Exercícios Resolvidos de Cinemática - Aceleração, Velocidade e Deslocamento

121 A bicyclist starts from rest and after traveling along a straight path a distance of 20 m reaches a speed of 30 kmh Determine his acceleration if it is constant Also how long does it take to reach the speed of 30 kmh v2 30 kmh 833 ms v22 v12 2 ac s2 s1 8332 0 2 ac 20 0 ac 174 ms2 Ans v2 v1 ac t 833 0 174t t 480 s Ans 122 A car starts from rest and reaches a speed of 80 fts after traveling 500 ft along a straight road Determine its constant acceleration and the time of travel v22 v12 2 ac s2 s1 802 0 2 ac 500 0 ac 640 fts2 Ans v2 v1 ac t 80 0 64t t 125 s Ans 123 A baseball is thrown downward from a 50ft tower with an initial speed of 18 fts Determine the speed at which it hits the ground and the time of travel v22 v12 2 ac s2 s1 v22 182 2 32250 0 v2 59532 595 fts Ans v2 v1 ac t 59532 18 322 t t 129 s Ans 124 A particle travels along a straight line such that in 2 s it moves from an initial position sA 05 m to a position sB 15 m Then in another 4 s it moves from sB to sC 25 m Determine the particles average velocity and average speed during the 6s time interval Total displacement sC sA 2 m Total distance traveled 05 15 15 25 6 m Total time traveled 2 4 6 s vavg 26 0333 ms Ans vsp 66 1 ms Ans 125 Traveling with an initial speed of 70 kmh a car accelerates at 6000 kmh2 along a straight road How long will it take to reach a speed of 120 kmh Also through what distance does the car travel during this time v v1 ac t 120 70 6000t t 833103 hr 30 s Ans v2 v12 2 ac s s1 1202 702 26000s 0 s 0792 km 792 m Ans 126 A freight train travels at v 601 et fts where t is the elapsed time in seconds Determine the distance traveled in three seconds and the acceleration at this time Prob 126 v 601 et 0s ds v dt 0t 601 et dt s 60t et 03 s 123 ft Ans a dvdt 60et At t 3 s a 60 e3 299 fts2 Ans 127 The position of a particle along a straight line is given by s t3 9t2 15t ft where t is in seconds Determine its maximum acceleration and maximum velocity during the time interval 0 t 10 s s t3 9t2 15t v dsdt 3t2 18t 15 a dvdt 6t 18 amax occurs at t 10 s amax 610 18 42 fts2 Ans vmax occurs when t 10 s vmax 3102 1810 15 135 fts Ans 128 From approximately what floor of a building must a car be dropped from an atrest position so that it reaches a speed of 807 fts 55 mih when it hits the ground Each floor is 12 ft higher than the one below it Note You may want to remember this when traveling 55 mih v2 v02 2 ac s s0 8072 0 2322s 0 s 10113 ft of floors 1011312 843 The car must be dropped from the 9th floor Ans 129 A car is to be hoisted by elevator to the fourth floor of a parking garage which is 48 ft above the ground If the elevator can accelerate at 06 fts decelerate at 03 fts and reach a maximum speed of 8 fts determine the shortest time to make the lift starting from rest and ending at rest v2 v12 2 ac s s1 vmax2 0 206y 0 0 vmax2 20348 y 0 0 12 y 06 48 y y 160 ft vmax 4382 fts 8 fts v v1 ac t 4382 0 06 t1 t1 7303 s 0 4382 03 t2 t2 1461 s t t1 t2 219 s Ans 1210 A particle travels in a straight line such that for a short time 2 s t 6 s its motion is described by v 4a fts where a is in fts2 If v 6 fts when t 2 s determine the particles acceleration when t 3 s a dvdt 4v v dv 4 dt 12 v2 18 4t 8 v2 8t 20 At t3 s choosing the positive root v 663 fts a 4663 0603 fts² Ans 1211 The acceleration of a particle as it moves along a straight line is given by a 2t 1 ms2 where t is in seconds If s 1 m and v 2 ms when t 0 determine the particles velocity and position when t 6 s Also determine the total distance the particle travels during this time period 2 dv 0 2t1 dt v t² t 2 1 ds 0 t² t 2 dt s 13 t³ 12 t² 2t 1 When t 6 s v 32 ms Ans s 67 m Ans Since v 0 then d 67 1 66 m Ans 1212 When a train is traveling along a straight track at 2 ms it begins to accelerate at a 60 v4 ms2 where v is in ms Determine its velocity v and the position 3 s after the acceleration Prob 1212 a dvdt dt dva 03 dt 1v dv 60 v4 3 1300 v5 32 v 3925 ms 393 ms Ans adt v dv ds v dva 160 v5 dv 0s ds 160 23925 v5 dv s 160 v6 6 9925 998 m Ans 1213 The position of a particle along a straight line is given by s 15t3 135 t2 225 t ft where t is in seconds Determine the position of the particle when t 6 s and the total distance it travels during the 6s time interval Hint Plot the path to determine the total distance traveled Position The position of the particle when t 6 s is st6 1563 13562 2256 270 ft Ans Total Distance Traveled The velocity of the particle can be determined by applying Eq 121 v dsdt 450t2 270 t 225 The times when the particle stops are 450 t² 270 t 225 0 t1s and t5s The position of the particle at t0 s 1 s and 5 s are st0 1503 13502 2250 0 st1 1513 13512 2251 105 ft st5 1553 13552 2255 375 ft From the particles path the total distance is stot 105 480 105 690 ft Ans 1214 The position of a particle on a straight line is given by s t3 9t2 15t ft where t is in seconds Determine the position of the particle when t 6 s and the total distance it travels during the 6s time interval Hint Plot the path to determine the total distance traveled s t3 9t2 15t v dsdt 3t2 18t 15 v 0 when t1 s and t5 s t0 s0 t1 s s7 ft t5 s s25 ft t6 s s18 ft Ans sT 7 7 25 25 18 46 ft Ans 1215 A particle travels to the right along a straight line with a velocity v 54 s ms where s is in meters Determine its position when t 6 s if s 5 m when t0 dsdt 54s 5s 4 s ds 5t dt 4s 05 s2 325 5t When t 6 s s2 8s 125 0 Solving for the positive root s 787 m Ans 1216 A particle travels to the right along a straight line with a velocity v 54 s ms where s is in meters Determine its deceleration when s 2 m v 5 4s v dv a ds dv 5 ds 4 s2 5 4 s 5 ds 4 s2 a ds a 25 4 s3 When s 2 m a 0116 ms2 Ans 1217 Two particles A and B start from rest at the origin s 0 and move along a straight line such that aA 6t 3 fts2 and aB 12t2 8 fts2 where t is in seconds Determine the distance between them when t 4 s and the total distance each has traveled in t 4 s Velocity The velocity of particles A and B can be determine using Eq 122 d vA aA dt integral from 0 to 4 of d vA integral from 0 to 4 of 6t 3 dt vA 3t2 3t d vB aB dt integral from 0 to 4 of d vB integral from 0 to 4 of 12t2 8 dt vB 4t3 8t The times when particle A stops are 3t2 3t 0 t0 s and t1 s The times when particle B stops are 4t3 8t0 t0 s and t square root 2 s Position The position of particles A and B can be determine using Eq 121 d sA vA dt integral from 0 to sA of d sA integral from 0 to sA of 3t2 3tdt sA t3 3 over 2 t2 d sB vB dt integral from 0 to sB of d sB integral from 0 to sB of 4t3 8tdt sB t4 4t2 The positions of particle A at t1 s and 4 s are sA t1s 13 3 over 2 12 0500 ft sA t4s 43 3 over 2 42 400 ft Particle A has traveled dA 205 400 410 ft Ans The positions of particle B at t square root 2 s and 4 s are sB t sqrt 2 sqrt 24 4sqrt 22 4 ft sB t4 44 442 192 ft Particle B has traveled dB 24 192 200 ft Ans At t4 s the distance between A and B is delta sAB 192 40 152 ft Ans 1218 A car starts from rest and moves along a straight line with an acceleration of a 3s13 ms2 where s is in meters Determine the cars acceleration when t4 s a 3s13 a ds v dv integral from 0 to s of 3s13 ds integral from 0 to v of v dv 3 over 2 3s23 1 over 2 v2 v 3s13 d s dt 3s13 integral from 0 to s13 ds integral from 0 to 3 dt 3 over 2 s13 3t s 2t32 s t4 2432 2262 226 m da t4 3226213 106 ms2 Ans 1219 A stone A is dropped from rest down a well and in 1s another stone B is dropped from rest Determine the distance between the stones another second later equation l s s1 v1 t 1 over 2 a t2 sA 0 0 1 over 2 32222 sA 644 ft sB 0 0 1 over 2 32212 sB 161 ft delta s 644 161 483 ft 1220 A stone A is dropped from rest down a well and in 1 s another stone B is dropped from rest Determine the time interval between the instant A strikes the water and the instant B strikes the water Also at what speed do they strike the water B is dropped one second after A so that delta t 1 s Ans l s s0 v0 t 1 over 2 a t2 80 0 0 1 over 2 322t2 t 22291 s l v v0 a t v 0 32222291 v 718 fts Ans Also v2 v02 2as v2 02 232280 v 718 fts Ans 1221 A particle travels in a straight line with accelerated motion such that a ks where s is the distance from the starting point and k is a proportionality constant which is to be determined For s 2 ft the velocity is 4 fts and for s 35 ft the velocity is 10 fts What is s when v0 a ks ads v dv k integral from s2 to s ds integral from v4 to v v dv k s2 over 2 22 over 2 v2 over 2 42 over 2 k s2 over 2 2 v2 over 2 8 Set s35 ft v10 fts Then k 1018 102 s2 Ans When v0 1018s2 over 2 2 8 s 156 ft Ans 1222 The acceleration of a rocket traveling upward is given by a 6 002s ms2 where s is in meters Determine the rockets velocity when s 2 km and the time needed to reach this altitude Initially v 0 and s 0 when t 0 1223 The acceleration of a rocket traveling upward is given by a 6 002s ms2 where s is in meters Determine the time needed for the rocket to reach an altitude of s 100 m Initially v 0 and s 0 when t 0 1224 At t 0 bullet A is fired vertically with an initial muzzle velocity of 450 ms When t 3 s bullet B is fired upward with a muzzle velocity of 600 ms Determine the time t after A is fired as to when bullet B passes bullet A At what altitude does this occur 1225 A particle moves along a straight line with an acceleration of a 53s13 s52 ms2 where s is in meters Determine the particles velocity when s 2 m if it starts from rest when s 1 m Use Simpsons rule to evaluate the integral 1226 Ball A is released from rest at a height of 40 ft at the same time that a second ball B is thrown upward 5 ft from the ground If the balls pass one another at a height of 20 ft determine the speed at which ball B was thrown upward 1227 A projectile initially at the origin moves vertically downward along a straightline path through a fluid medium such that its velocity is defined as v 38et t12 ms where t is in seconds Plot the position s of the projectile during the first 2 s Use the RungeKutta method to evaluate s with incremental values of h 025 s 1228 The acceleration of a particle along a straight line is defined by a 2t 9 ms2 where t is in seconds At t 0 s 1 m and v 10 ms When t 9 s determine a the particles position b the total distance traveled and c the velocity 1229 A particle is moving along a straight line such that when it is at the origin it has a velocity of 4 ms If it begins to decelerate at the rate of a 15v12 ms2 where v is in ms determine the distance it travels before it stops a dvdt 15v12 4v12 dv 015 dt 2v12 15t0 2v122 15t v 2 075t2 ms 1 0s ds 02 075t2 dt 04 3t 05625t2 dt s 4t 15t2 01875t3 2 From Eq 1 the particle will stop when 0 2 075t2 t 2667 s st2667 42667 1526672 0187526673 356 m Ans 1230 A particle moves along a straight line with an acceleration of a 53s13 s52 ms2 where s is in meters Determine the particles velocity when s 2 m if it starts from rest when s 1 m Use Simpsons rule to evaluate the integral a 53s13 s52 a ds v dv 12 5 ds3s13 s52 0 v dv 08351 12 v2 v 129 ms Ans 1231 Determine the time required for a car to travel 1 km along a road if the car starts from rest reaches a maximum speed at some intermediate point and then stops at the end of the road The car can accelerate at 15 ms2 and decelerate at 2 ms2 Using formulas of constant acceleration v2 15 t1 x 1215t12 0 v2 2 t2 1000 x v2 t2 122t22 Combining equations t1 133 t2 v2 2 t2 x 133 t2 1000 133 t22 2 t22 t22 t2 20702 s t1 27603 s t t1 t2 483 s Ans 1232 When two cars A and B are next to one another they are traveling in the same direction with speeds vA and vB respectively If B maintains its constant speed while A begins to decelerate at aA determine the distance d between the cars at the instant A stops Motion of carA v v0 aA t 0 vA aA t t vA aA v2 v02 2 aA s s0 0 v12 2aAsA 0 sA v12 2 aA Motion of carB sB vB t vBvAaA vA vB aA The distance between cars A and B is sBA sB sA vA vBaA v122 aA 2 vA vB v122 aA Ans 1233 If the effects of atmospheric resistance are accounted for a freely falling body has an acceleration defined by the equation a 9811 v2104 ms2 where v is in ms and the positive direction is downward If the body is released from rest at a very high altitude determine a the velocity when t 5 s and b the bodys terminal or maximum attainable velocity as t a a dvdt 9811 v2104 dv 104 v2 0 98110 4 dt 1 1100 tanh1 v1000 981 104 t tanh1 v100 981 102 t 2 v 100 tanh 981 102 5 100 tanh04905 455 ms Ans b From Eq 2 with t v 100 tanh 100 ms Ans Also note that Eq 1 can be written as 104 0 dv 104 v2 981 t 10412102 ln102 v 102 v 0 981 t 50 ln100 v 100 v ln1 981 t When t 5 s 100 v 100 v e0981 2667 100 v 2667 2667v v 16673667 455 ms Ans 1234 As a body is projected to a high altitude above the earths surface the variation of the acceleration of gravity with respect to altitude y must be taken into account Neglecting air resistance this acceleration is determined from the formula a g0 R2R y2 where g0 is the constant gravitational acceleration at sea level R is the radius of the earth and the positive direction is measured upward If g0 981 ms2 and R 6356 km determine the minimum initial velocity escape velocity at which a projectile should be shot vertically from the earths surface so that it does not fall back to the earth Hint This requires that v 0 as y v dv a dy v dv g0 R2 dy R y2 v220 g0 R2 R y0 v 2g0 R 29816356103 11167 ms 112 kms Ans 1235 Accounting for the variation of gravitational acceleration a with respect to altitude y see Prob 1234 derive an equation that relates the velocity of a freely falling particle to its altitude Assume that the particle is released from rest at an altitude y0 from the earths surface With what velocity does the particle strike the earth if it is released from rest at an altitude y0 500 km Use the numerical data in Prob 1234 From Prob 1234 I g go R2Ry2 Since g dy v dv then go R2 y0 dyRy2 0v v dv goR2 1Ryy0y v22 goR2 1Ry 1Ry0 v22 Thus v R2goy0 y R yR y0 When y0 500 km y 0 v 63561032981500103 63566356 500106 v 3016 ms 302 kms Ans 1236 When a particle falls through the air its initial acceleration a g diminishes until it is zero and thereafter it falls at a constant or terminal velocity vf If this variation of the acceleration can be expressed as a gvf2vf2 v2 determine the time needed for the velocity to become v vf Initially the particle falls from rest dvdt a gvf2vf2 v2 0v dvvf2v2 gvf2 0t dt 12vf lnvf vvf v0v gvf2 t t vf2g lnvf vvf v Ans 1237 An airplane starts from rest travels 5000 ft down a runway and after uniform acceleration takes off with a speed of 162 mih It then climbs in a straight line with a uniform acceleration of 3 fts2 until it reaches a constant speed of 220 mih Draw the st vt and at graphs that describe the motion v1 0 v2 162 mih 1 h 5280 fth 3600 s1 mi 2376 fts v22 v12 2a1s2 s1 23762 02 2a15000 0 a1 564538 fts2 v2 v1 a1 t 2376 0 564538 t t 4209 421 s v3 220 mih 1 h 5280 fth 3600 s1 mi 32267 fts v32 v22 2 a2s3 s2 322672 23762 23s 5000 s 12 94334 ft v3 v2 a2 t 32267 2376 3 t t 284 s 1238 The elevator starts from rest at the first floor of the building It can accelerate at 5 fts2 and then decelerate at 2 fts2 Determine the shortest time it takes to reach a floor 40 ft above the ground The elevator starts from rest and then stops Draw the at vt and st graphs for the motion 40 ft t v2 v1 a1 t vmax 0 5 t1 t v3 v1 a1 t 0 vmax 2 t2 Thus t1 04 t2 t s2 s1 v1 t 12 a1 t12 h 0 0 12 5t12 25 t12 t 40 h 0 vmax t2 12 2 t22 t v2 v12 2 a2s s1 vmax2 0 25h 0 vmax2 10 h 0 vmax2 2240 h vmax2 160 4 h Thus 10 h 160 4 h h 11429 ft vmax 1069 fts t1 2138 s t2 5345 s t t1 t2 748 s Ans 1239 A freight train starts from rest and travels with a constant acceleration of 05 fts² After a time t it maintains a constant speed so that when t 160 s it has traveled 2000 ft Determine the time t and draw the vt graph for the motion Total Distance Traveled The distance for part one of the motion can be related to time t t by applying Eq 125 with s₀ 0 and v₀ 0 s s₀ v₀ t ½ a t² s₁ 0 0 ½ 05t² 025 t² The velocity at time t can be obtained by applying Eq 124 with v₀ 0 v v₀ a t 0 05 t 05 t The time for the second stage of motion is s₂ 160 t and the train is traveling at a constant velocity of v 05 t Eq 11 Thus the distance for this part of motion is s₂ v t₂ 05 t 160 t 80 t 05 t² If the total distance traveled is s tot 2000 then s tot s₁ s₂ 2000 025 t² 80 t 05 t² 025 t² 80 t 2000 0 Choose a root that is less than 160 s then t 2734 s 273 s Ans vt Graph The equation for the velocity is given by Eq 1 When t t 2734 s v 05 2734 137 fts 1240 If the position of a particle is defined by s 2 sin π5 t 4 m where t is in seconds construct the st vt and at graphs for 0 t 10 s 1241 The vt graph for a particle moving through an electric field from one plate to another has the shape shown in the figure The acceleration and deceleration that occur are constant and both have a magnitude of 4 ms² If the plates are spaced 200 mm apart determine the maximum velocity vmax and the time t for the particle to travel from one plate to the other Also draw the st graph When t t2 the particle is at s 100 mm aₙ 4 ms² s2 100 mm 01 m v² v₀² 2 aₙ s s₀ vmax² 0 2 4 01 0 vmax 089442 ms Ans v v₀ aₙ t 089442 0 4 t2 t 044721 s Ans s s₀ v₀ t ½ aₙ t² s 0 0 ½ 4 t² s 2 t² When t 0447212 02236 0224 s s 01 m ₀₈₉₄ dv ₀₂₂₃₅ 4 dt v 4 t 1788 ₀₁ ds ₀₂₂₃₅ 4 t 1788 dt s 2 t² 1788 t 02 When t 0447 s s 02 m 1242 The vt graph for a particle moving through an electric field from one plate to another has the shape shown in the figure where t 02 s and vmax 10 ms Draw the st and at graphs for the particle When t t2 the particle is at s 05 m For 0 t 01 s v 100t a dvdt 100 ds v dt ₀ᵗ ds ₀ᵗ 100 t dt s 50t² When t 01 s s 05 m For 01 s t 02 s v 100t 20 a dvdt 100 ds v dt ₀₅ ds ₀₁ 100 t 20 dt s 05 50 t² 20 t 15 s 50 t² 20 t 1 When t 02 s s 1 m 1243 The as graph for a jeep traveling along a straight road is given for the first 300 m of its motion Construct the vs graph At s 0 v 0 as Graph The function of acceleration ain terms of s for the interval 0 m s 200 m is a0 20 a 001s ms2 s0 2000 For the interval 200 m s 300 m a2 02 a 002s 6 ms2 s200 300200 vs Graph The function of velocity v in terms of s can be obtained by applying vdv ads For the interval 0 m s 200 m vdv ads 0 vdv 0 001sds v 01s ms At s 200 m v 0100200 200 ms For the interval 200 m s 300 m vdv ads 200 ms vdv 200 m 002s 6 ds v 002s2 12s 1200 ms At s 300 m v 0023002 12300 1200 245 ms 1244 A motorcycle starts from rest at s 0 and travels along a straight road with the speed shown by the vt graph Determine the motorcycles acceleration and position when t 8 s and t 12 s At t 8 s Ans a dv dt 0 s v dt s 0 12 45 845 30 s 30 m Ans At t 12 s a dv dt 55 1 ms2 Ans s v dt s 0 12 45 1045 12 15105 12 355355 s 48 m Ans 1245 An airplane lands on the straight runway originally traveling at 110 fts when s 0 If it is subjected to the decelerations shown determine the time t needed to stop the plane and construct the st graph for the motion vo 110 fts v a dt 0 110 3155 82015 3t 20 t 333 s Ans 1246 A race cars starting from rest travels along a straight road and for 10 s has the acceleration shown Construct the vt graph that describes the motion and find the distance traveled in 10 s v t Graph The velocity function in terms of time t can be obtained by applying formula a dv dt For time interval 0 t 6 s dv adt 0 dv 0 16 t2 dt v 118 t3 ms At t 6 s v 118 63 120 ms For time interval 6 s t 10 s dv adt 6 120 ms dv 6 6dt v 6t 24 ms At t 10 s v 610 24 360 ms Position The position in terms of time t can be obtained by applying v ds dt For time interval 0 s 6 s ds vdt 0 ds 0 118 t3 dt s 172 t4 m When t 6 s s 172 64 180 m For time interval 6 s t 10 s ds vdt 180 m ds 6 6t 24 dt s 3t2 24t 54 m When t 10 s s 3102 2410 54 114 m Ans 1247 The vt graph for the motion of a train as it moves from station A to station B is shown Draw the at graph and determine the average speed and the distance between the stations For 0 t 30 s a ΔvΔt 4030 133 fts2 For 30 t 90 s a ΔvΔt 0 For 90 t 120 s a ΔvΔt 0 40 120 90 133 fts2 Δs v dt s 0 12 4030 4090 30 12 40120 90 s 3600 ft Ans vpavg ΔsΔt 3600120 30 fts Ans 1248 The velocity of a car is plotted as shown Determine the total distance the car moves until it stops t 80 s Construct the at graph Distance Traveled The total distance traveled can be obtained by computing the area under the vt graph s 1040 12 1080 40 600 m Ans at Graph The acceleration in terms of time t can be obtained by applying a dvdt For time interval 0 t 40 s a dvdt 0 For time interval 40 s t 80 s v 10 t 40 0 10 80 40 v 14 t 20 ms a dvdt 14 0250 ms2 1249 The vt graph for the motion of a car as it moves along a straight road is shown Draw the at graph and determine the maximum acceleration during the 30s time interval The car starts from rest at s 0 For t 10 s v 04t2 a dvdt 08 t At t 10 s a 8 fts2 For 10 t 30 s v t 30 a dvdt 1 amax 8 fts2 Ans 1250 The vt graph for the motion of a car as it moves along a straight road is shown Draw the st graph and determine the average speed and the distance traveled for the 30s time interval The car starts from rest at s 0 For t 10 s v 04t2 ds v dt 0t ds 0t 04 t2 dt s 01333 t3 At t10 s s 1333 ft For 10 t 30 s v t 30 ds v dt 1333t ds 10t t 30 dt s 05 t2 30 t 2167 At t 30 s s 1133 ft vpavg ΔsΔt 113330 378 fts Ans sT 1133 ft 113103 ft Ans 1251 A car travels along a straight road with the speed shown by the vt graph Determine the total distance the car travels until it stops when t 48 s Also plot the st and at graphs 0 i 30 v 15 a dvdt 15 0t ds 0t 15 t dt s 110 t2 When t 30 s s 90 m v 13 t 48 a dvdt 13 90t ds 30t 13 t 48 dt s 16 t2 16 t 240 When t 48 s s 144 m Ans Also from vt graph Δ s v dt s 0 12 648 144 m Ans 1252 A man riding upward in a freight elevator accidentally drops a package off the elevator when it is 100 ft from the ground If the elevator maintains a constant upward speed of 4 fts determine how high the elevator is from the ground the instant the package hits the ground Draw the vt curve for the package during the time it is in motion Assume that the package was released with the same upward speed as the elevator For package v2 v₀2 2aₐs₂ s₀ v2 42 2322100 0 v 8035 fts v v₀ aₐ t 8035 4 322 t t 2620 s For elevator s₂ s₀ vt s 100 42620 s 110 ft Ans 1253 Two cars start from rest side by side and travel along a straight road Car A accelerates at 4 ms² for 10 s and then maintains a constant speed Car B accelerates at 5 ms² until reaching a constant speed of 25 ms and then maintains this speed Construct the at vt and st graphs for each car until t 15 s What is the distance between the two cars when t 15 s Car A v v₀ aₐ t vₐ 0 4t At t 10 s vₐ 40 ms s s₀ v₀ t 12 aₐ t² sₐ 0 0 12 4t² 2 t² At t 10 s sₐ 200 m t 10 s ds v dt 200t ds 10t 40 dt sₐ 40 t 200 At t 15 s sₐ 400 m Car B v v₀ aₐ t vb 0 5 t When vb 25 ms t 255 5 s s s₀ v₀ t 12 aₐ t² sb 0 0 12 5 t² 25 t² At t 5 s sb 625 m t 5 s ds v dt 625t ds 5t 25 dt sb 625 25 t 125 sb 25 t 625 When t 15 s sb 3125 Distance between the cars is Δ s sₐ sb 400 3125 875 m Ans Car A is ahead of car B 1254 A twostage rocket is fired vertically from rest at s 0 with an acceleration as shown After 30 s the first stage A burns out and the second stage B ignites Plot the vt and st graphs which describe the motion of the second stage for 0 t 60 s For 0 t 30 s 0t dv 0t 001 t² dt v 000333 t³ When t 30 s v 90 ms 0s ds 0s 000333 t³ dt s 0000833 t⁴ When t 30 s s 675 m For 30 s t 60 s 90t dv 30t 15 dt v 15 t 360 When t 60 s v 540 ms 675t ds 30t 15 t 360 dt s 75 t² 360 t 4725 When t 60 s s 10 125 m 1255 The as graph for a boat moving along a straight path is given If the boat starts at s 0 when v 0 determine its speed when it is at s 75 ft and 125 ft respectively Use Simpsons rule with n 100 to evaluate v at s 125 ft Velocity The velocity v in terms of s can be obtained by applying v dv a ds For the interval 0 ft s 100 ft v dv a ds 0v v dv 0s 5 ds v 10 s fts At s 75 ft v 1075 274 fts Ans At s 100 ft v 10100 3162 fts Ans For the interval 100 ft s 125 ft v dv a ds 3162 ftsv v dv 100 ft125 ft s 6s 10⁵³ ds a fts² a s 6s 10⁵³ Evaluating the integral on the right using Simpsons rule we have v²2 3162 fts v 288853 v 823 fts Ans 1256 The jet plane starts from rest at s 0 and is subjected to the acceleration shown Determine the speed of the plane when it has traveled 200 ft Also how much time is required for it to travel 200 ft a fts2 75 a 75 015s 500 sft a 75 015s v dv 75 015s ds v 150s 015s2 At s 200 ft v 150200 0152002 155 fts Ans v dsdt ds ds 150s 015s2 t 2582 sin103s 150150 0200 239 s Ans 1257 The vt graph of a car while traveling along a road is shown Draw the st and at graphs for the motion v ms 20 5 20 30 t s 0 t 5 a ΔvΔt 205 4 ms2 5 t 20 a ΔvΔt 20 2020 5 0 ms2 20 t 30 a ΔvΔt 0 2030 20 2 ms2 From the vt graph at t1 5s t2 20 s and t3 30 s s1 A1 12 520 50 m s2 A1 A2 50 2020 5 350 m s3 A1 A2 A3 350 12 30 2020 450 m The equations defining the portions of the st graph are 0 t 5 s v 4t ds v dt 0t ds 0t 4t dt s 2t2 5 t 20 s v 20 ds v dt 50s ds 520 20 dt s 20t 50 20 t 30 s v 230 t ds v dt 350s ds 20t 230 t dt s t2 60t 450 a ms2 4 2 0 2 5 20 30 t s s m 450 350 50 s t2 60t 450 s 2t2 1258 A motorcyclist at A is traveling at 60 fts when he wishes to pass the truck T which is traveling at a constant speed of 60 fts To do so the motorcyclist accelerates at 6 fts2 until reaching a maximum speed of 85 fts If he then maintains this speed determine the time needed for him to reach a point located 100 ft in front of the truck Draw the vt and st graphs for the motorcycle during this time vm1 60 fts vm2 85 fts 40 ft 55 ft 100 ft Motorcycle Time to reach 85 fts v v0 act 85 60 6t t 4167 s v2 v02 2ac s s0 Distance traveled 852 602 26sm 0 sm 30208 ft In t 4167 s truck travels s1 604167 250 ft Further distance for motorcycle to travel 405525010030208 14292 ft Motorcycle s s0 v0t s 14292 0 85t Truck s 0 60t Thus t 5717 s t 4167 5717 988 s Ans Total distance motorcycle travels sT 30208 855717 788 ft v fts 85 60 417 t s truck motorcycle s ft 30208 85 fts 60 fts 40 ft 250 ft 55 ft 345 ft 85 fts 60 fts 4292 ft 1259 The vs graph for a gocart traveling on a straight road is shown Determine the acceleration of the gocart at s 50 m and s 150 m Draw the as graph For 0 s 100 v 008 s dv 008 ds a ds 008 s008 ds a 64103 s At s 50 m a 032 ms2 Ans For 100 s 200 v 008 s 16 dv 008 ds a ds 008 s 16008 ds a 008008 s 16 At s 150 m a 032 ms2 Ans Also v dv a ds a vdvds At s 50 m a 48100 032 ms2 Ans At s 150 m a 48100 032 ms2 Ans 1260 The vs graph for the car is given for the first 500 ft of its motion Construct the as graph for 0 s 500 ft How long does it take to travel the 500ft distance The car starts at s 0 when t 0 as Graph The acceleration a in terms of s can be obtained by applying v du a ds a v dvds 01 s 1001 001s 1 fts2 At s 0 and s 500 ft a 0010 1 100 fts2 and a 001500 1 600 fts2 respectively Position The position s in terms of time t can be obtained by applying v dsdt dt dsv ₀t dt ₀s ds01 s 10 t 10 ln001 s 1 When s 500 ft t 10 ln001500 1 179 s Ans 1261 The as graph for a train traveling along a straight track is given for the first 400 m of its motion Plot the vs graph v 0 at s 0 0 s 200 a 1100 s a ds v dv ₀1100 s ds ₀v v dv 1200 s2 12 v2 v 01 s At s 200 v 20 ms 200 s 400 a 2 a ds v dv 2002 ds 20v v dv 2 s 200 12 v2 400 v2 4 s 400 At s 400 m v 4400 400 346 ms 1262 The vs graph for an airplane traveling on a straight runway is shown Determine the acceleration of the plane at s 100 m and s 150 m Draw the as graph For 0 s 100 m a ds v dv a ds 04 s 04 ds a 016 s At s 100 m a 16 ms2 Ans For 100 m s 200 m a ds v dv a ds 30 01 s01 ds a 3 001 s At s 150 m a 45 ms2 Ans At s 100 m a 4 ms2 At s 200 m a 5 ms2 1263 Starting from rest at s 0 a boat travels in a straight line with an acceleration as shown by the as graph Determine the boats speed when s 40 90 and 200 ft Since a ds v dv a ds 12 v2 v 2 a ds a ds area under as graph For s 40 ft v 240 127 fts Ans For s 90 ft v 250 440 228 fts Ans For s 200 ft v 250 4100 12504 2 v 361 fts Ans Also For 0 s 50 ft a 2 v dv a ds ₀2 ds ₀v v dv v 4 s When s 40 ft v 440 127 fts Ans When s 50 ft v 450 1414 fts For 50 ft s 150 ft a 4 504 ds 1414v v dv v 8 s 200 When s 90 ft v 890 200 228 fts Ans When s 150 ft v 890 200 3162 fts For 150 ft s 250 ft a 4100 s 10 150125 s 10 ds 3162v v dv v 125 s2 20 s 110012 When s 200 ft v 1252002 20200 110012 v 361 fts Ans 1264 The test car starts from rest and is subjected to a constant acceleration of ac 15 fts2 for 0 t 10 s The brakes are then applied which causes a deceleration at the rate shown until the car stops Determine the cars maximum speed and the time t when it stops vmax A1 1510 150 fts Ans From the graph for t 10 s a 12 t 10 dv a dt 150 dv 10 12 t 10 dt v 150 1212 t210t10150 v 150 14 t2 5t 14 102 12 102 14 t2 5t 125 When the car stops v 0 14 t2 5t 125 1 Solving for the positive root t 345 s Ans 1265 The as graph for a race car moving along a straight track has been experimentally determined If the car starts from rest at s 0 determine its speed when s 50 ft 150 ft and 200 ft respectively Velocity The velocity v in terms of s can be obtained by applying v dv a ds For the interval 0 ft s 150 ft v dv a ds 0v v dv 0s 5 ds v 10s ms At s 50 ft v 1050 224 fts Ans At s 150 ft v 10150 387 fts Ans For the interval 150 ft s 200 ft a 5s 150 10 5200 150 a 110 s 10 fts2 v dv a ds 387 ftsv v dv 150fts 110 s 10 ds v 110 s2 20s 2250 fts At s 200 ft v 110 2002 20200 2250 474 fts Ans a fts2 15 10 1 2 0 10 t s Using the a t graph we can obtain the same result by requiring A1 A2 1510 12 at 10 0 150 1212 t 10 t 10 0 14 t2 5t 125 0 Which is the same as Eq 1 a fts2 10 5 150 200 s ft 1266 A particle originally at rest and located at point 3 ft 2 ft 5 ft is subjected to an acceleration of a 6ti 12 t2 k fts2 Determine the particles position x y z at t 1 s Velocity The velocity express in Cartesian vector form can be obtained by applying Eq 12 9 dv a dt 0t dv 0t 6ti 12 t2 k dt v 3 t2 i 4 t3 k fts Position The position express in Cartesian vector form can be obtained by applying Eq 12 7 dr v dt rtr dr 0t 3 t2 i 4 t3 k dt r 3t 2j 5k t3 i t4 k r t3 3 i 2j t4 5 k ft When t 1 s r 13 3 i 2j 14 5 k 4i 2j 6k ft The coordinates of the particle are 4 2 6 ft Ans 1267 The velocity of a particle is given by v 16 t2 i 4 t3 j 5t 2k ms where t is in seconds If the particle is at the origin when t 0 determine the magnitude of the particles acceleration when t 2 s Also what is the x y z coordinate position of the particle at this instant Velocity The velocity express in Cartesian vector form can be obtained by applying Eq 12 9 Acceleration The acceleration express in Cartesian vector form can be obtained by applying Eq 12 9 dvdt 32t 12 t2 j 5k ms2 When t 2 s a 322 i 12 22 j 5k 64i 48 j 5k ms2 The magnitude of the acceleration is a ax2 ay2 az2 642 482 52 802 ms2 Ans Position The position express in Cartesian vector form can be obtained by applying Eq 12 7 dr v dt 0t dr 0t 16 t2 i 4 t3 j 5 t 2 k dt r 163 t3 i t4 j 52 t2 2 k m When t 2 s r 163 23 i 24 j 52 22 22 k 427 i 160 j 140 k m Thus the coordinate of the particle is 427 160 140 ft Ans 1268 A particle is traveling with a velocity of v 3 t e02 t i 4 e08 t3 j ms where t is in seconds Determine the magnitude of the particles displacement from t 0 to t 3 s Use Simpsons rule with n 100 to evaluate the integrals What is the magnitude of the particles acceleration when t 2 s ds v dt Δsx 03 3 t e02 t dt 7341 Δsy 04 4 e08 t3 dt 3963 Thus Δs 73412 39632 834 m Ans ax vx 3 12 t1 e02 t 3 t e02 t 02t2 01422 ay vy 4 e08 t3 08 2t2 05218 a 014222 052182 0541 ms2 Ans 1269 The position of a particle is defined by r 5 cos 2t i 4 sin 2t j m where t is in seconds and the arguments for the sine and cosine are given in radians Determine the magnitudes of the velocity and acceleration of the particle when t 1 s Also prove that the path of the particle is elliptical Velocity The velocity expressed in Cartesian vector form can be obtained by applying Eq 12 7 v drdt 10 sin 2t i 8 cos 2t j ms When t 1 s v 10 sin 21 i 8 cos 21 j 9093 i 3329 j ms Thus the magnitude of the velocity is v vx2 vy2 90932 33292 968 ms Ans Acceleration The acceleration express in Cartesian vector form can be obtained by applying Eq 12 9 a dvdt 20 cos 2i 16 sin 2t j ms2 When t 1 s a 20 cos 21 i 16 sin 21 j 8323 i 14549 j ms2 Thus the magnitude of the acceleration is a ax2 ay2 83232 145492 168 ms2 Ans Travelling Path Here x 5 cos 2t and y 4 sin 2t Then x225 cos2 2t 1 y216 sin2 2t 2 Adding Eqs 1 and 2 yields x225 y216 cos2 2t sin2 2t However cos2 2t sin2 2t 1 Thus x225 y216 1 Equation of an Ellipse QED 1270 The car travels from A to B and then from B to C as shown in the figure Determine the magnitude of the displacement of the car and the distance traveled Displacement Δr 2i 3j km Δr 2² 3² 361 km Ans Distance traveled d 2 3 5 km Ans 1271 A particle travels along the curve from A to B in 2 s It takes 4 s for it to go from B to C and then 3 s to go from C to D Determine its average speed when it goes from A to D sT 14 2π10 15 142π5 3856 vsp sTt 3856243 428 ms Ans 1272 A car travels east 2 km for 5 minutes then north 3 km for 8 minutes and then west 4 km for 10 minutes Determine the total distance traveled and the magnitude of displacement of the car Also what is the magnitude of the average velocity and the average speed Total Distance Traveled and Displacement The total distance traveled is s 2 3 4 9 km Ans and the magnitude of the displacement is Δr 2² 3² 3606 km 361 km Ans Average Velocity and Speed The total time is Δt 5 8 10 23 min 1380 s The magnitude of average velocity is vavg ΔrΔt 360610³1380 261 ms Ans and the average speed is vsp avg sΔt 910³1380 652 ms Ans 1273 A car traveling along the straight portions of the road has the velocities indicated in the figure when it arrives at points A B and C If it takes 3 s to go from A to B and then 5 s to go from B to C determine the average acceleration between points A and B and between points A and C vA 20 i vB 2121 i 2121 j vC 40 i aAB ΔvΔt 21211 2121 j 20 i 3 aAB 0404 i 707 j ms² Ans aAC ΔvΔt 40 i 20 i 8 aAC 2501 ms² Ans 1274 A particle moves along the curve y e²ˣ such that its velocity has a constant magnitude of v 4 fts Determine the x and y components of velocity when the particle is at y 5 ft Velocity Taking the first derivative of the path y e²ˣ we have ẏ 2e²ˣ x 1 However x vx and ẏ vy Thus Eq 1 becomes vy 2e²ˣ vx 2 Here v 4 fts Then v² vₓ² vy² vₓ² vy² 16 3 Solving Eqs 2 and 3 yields vx 4114e⁴ˣ and vy 8e⁴ˣ14e⁴ˣ At y 5 ft 5 e²ˣ x 08047 ft Thus vx 4114e⁴08047 0398 fts Ans vy 8e⁴0804714e⁴08047 398 fts Ans 1275 The path of a particle is defined by y² 4kx and the component of velocity along the y axis is vy ct where both k and c are constants Determine the x and y components of acceleration y² 4kx 2y vy 4kx 2vy² 2y ay 4kx vy ct ay c Ans 2ct² 2yc 4kx ax c2k y ct² Ans 1276 A particle is moving along the curve y x x²400 where x and y are in ft If the velocity component in the x direction is vx 2 fts and remains constant determine the magnitudes of the velocity and acceleration when x 20 ft Velocity Taking the first derivative of the path y x x²400 we have ẏ x 1400 2xx x x200 x 1 However x vx and ẏ vy Thus Eq 1 becomes vy vx x200 vx 2 Here vx 2 fts at x 20 ft Then From Eq 2 vy 2 20200 2 180 fts Also v vₓ² vy² 2² 180² 269 fts Ans Acceleration Taking the second derivative of the path y x x²400 we have ÿ x 1200 x² xx 3 However x ax and ÿ ay Thus Eq 3 becomes ay ax 1200 vₓ² xax 4 Since vₓ 2 fts is constant hence ax 0 at x 20 ft Then From Eq 4 ay 0 1200 2² 200 0020 fts Also a aₓ² ay² 0² 0020² 00200 fts² Ans 1277 The motorcycle travels with constant speed v0 along the path that for a short distance takes the form of a sine curve Determine the x and y components of its velocity at any instant on the curve y c sinπL x y πL c cos πL x x vy πL c vx cos πL x v0² vy² vx² v0² v0²1 πL c² cos²πL x vx v0 1 πL c² cos²πL x¹² Ans vy v0 π c L cos πL x 1 πL c² cos²πL x¹² Ans 1278 The particle travels along the path defined by the parabola y 05x² If the component of velocity along the x axis is vx 5t fts where t is in seconds determine the particles distance from the origin O and the magnitude of its acceleration when t 1 s When t 0 x 0 y 0 Position The x position of the particle can be obtained by applying the vx dxdt dx vx dt ₀ˣ dx ₀ᵗ 5t dt x 250t² ft Thus y 05 250 t²² 3125 t⁴ ft At t1 s x 25 1² 250 ft and y 3125 1⁴ 3125 ft The particles distance from the origin at this moment is d 2500² 31250² 400 ft Ans Acceleration Taking the first derivative of the path y 05x² we have ẏ xẋ The second derivative of the path gives ÿ ẋ² xẍ 1 However ẋ vx ẍ ax and ÿ ay Thus Eq1 becomes ay vx² x ax 2 When t 1 s vx 51 5 fts ax dvxdt 5 fts and x 250 ft Then from Eq2 ay 5² 250 5 375 fts² Also a ax² ay² 5² 375² 378 fts² Ans 1279 When a rocket reaches an altitude of 40 m it begins to travel along the parabolic path y 40² 160 x where the coordinates are measured in meters If the component of velocity in the vertical direction is constant at vy 180 ms determine the magnitudes of the rockets velocity and acceleration when it reaches an altitude of 80 m vy 180 ms y 40² 160 x 2y 40vy 160 vx 1 280 40180 160 vx vx 90 ms v 90² 180² 201 ms Ans ay dvydt 0 From Eq 1 2 vx² 2y 40ay 160 ax 2180² 0 160 ax ax 405 ms² a 405 ms² Ans 1280 The girl always throws the toys at an angle of 30 from point A as shown Determine the time between throws so that both toys strike the edges of the pool B and C at the same instant With what speed must she throw each toy To strike B s s₀ v₀ t 25 0 vx cos 30 t s s₀ v₀ t 12 ax t² 025 1 vA sin 30 t 12 981 t² Solving t 06687 s vA B 432 ms Ans To strike C s s₀ v₀ t 4 0 vx cos 30 t s s₀ v₀ t 12 ax t² 025 1 vA sin 30 t 12 981 t² Solving t 0790 s vAC 585 ms Ans Time between throws Δt 0790 s 06687 s 0121 s Ans 1283 Determine the maximum height on the wall to which the firefighter can project water from the hose if the speed of the water at the nozzle is vC 48 fts v vo at 0 48 sinθ 322t s so vo t 30 0 48 cosθt 48 sinθ 322 30 48 cosθ sinθcosθ 041927 sin2θ 083854 θ 285 t 07111 s s so vo t 12 at² h 3 0 48 sin285 07111 12 32207111² h 111 ft Ans 1284 Determine the smallest angle θ measured above the horizontal that the hose should be directed so that the water stream strikes the bottom of the wall at B The speed of the water at the nozzle is vC 48 fts s so vot 30 0 48 cosθ t t 30 48 cosθ s so vot 12 at² 0 3 48sinθ t 12 322 t² 0 3 48 sinθ 30 48 cosθ 161 30 48 cosθ² 0 3cos² θ 30sinθcosθ 62891 3 cos² θ 15 sin2θ 62891 Solving θ 641 Ans 1281 The nozzle of a garden hose discharges water at the rate of 15 ms If the nozzle is held at ground level and directed θ 30 from the ground determine the maximum height reached by the water and the horizontal distance from the nozzle to where the water strikes the ground v0x 15 cos30 1299 ms v0y 15 sin30 75 ms Maximum height v² v0² 2ac s s0 0 75² 2981h 0 h 287 m Ans Time of travel to top of path v v0 act 0 75 981f t 07645 s Total time along path t 207645 1529 s Range sx vxt 12991529 199 m Ans 1282 The balloon A is ascending at the rate vA 12 kmh and is being carried horizontally by the wind at vw 20 kmh If a ballast bag is dropped from the balloon at the instant h 50 m determine the time needed for it to strike the ground Assume that the bag was released from the balloon with the same velocity as the balloon Also with what speed does the bag strike the ground vy² v0² 2 acs s0 vy² 333² 298150 0 vy 3150 ms v vo at 3150 333 981 t t 355 s Ans vx 20 kmh 5556 ms v 3150² 5556² 320 ms Ans 1285 From a videotape it was observed that a pro football player kicked a football 126 ft during a measured time of 36 seconds Determine the initial speed of the ball and the angle θ at which it was kicked s s0 vo t 126 0 vox 36 vox 35 fts s s0 vot 12 act² 0 0 voy 36 12 32236² voy 5796 fts vo 35² 5796² 677 fts Ans θ tan¹579635 589 Ans 1286 During a race the dirt bike was observed to leap up off the small hill at A at an angle of 60 with the horizontal If the point of landing is 20 ft away determine the approximate speed at which the bike was traveling just before it left the ground Neglect the size of the bike for the calculation s s0 vt 20 0 vA cos60 t s s0 vt 12 at² 0 0 vA sin60 t 12 322 t² Solving t 14668 s vA 273 fts Ans 1287 Measurements of a shot recorded on a videotape during a basketball game are shown The ball passed through the hoop even though it barely cleared the hands of the player B who attempted to block it Neglecting the size of the ball determine the magnitude vA of its initial velocity and the height h of the ball when it passes over player B 1288 The snowmobile is traveling at 10 ms when it leaves the embankment at A Determine the time of flight from A to B and the range R of the trajectory 1289 The snowmobile is traveling at 10 ms when it leaves the embankment at A Determine the speed at which it strikes the ground at B and its maximum acceleration along the trajectory AB 1290 A golf ball is struck with a velocity of 80 fts as shown Determine the distance d to where it will land 1291 It is observed that the skier leaves the ramp A at an angle θA 25 with the horizontal If he strikes the ground at B determine his initial speed vA and the time of flight tAB 1292 The man stands 60 ft from the wall and throws a ball at it with a speed v₀ 50 fts Determine the angle θ at which he should release the ball so that it strikes the wall at the highest point possible What is this height The room has a ceiling height of 20 ft v₀ 50 fts 20 ft 5 ft 60 ft vx 50 cosθ s s₀ v₀t x 0 50cosθ t 1 v v₀ at vy 50 sinθ 322 t 2 s s₀ v₀ t 12 a t² y 0 50 sinθ t 161 t² 3 v² v₀² 2as s₀ vy² 50 sinθ² 2322s 0 vy² 2500 sin² θ 644 s 4 Require vy 0 at s 20 5 15 ft 0 2500 sin² θ 644 15 θ 38433 384 Ans From Eq 2 0 50 sin 38433 322 t t 09652 s From Eq 1 x 50 cos 38433 09652 378 ft Time for ball to hit wall From Eq 1 60 50 cos 38433 t t 153193 s From Eq 3 y 50 sin 38433 153193 161 153193² y 9830 ft h 9830 5 148 ft Ans 1293 The ball at A is kicked with a speed vA 80 fts and at an angle θA 30 Determine the point x y where it strikes the ground Assume the ground has the shape of a parabola as shown vAx 80 cos 30 6928 fts vAy 80 sin 30 40 fts s s₀ v₀ t x 0 6928 t 1 s s₀ v₀ t 12 a t² y 0 40 t 12 322 t² 2 y 004 x² From Eqs 1 and 2 y 05774 x 0003354 x² 004 x² 05774 x 0003354 x² 004335 x² 05774 x x 133 ft Ans Thus y 004 133² 709 ft Ans 1294 The ball at A is kicked such that θA 30 If it strikes the ground at B having coordinates x 15 ft y 9 ft determine the speed at which it is kicked and the speed at which it strikes the ground s s₀ v₀ t 15 0 vA cos 30 t s s₀ v₀ t 12 a t² 9 0 vA sin 30 t 12 322 t² vA 165 fts Ans t 1047 s vBx 1654 cos 30 1432 fts v v₀ at vBy 1654 sin 30 3221047 2545 fts vB sqrt1432² 2545² 292 fts Ans 1295 Determine the horizontal velocity vA of a tennis ball at A so that it just clears the net at B Also find the distance s where the ball strikes the ground Vertical Motion The vertical component of initial velocity is v₀y 0 For the ball to travel from A to B the initial and final vertical positions are s₀y 75 ft and sy 3 ft respectively sy s₀y v₀y t 12 ay t² 3 75 0 12 322 t₁² t₁ 05287 s For the ball to travel from A to C the initial and final vertical positions are s₀y 75 ft and sy 0 respectively sy s₀y v₀y t 12 ay t² 0 75 0 12 322 t₂² t₂ 06825 s Horizontal Motion The horizontal component of velocity is v₀x vA For the ball travel from A to B the initial and final horizontal positions are s₀x 0 and sx 21 ft respectively The time is t t₁ 05287 s sx s₀x v₀x t 21 0 vA 05287 vA 3972 fts 397 fts Ans For the ball travel from A to C the initial and final horizontal positions are s₀x 0 and sx 21 s ft respectively The time is t t₂ 06825 s sx s₀x v₀x t 21 s 0 3972 06825 s 611 ft Ans 1296 A boy at O throws a ball in the air with a speed v0 at an angle θ1 If he then throws another ball at the same speed v0 at an angle θ2 θ1 determine the time between the throws so the balls collide in mid air at B 1297 The man at A wishes to throw two darts at the target at B so that they arrive at the same time If each dart is thrown with a speed of 10 ms determine the angles θC and θD at which they should be thrown and the time between each throw Note that the first dart must be thrown at θCθD then the second dart is thrown at θD Time of flight s s0 v0 t x1 0 v0 cosθ1 t1 1 x2 0 v0 cosθ2 t2 2 Thus x1 x2 Δt t1 t2 x1 v0 cos θ2 cos θ1 cos θ1 cos θ2 3 s s0 v0 t 12 a t2 y 0 v0 sinθ1t1 12 g t2 4 Use Eq 1 y x1 tanθ1 12 g x12 v02 cos2 θ1 In the same way y x2 tanθ2 12 g x22 v02 cos2 θ2 Equating and solving for x1 x2 x x 2v2g cos2 θ1 cos2 θ2tan θ1 tan θ2 cos2 θ2 cos2 θ1 Substituting into Eq 3 yields Δt 2vo g cos θ1 cos θ2tan θ1 tan θ2 cos θ2 cos θ1 Δt 2vo g sinθ1 θ2 cos θ2 cos θ1 Ans s s0 v0 t 5 0 10 cos θ t 1 v v0 at 10 sin θ 10 sin θ 981 t t 210 sin θ 981 2039 sin θ From Eq 1 5 2039 sin θ cos θ Since sin 2 θ 2 sinθ cos θ sin 2θ 04905 The two roots are θD 147 Ans θC 753 Ans From Eq1 tD 0517 s tC 197 s So that Δt tC tD 145 s Ans 1298 The ball is thrown from the tower with a velocity of 20 fts as shown Determine the x and y coordinates to where the ball strikes the slope Also determine the speed at which the ball hits the ground Assume ball hits slope s s0 v0 t x 0 35 20t 12t s s0 v0 t 12 a t2 y 80 45 20t 12 322t2 80 16t 161 t2 Equation of slope y y1 mx x1 y 0 12 x 20 y 05x 10 Thus 80 16t 161 t2 0512t 10 161 t2 10t 90 0 Choosing the positive root t 26952 s x 1226952 323 ft Ans Since 323 ft 20 ft assumption is valid y 80 1626952 161269522 617 ft Ans vx v0x 35 20 12 fts vy v0y ac t 45 20 32226952 70785 fts v 122 707852 718 fts Ans 1299 The baseball player A hits the baseball at vA 40 fts and θA 60 from the horizontal When the ball is directly overhead of player B he begins to run under it Determine the constant speed at which B must run and the distance d in order to make the catch at the same elevation at which the ball was hit vA 40 fts Vertical Motion The vertical component of initial velocity for the football is v0y 40 sin 60 3464 fts The initial and final vertical positions are s0y 0 and sy 0 respectively sy s0y v0y t 12 acy t2 0 0 3464t 12 322 t2 t 2152 s Horizontal Motion The horizontal component of velocity for the baseball is v0x 40 cos 60 200 fts The initial and final horizontal positions are s0x 0 and sx R respectively sx s0x v0x t R 0 2002152 4303 ft The distance for which player B must travel in order to catch the baseball is d R 15 4303 15 280 ft Ans Player B is required to run at a same speed as the horizontal component of velocity of the baseball in order to catch it vB 40 cos 60 200 fts Ans 12100 A car is traveling along a circular curve that has a radius of 50 m If its speed is 16 ms and is increasing uniformly at 8 ms² determine the magnitude of its acceleration at this instant v 16 ms aₜ 8 ms² ρ 50 m aₙ v² ρ 16² 50 512 ms² a 8² 512² 950 ms² Ans 12101 A car moves along a circular track of radius 250 ft such that its speed for a short period of time 0 t 4 s is v 3t t² fts where t is in seconds Determine the magnitude of its acceleration when t 3 s How far has it traveled in t 3 s v 3t t² aₜ dvdt 3 6t When t 3 s aₜ 3 63 21 fts² aₙ 333²² 250 518 fts² a 21² 518² 216 fts² Ans ds ₀³ 3t t² dt Δs 32 t² t³ ₀³ Δs 405 ft Ans 12102 At a given instant the jet plane has a speed of 400 fts and an acceleration of 70 fts² acting in the direction shown Determine the rate of increase in the planes speed and the radius of curvature ρ of the path aₜ 70 cos 60 350 fts² Ans aₙ 400² ρ 70 sin 60 ρ 26410³ ft Ans 12103 A boat is traveling along a circular curve having a radius of 100 ft If its speed at t 0 is 15 fts and is increasing at v 08t fts² determine the magnitude of its acceleration at the instant t 5 s ₀⁵ dv ₀⁵ 08 t dt v 25 fts aₙ v² ρ 25² 100 625 fts² At t 5 s aₜ v 085 4 fts² a aₜ² aₙ² 4² 625² 742 fts² Ans 12104 A boat is traveling along a circular path having a radius of 20 m Determine the magnitude of the boats acceleration when the speed is v 5 ms and the rate of increase in the speed is v 2 ms² aₜ 2 ms² aₙ v² ρ 5² 20 125 ms² a aₜ² aₙ² 2² 125² 236 ms² Ans 12105 Starting from rest a bicyclist travels around a horizontal circular path ρ 10 m at a speed of v 009t² 01t ms where t is in seconds Determine the magnitudes of his velocity and acceleration when he has traveled s 3 m ds 009t² 01t dt s 003t³ 005t² When s 3 m 3 003t³ 005t² Solving t 4147 s v dsdt 009t² 01t v 0094147² 014147 196 ms Ans aₜ dvdt 018t 01 t4147 08465 ms² aₙ v² ρ 196² 10 03852 ms² a aₜ² aₙ² 08465² 03852² 0930 ms² Ans 12106 The jet plane travels along the vertical parabolic path When it is at point A it has a speed of 200 ms which is increasing at the rate of 08 ms² Determine the magnitude of acceleration of the plane when it is at point A y 04 x² dydx x5 km 4 d²ydx² 08 ρ 1 4²32 08 8762 km aₜ 08 ms² aₙ 0200² 8762 045710³ kms² aₙ 0457 ms² a 08² 0457² 0921 ms² Ans 12107 Starting from rest motorboat travels around the circular path ρ 50 m at a speed v 08t ms where t is in seconds Determine the magnitudes of the boats velocity and acceleration when it has traveled 20 m Velocity The time for which the boat to travel 20 m must be determined first ds v dt ₀²⁰m ds ₀ t 08 t dt t 7071 s The magnitude of the boats velocity is v 087071 5657 ms 566 ms Ans Acceleration The tangential acceleration is aₜ v 08 ms² To determine the normal acceleration apply Eq 1220 aₙ v² ρ 5657² 50 0640 ms² Thus the magnitude of acceleration is a aₜ² aₙ² 08² 0640² 102 ms² Ans