω ω ω ω ω 2 0 0 0 3 2 n x v x ω Q E D PROBLEM 1133 A motorist enters a freeway at 45 kmh and accelerates uniformly to 99 kmh From the odometer in the car the motorist knows that she traveled 02 km while accelerating Determine a the acceleration of the car b the time required to reach 99 kmh SOLUTION a Acceleration of the car v12 v02 2ax1 x0 a v12 v02 2x1 x0 Data v0 45 kmh 125 ms v1 99 kmh 275 ms x0 0 x1 02 km 200 m a 2752 1252 2200 0 a 1500 ms² b Time to reach 99 kmh v1 v0 at1 t0 t1 t0 v1 v0 a 275 125 1500 1000 s PROBLEM 1134 A truck travels 220 m in 10 s while being decelerated at a constant rate of 06 ms² Determine a its initial velocity b its final velocity c the distance traveled during the first 15 s SOLUTION a Initial velocity x x0 v0t 1 2 at² v0 x x0 t 1 2 at v0 220 10 1 2 0610 259 ms b Final velocity v v0 at v 250 0610 1900 ms c Distance traveled during first 15 s x x0 v0t 1 2 at² 0 25015 1 2 0615² 368 m Assuming a uniform acceleration of 11 fts² and knowing that the speed of a car as it passes A is 30 mih determine a the time required for the car to reach B b the speed of the car as it passes B 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 896 ft 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 g 322 fts² determine a the speed vi of the rocket at the end of powered flight b the maximum altitude reached by the rocket A sprinter in a 100m race accelerates uniformly for the first 35 m and then runs with constant velocity If the sprinters time for the first 35 m in 54 s determine a his acceleration b his final velocity c his time for the race PROBLEM 1138 A small package is released from rest at A and moves along the skate wheel conveyor ABCD The package has a uniform acceleration of 48 ms² as it moves down sections AB and CD and its velocity is constant between B and C If the velocity of the package at D is 72 ms determine a the distance d between C and D b the time required for the package to reach D PROBLEM 1138 Continued Now for B C we have xC xB vBCtBC or 3 m 53666 mstBC or tBC 055901 s Finally tD tAB tBC tCD 111804 055901 038196 s tD 206 s PROBLEM 1139 A police officer in a patrol car parked in a 70 kmh speed zone observes a passing automobile traveling at a slow constant speed Believing that the driver of the automobile might be intoxicated the officer starts his car accelerates uniformly to 90 kmh in 8 s and maintaining a constant velocity of 90 kmh overtakes the motorist 42 s after the automobile passed him Knowing that 18 s elapsed before the officer began pursuing the motorist determine a the distance the officer traveled before overtaking the motorist b the motorists speed PROBLEM 1140 As relay runner A enters the 20mlong exchange zone with a speed of 129 ms he begins to slow down He hands the baton to runner B 182 s later as they leave the exchange zone with the same velocity Determine a the uniform acceleration of each of the runners b when runner B should begin to run PROBLEM 1141 Automobiles A and B are traveling in adjacent highway lanes and at t 0 have the positions and speeds shown Knowing that automobile A has a constant acceleration of 18 fts² and that B has a constant deceleration of 12 fts² determine a when and where A will overtake B b the speed of each automobile at that time PROBLEM 1141 Continued Velocities when t₁ 1505 s vA vA0 aAt aA aB α xA0 xB0 0 vA0 vB0 0 xA 0 vA0t 12 aAt² PROBLEM 1144 Continued c We have vB vB0 aBf At t tAB vB 175 ms030045 ms²20822 s 23756 ms vB 855 kmh PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 58 PROBLEM 1145 Car A is parked along the northbound lane of a highway and car B is traveling in the southbound lane at a constant speed of 60 mih At t 0 A starts and accelerates at a constant rate aA while at t 5 s B begins to slow down with a constant deceleration of magnitude aB6 Knowing that when the cars pass each other x 294 ft and vA vB determine a the acceleration aA b when the vehicles pass each other c the distance d between the vehicles at t 0 SOLUTION For t 0 vA 0 aAft xA 0 0 12 aAt² 0 t 5 s xB 0 vB0t vB0 60 mih 88 fts At t 5 s xB 88 fts5 s 440 ft For t 5 s vB vB0 aBt s aB 16 aA xB xBs vB0t s 12 aBt s² Assume t 5 s when the cars pass each other At that time tAB vA vB aAfAB 88 fts aA6 tAB 5 xA 294 ft 294 ft 12 aAfAB² PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 59 PROBLEM 1145 Continued Then aA12 tAB² tAB s 88 or 44aA²tAB 343tAB 245 0 Solving tAB 0795 s and tAB 700 s a With tAB 5 s 294 ft 12 aA700 s² or b From above Note An acceptable solution cannot be found if it is assumed that tAB 5 s c We have d x xBtAB 294 ft 440 ft 88 fts700 s 12 161200 fts²700 s² d 906 ft PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 60 Two blocks A and B are placed on an incline as shown At t 0 A is projected up the incline with an initial velocity of 27 fts and B is released from rest The blocks pass each other 1 s later and B reaches the bottom of the incline when t 34 s Knowing that the maximum distance from the bottom of the incline reached by block A is 21 ft and that the accelerations of A and B due to gravity and friction are constant and are directed down the incline determine a the accelerations of A and B b the distance d c the speed of A when the blocks pass each other a We have vA2 vA02 2aAxA 0 When xA xAm vA 0 Then 0 27 fts2 2aA21 ft or aA 173571 fts2 or Now xA 0 vA0t 12aAt2 and xB 0 0 12aBt2 At t 1 s the blocks pass each other xA1 xB1 d At t 34 s xB d Thus xA1 xB1 xB34 or 27 fts1 s 12173571 fts21 s2 12 aB1 s2 12 aB34 s2 or aB 34700 fts2 b At t 34 s xB d d 12 34700 fts234 s2 or c We have vA vA0 aAt At t 1 s vA 27 fts 173571 fts1 s or vA 964 fts PROBLEM 1147 Slider block A moves to the left with a constant velocity of 6 ms Determine a the velocity of block B b the velocity of portion D of the cable c the relative velocity of portion C of the cable with respect to portion D SOLUTION From the diagram we have xA 3yB constant Then vA 3vB 0 1 and aA 3aB 0 2 a Substituting into Eq 1 6 ms 3vB 0 or vB 2 ms b From the diagram yB yD constant Then vB vD 0 c From the diagram xA yC constant Then vA vC 0 Now vCID vC vD 6 ms 2 ms 8 ms vCID 8 ms BLOCK B starts from rest and moves downward with a constant acceleration Knowing that after slider block A has moved 400 mm its velocity is 4 ms determine a the accelerations of A and B b the velocity and the change in position of B after 2 s SOLUTION From the diagram we have xA 3yB constant Then vA 3vB 0 1 and aA 3aB 0 2 Eq 2 aA 3aB 0 and aB is constant and positive aA is constant and negative Also Eq 1 and vB0 0 vA0 0 Then vB2 0 2aAxA xA0 When ΔxA 04 m 4 ms2 2aA04 m or Then substituting into Eq 2 20 ms2 3aB 0 or aB 203 ms2 aB 667 ms2 We have vB 0 agt At t 2 s vB 203 ms22 s or Also yB yB0 0 12 agt2 At t 2 s yB yB0 12 203 ms22 s2 or yB yB0 1333 m PROBLEM 1149 The elevator shown in the figure moves downward with a constant velocity of 15 fts Determine a the velocity of the cable C b the velocity of the counterweight W c the relative velocity of the cable C with respect to the elevator d the relative velocity of the counterweight W with respect to the elevator SOLUTION Choose the positive direction downward a Velocity of cable C yC 2yE constant vC 2vE 0 vE 15 fts vC 30 fts b Velocity of counterweight W yW yE constant yW vE 0 yW vE 15 fts c Relative velocity of C with respect to E vCE vC vE 30 fts 15 fts 45 fts d Relative velocity of W with respect to E vWE vW vE 15 fts 15 fts 30 fts PROBLEM 1150 The elevator shown starts from rest and moves upward with a constant acceleration If the counterweight W moves through 30 ft in 5 s determine a the accelerations of the elevator and the cable C b the velocity of the elevator after 5 s SOLUTION We choose Positive direction downward for motion of counterweight ywr 12 aywt2 At t 5 s ywr 30 ft 30 ft 12 ayw5 s2 ayw 24 fts² a Accelerations of E and C Since ywr yE constant ywr vE 0 and ayw aE 0 Thus aE ayw 24 fts² Also yC 2yE constant vC 2vE 0 and aC 2aE 0 Thus aC 2aE 224 fts² 48 fts² aC 480 fts² b Velocity of elevator after 5 s vE vE₀ aEtE 0 24 fts²5 s 12 fts vE₁ 1200 fts PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 68 PROBLEM 1151 Collar A starts from rest and moves upward with a constant acceleration Knowing that after 8 s the relative velocity of collar B with respect to collar A is 24 ins determine a the accelerations of A and B b the velocity and the change in position of B after 6 s SOLUTION From the diagram 2 constant A B B A y y y y Then 2 0 A B v v 1 and 2 0 A B a a 2 a Eq 1 and 0 0 0 A B v v Also Eq 2 and A a is constant and negative B a is constant and positive Then 0 0 A A B B v a t v a t Now B A B A B A v v v a a t From Eq 2 1 2 B A a a So that 3 2 B A A v a t PROBLEM 1151 Continued At t 8 s 24 ins 32 aB8 s or and then aB 12 2 ins² or b At t 6 s vB 1 ins6 s or Now yB yB₀ 0 12 aBt² At t 6 s yB yB₀ 12 1 ins²6 s² or yB yB₀ 18 in PROBLEM 1152 In the position shown collar B moves downward with a velocity of 12 ins Determine a the velocity of collar A b the velocity of portion C of the cable c the relative velocity of portion C of the cable with respect to collar B SOLUTION From the diagram 2yA yB yB yA constant Then vA 2vB 0 1 and aA 2aB 0 2 a Substituting into Eq 1 vA 212 ins 0 or vA 24 ins b From the diagram 2yA yC constant Then 2vA vC 0 Substituting 224 ins vC 0 or c We have vCB vC vB 48 ins 12 ins or vCB 36 ins PROBLEM 1153 Slider block B moves to the right with a constant velocity of 300 mms Determine a the velocity of slider block A b the velocity of portion C of the cable c the velocity of portion D of the cable d the relative velocity of portion C of the cable with respect to slider block A PROBLEM 1153 Continued PROBLEM 1154 At the instant shown slider block B is moving with a constant acceleration and its speed is 150 mms Knowing that after slider block A has moved 240 mm to the right its velocity is 60 mms determine a the accelerations of A and B b the acceleration of portion D of the cable c the velocity and change in position of slider block B after 4 s Then substituting into Eq 2 2aB 3403 mms² 0 or aB 20 mms² b From the solution to Problem 1153 vD vA 0 Then aD aA 0 Substituting aD 403 mms² 0 or c We have vB vB₀ aBf At t 4 s vB 150 mms 200 mms²4 s or Also yB vB₀t vB₀t 12 aBf² At t 4 s yB yB₀ 150 mms4 s 12 200 mms²4 s² or yB yB₀ 440 mm PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 74 PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 83 PROBLEM 1159 The system shown starts from rest and each component moves with a constant acceleration If the relative acceleration of block C with respect to collar B is 60 2 mms upward and the relative acceleration of block D with respect to block A is 2 110 mms downward determine a the velocity of block C after 3 s b the change in position of block D after 5 s SOLUTION From the diagram Cable 1 2 2 constant A B C y y y Then 2 2 0 A B C v v v 1 and 2 2 0 A B C a a a 2 Cable 2 constant D A D B y y y y Then 2 0 A B D v v v 3 and 2 0 A B D a a a 4 Given At 0 0 t v all accelerations constant 2 60 mms aC B 2 110 mms aD A a We have 60 or 60 C B C B B C a a a a a and 110 or 110 D A D A A D a a a a a Substituting into Eqs 2 and 4 Eq 2 2 110 2 60 0 D C C a a a or 3 100 C D a a 5 Eq 4 110 60 2 0 D C D a a a or 50 C D a a 6 PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 100 PROBLEM 1169 Two road rally checkpoints A and B are located on the same highway and are 12 km apart The speed limits for the first 8 km and the last 4 km of the section of highway are 100 kmh and 70 kmh respectively Drivers must stop at each checkpoint and the specified time between Points A and B is 8 min 20 s Knowing that a driver accelerates and decelerates at the same constant rate determine the magnitude of her acceleration if she travels at the speed limit as much as possible SOLUTION Given max max max 100 kmh 70 kmh 0 min 20 s constant as such as possible AC CB A B AB v v v v t B a v v The v t curve is first drawn as shown where the magnitudes of the slopes accelerations of the three inclined lines are equal Note 5 8 min 20 s 36 h At 1 8 km t x 5 h 12 km 36 x Denoting the magnitude of the accelerations by a we have 100 30 70 a b c a t a t a t where a is in kmh2 and the times are in h Now 1 A 8 km 1 1 1 100 100 30 8 2 2 a b t t t PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 117 PROBLEM 1179 During a manufacturing process a conveyor belt starts from rest and travels a total of 12 ft before temporarily coming to rest Knowing that the jerk or rate of change of acceleration is limited to 48 fts2 per second determine a the shortest time required for the belt to move 12 ft b the maximum and average values of the velocity of the belt during that time SOLUTION Given At max 0 0 0 12 ft t x v x when 2 max max 0 48 fts da x x v dt a Observing that vmax must occur at 1 2 min t t the a t curve must have the shape shown Note that the magnitude of the slope of each portion of the curve is 48 fts2s We have at t t Δ max max 1 1 0 2 2 v t a a t Δ Δ 2 t t Δ max max max max 1 1 2 2 v a t t a a t Δ Δ Δ Using symmetry the v t is then drawn as shown Noting that 1 2 3 4 A A A A and that the area under the v t curve is equal to xmax we have max max 2 max max max max 2 2 t v x v a t a t x Δ Δ Δ PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 153 PROBLEM 11106 A basketball player shoots when she is 16 ft from the backboard Knowing that the ball has an initial velocity v0 at an angle of 30 with the horizontal determine the value of v0 when d is equal to a 9 in b 17 in SOLUTION First note 0 0 0 0 cos 30 sin 30 x y v v v v Horizontal motion Uniform 0 0 x x v t At B 0 0 16 16 cos 30 or cos 30 B d d v t t v Vertical motion Uniformly accelerated motion 2 2 0 1 0 322 fts 2 y y v t gt g At B 2 0 1 32 sin 30 2 B B v t gt Substituting for tB 2 0 0 0 16 1 16 32 sin 30 cos 30 2 cos 30 d d v g v v or 2 2 0 2 16 ft 1 3 16 32 3 g d v d d PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 155 PROBLEM 11107 A group of children are throwing balls through a 072minnerdiameter tire hanging from a tree A child throws a ball with an initial velocity v0 at an angle of 3 with the horizontal Determine the range of values of v0 for which the ball will go through the tire SOLUTION First note 0 0 0 0 cos 3 sin 3 x y v v v v Horizontal motion Uniform 0 0 x x v t When x 6 m 0 6 0 6 6 cos 3 or cos 3 v t t v Vertical motion Uniformly accelerated motion 2 2 0 1 0 981 ms 2 y y v t gt g When the ball reaches the tire 6 t t 0 0 2 0 6 sin 3 cos 3 1 6 2 cos 3 yB C v v g v or 2 0 2 18981 cos 3 6 tan 3 B C v y or 2 0 177065 0314447 B C v y At B y 053 m 2 0 177065 0314447 053 v or 0 1448 ms B v PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 157 PROBLEM 11108 The nozzle at A discharges cooling water with an initial velocity v0 at an angle of 6 with the horizontal onto a grinding wheel 350 mm in diameter Determine the range of values of the initial velocity for which the water will land on the grinding wheel between Points B and C SOLUTION First note 0 0 0 0 cos 6 sin 6 x y v v v v Horizontal motion Uniform 0 0 x x x v t Vertical motion Uniformly accelerated motion 2 2 0 0 1 981 ms 2 y y y v t gt g At Point B 0175 m sin 10 0175 m cos10 x y 0 0175 sin 10 0020 cos 6 x v t or 0 0050388 cos 6 Bt v 2 0 1 0175 cos10 0205 sin 6 2 B B y v t gt Substituting for Bt 2 0 0 0 0050388 1 0050388 0032659 sin 6 981 cos 6 2 cos 6 v v v PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 159 PROBLEM 11109 While holding one of its ends a worker lobs a coil of rope over the lowest limb of a tree If he throws the rope with an initial velocity v0 at an angle of 65 with the horizontal determine the range of values of v0 for which the rope will go over only the lowest limb SOLUTION First note 0 0 0 0 cos 65 sin 65 x y v v v v Horizontal motion Uniform 0 0 x x v t At either B or C x 5 m 0 cos 65 B C s v t or 0 5 cos 65 tB C v Vertical motion Uniformly accelerated motion 2 2 0 1 0 981 ms 2 y y v t gt g At the tree limbs B C t t 2 0 0 0 5 1 5 sin 65 cos 65 2 cos 65 yB C v g v v PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 162 PROBLEM 11111 A model rocket is launched from Point A with an initial velocity v0 of 250 fts If the rockets descent parachute does not deploy and the rocket lands 400 ft from A determine a the angle α that v0 forms with the vertical b the maximum height above Point A reached by the rocket and c the duration of the flight SOLUTION Set the origin at Point A 0 0 0 0 x y Horizontal motion 0 0 sin sin x x v t v t α α 1 Vertical motion 2 0 1 cos 2 y v t gt α 2 0 1 1 cos 2 y gt v t α 2 2 2 2 2 2 2 0 1 1 sin cos 1 2 x y gt v t α α 2 2 2 2 4 2 2 0 1 4 x y gyt g t v t 2 4 2 2 2 2 0 1 0 4 g t v gy t x y 3 At Point B 2 2 400 ft 400 cos 30 ft 400 sin 30 200 ft x y x y 2 4 2 2 2 4 2 1 322 250 322 200 400 0 4 25921 68940 160000 0 t t t t 2 2 2 26362 s and 234147 s 162364 s and 153019 s t t PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 164 PROBLEM 11112 The initial velocity v0 of a hockey puck is 105 mih Determine a the largest value less than 45 of the angle α for which the puck will enter the net b the corresponding time required for the puck to reach the net SOLUTION First note 0 105 mih 154 fts v and 0 0 0 0 cos 154 fts cos sin 154 fts sin x y v v v v α α α α a Horizontal motion Uniform 0 0 154 cos x x v t α t At the front of the net x 16 ft Then 16 154 cos αt or enter 8 77 cos t α Vertical motion Uniformly accelerated motion 2 0 2 2 1 0 2 1 154 sin 322 fts 2 y y v t gt t gt g α At the front of the net 2 front enter enter 2 2 1 154 sin 2 8 1 8 154 sin 77 cos 2 77 cos 32 16 tan 5929 cos y t gt g g α α α α α α Now 2 2 2 1 sec 1 tan cos α α α PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 166 PROBLEM 11113 The pitcher in a softball game throws a ball with an initial velocity v0 of 72 kmh at an angle α with the horizontal If the height of the ball at Point B is 068 m determine a the angle α b the angle θ that the velocity of the ball at Point B forms with the horizontal SOLUTION First note 0 72 kmh 20 ms v and 0 0 0 0 cos 20 ms cos sin 20 ms sin x y v v v v α α α α a Horizontal motion Uniform 0 0 20 cos x x v t α t At Point B 7 14 20 cos or 10 cos B t t α α Vertical motion Uniformly accelerated motion 2 2 2 0 1 1 0 20 sin 981 ms 2 2 y y v t gt t gt g α At Point B 2 1 008 20 sin 2 B B t gt α Substituting for Bt 2 7 1 7 008 20 sin 10 cos 2 g 10 cos α α α or 2 1 49 8 1400 tan 2 cos g α α Now 2 2 2 1 sec 1 tan cos α α α PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 168 PROBLEM 11114 A mountain climber plans to jump from A to B over a crevasse Determine the smallest value of the climbers initial velocity v0 and the corresponding value of angle α so that he lands at B SOLUTION First note 0 0 0 0 cos sin x y v v v v α α Horizontal motion Uniform 0 0 0 cos x x v t v α t At Point B 0 18 cos v α t or 0 18 cos Bt v α Vertical motion Uniformly accelerated motion 2 0 2 2 0 1 0 2 1 sin 981 ms 2 y y v t gt v t gt g α At Point B 2 0 1 14 sin 2 B B v t gt α Substituting for Bt 2 0 0 0 18 1 18 14 sin cos 2 cos v g v v α α α or 2 0 2 2 162 cos 18 tan 14 162 09 sin 2 14 cos g v g α α α α PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 172 PROBLEM 11116 A worker uses highpressure water to clean the inside of a long drainpipe If the water is discharged with an initial velocity v0 of 115 ms determine a the distance d to the farthest Point B on the top of the pipe that the worker can wash from his position at A b the corresponding angle α SOLUTION First note 0 0 0 0 cos 115 ms cos sin 115 ms sin x y v v v v α α α α By observation dmax occurs when max 11 m y Vertical motion Uniformly accelerated motion 2 0 0 2 1 0 2 1 115 sin 115 sin 2 y y y v v gt y v t gt gt t gt α α When max at 0 y B y y B v Then 0 115 sin vy B gt α or 2 115 sin 981 ms Bt g g α and 2 1 115 sin 2 B B B y t gt α Substituting for Bt and noting 11 m yB 2 2 2 115 sin 1 115 sin 11 115 sin 2 1 115 sin 2 g g g g α α α α or 2 2 22 981 sin 238265 115 α α PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 187 PROBLEM 11127 Conveyor belt A which forms a 20 angle with the horizontal moves at a constant speed of 4 fts and is used to load an airplane Knowing that a worker tosses duffel bag B with an initial velocity of 25 fts at an angle of 30 with the horizontal determine the velocity of the bag relative to the belt as it lands on the belt SOLUTION First determine the velocity of the bag as it lands on the belt Now 0 0 0 0 cos 30 25 ftscos 30 sin 30 25 ftssin 30 B x B B y B v v v v Horizontal motion Uniform 0 0 B x x v t 0 B x B x v v 25 cos 30 t 25 cos 30 Vertical motion Uniformly accelerated motion 2 0 0 1 2 B y y y v t gt 0 B y B y v v gt 2 1 15 25 sin 30 2 t gt 25 sin 30 gt The equation of the line collinear with the top surface of the belt is tan 20 y x Thus when the bag reaches the belt 2 1 15 25 sin 30 25 cos 30 tan 20 2 t gt t or 1 322 2 25cos 30 tan 20 sin 30 15 0 2 t t or 161 2 046198 15 0 t t Solving 031992 s and 029122 s t t Reject PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 191 PROBLEM 11130 When a small boat travels north at 5 kmh a flag mounted on its stern forms an angle θ 50 with the centerline of the boat as shown A short time later when the boat travels east at 20 kmh angle θ is again 50 Determine the speed and the direction of the wind SOLUTION We have W B W B v v v Using this equation the two cases are then graphically represented as shown With vW now defined the above diagram is redrawn for the two cases for clarity Noting that 180 50 90 40 θ α α 180 50 130 φ α α We have 5 20 sin 50 sin 40 sin 50 sin 130 W W v v α α PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 204 PROBLEM 11141 A motorist traveling along a straight portion of a highway is decreasing the speed of his automobile at a constant rate before exiting from the highway onto a circular exit ramp with a radius of 560ft He continues to decelerate at the same constant rate so that 10 s after entering the ramp his speed has decreased to 20 mih a speed which he then maintains Knowing that at this constant speed the total acceleration of the automobile is equal to onequarter of its value prior to entering the ramp determine the maximum value of the total acceleration of the automobile SOLUTION First note 10 88 20 mih 3 fts v While the car is on the straight portion of the highway straight t a a a and for the circular exit ramp 2 2 t n a a a where 2 n v a ρ By observation amax occurs when v is maximum which is at t 0 when the car first enters the ramp For uniformly decelerated motion 0 t v v a t and at t 10 s 2 10 constant n v v a a ρ st 1 4 a a Then 2 88 2 3 10 straight fts 1 4 560 ft t t v a a a ρ or ta 61460 fts2 The car is decelerating hence the minus sign PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 213 PROBLEM 11148 A child throws a ball from Point A with an initial velocity A v of 20 ms at an angle of 25 with the horizontal Determine the velocity of the ball at the points of the trajectory described by the ball where the radius of curvature is equal to threequarters of its value at A SOLUTION Assume that Point B and C are the points of interest where and B C B C y y v v Now 2 A A n A v a ρ or 2 cos 25 A A v g ρ Then 2 3 3 4 4 cos 25 A B A v g ρ ρ We have 2 B B n B v a ρ where cos aB n g θ so that 2 2 3 4 cos25 cos A B v v g g θ or 2 2 3 cos 4 cos25 B A v θ v 1 Noting that the horizontal motion is uniform we have A x B x v v where cos25 cos A x A B x B v v v v θ Then cos25 cos A B v v θ or cos cos25 A B v θ v PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 216 PROBLEM 11150 A projectile is fired from Point A with an initial velocity 0 v which forms an angle α with the horizontal Express the radius of curvature of the trajectory of the projectile at Point C in terms of 0 x v α and g SOLUTION We have 2 C C n C v a ρ or 2 cos C C v g ρ θ Noting that the horizontal motion is uniform we have 0 0 0 cos A x C x x v v x v t v α t where 0 cos cos A x C x C v v v v α θ Then 0 0 cos cos and cos C C x v v v v α θ α 1 or 0 cos cos C v v θ α so that 3 0 cos C C v g v ρ α For the uniformly accelerated vertical motion have 0 0 sin C y y v v gt v gt α From above 0 0 cos or cos x x v t t v α α Then 0 0 sin cos C y x v v g v α α 2 Now 2 2 2 C C x C y v v v PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 219 PROBLEM 11152 Determine the radius of curvature of the path described by the particle of Problem 1196 when t 0 A 3 and B 1 SOLUTION With 3 1 A B We have 2 3 cos 3 1 sin t t t t t r i j k Now 2 3 3cos sin sin cos 1 t d t t t t t t dt t r v i j k and 2 2 1 2 2 12 1 3 sin sin cos 3 1 cos cos sin 1 32sin cos 3 1 2cos sin t t d t t t t t t dt t t t t t t t t t t t t v a i j k i j k Then 2 2 2 2 2 9cos sin 9 sin cos 1 t v t t t t t t t Expanding and simplifying yields 2 4 2 2 4 2 3 19 1 8cos sin 8 sin 2 v t t t t t t t t Then 4 2 2 4 2 3 1 2 19 1 8cos sin 8 sin 2 v t t t t t t t t and 3 3 2 4 2 3 4 2 2 4 2 3 1 2 4 38 8 2cos sin 4 sin 2 sin cos 83 1sin 2 2 cos2 2 19 1 8cos sin 8 sin 2 dv t t t t t t t t t t t t t t dt t t t t t t t t Now 2 2 2 2 2 2 t n dv v a a a dt ρ PROPRIETARY MATERIAL 2009 The McGrawHill 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 McGrawHill for their individual course preparation If you are a student using this Manual you are using it without permission 256 PROBLEM 11180 For the conic helix of Problem 1195 determine the angle that the osculating plane forms with the y axis SOLUTION First note that the vectors v and a lie in the osculating plane Now cos sin n n Rt t ct Rt t ω ω r i j k Then cos sin sin cos n n n n n n dr R t t t c R t t t dt ω ω ω ω ω ω v i j k and 2 2 sin sin cos cos cos sin 2sin cos 2cos sin n n n n n n n n n n n n n n n n n n n dv dt R t t t t R t t t t R t t t t t t ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω a i k i k It then follows that the vector v a is perpendicular to the osculating plane cos sin sin cos 2sin cos 0 2cos sin n n n n n n n n n n n n n R R t t t c R t t t t t t t t t ω ω ω ω ω ω ω ω ω ω ω ω ω i j k v a 2 2 2cos sin sin cos 2sin cos cos sin 2cos sin 2sin cos 2cos sin 2 2sin cos n n n n n n n n n n n n n n n n n n n n n n n n n n n R c t t t R t t t t t t t t t t t t c t t t R c t t t R t c t t t ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω i j k i j k