PROBLEM SOLUTIONS 3 1 A Mosinee School District

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Chapter 3 PROBLEM SOLUTIONS 3 1 We are given that When two vectors are added graphically the second vector is positioned with its tail at the tip of the first

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A as shown in the diagram You should find that,A B 5 0 units at 53. 3 4 Sketches of the scale drawings needed for parts a through d are given below Following the sketches is a brief. comment on each part with its answer, a Drawing the vectors to scale and maintaining their respective directions yields a resultant of. 5 2 m at 60, b Maintain the direction of A but reverse the direction of B to produce B The resultant is. 3 0 m at 30, c Maintain the direction of B but reverse the direction of A to produce A The resultant is. 3 0 m at 150, d Maintain the direction of A reverse the direction of B and double its magnitude to produce 2 B The re.
sultant is 5 2 m at 60, 3 5 Your sketch should be drawn to scale similar to that pictured below The length of R and the angle can be. measured to find with use of your scale factor the magnitude and direction of the resultant displacement The result. 3 7 Using a vector diagram drawn to scale like that shown. at the right the final displacement of the plane can be. found to be R plane 310 km at 57 N of E The requested dis. placement of the base from point B is plane,R plane which has the same magnitude but the. opposite direction Thus the answer is,R plane 310 km at 57 S of W. 3 8 Your vector diagram should look like the,one shown at the right The initial. displacement A 100 m due west and the re,sultant R 175 m at 15 0 N of W are.
both known In order to reach the end point,of the run following the initial displacement the. jogger must follow the path shown as B,The length of B and the angle can be. measured The results should be 83 m at 33 N of W, 3 9 The displacement vectors A 8 00 m westward westward and. B 13 0 m north north can be drawn to scale as at the right. The vector C represents the displacement that the man. in the maze must undergo to return to his starting point. The scale used to draw the sketch can be used to,find C to be 15 m at 58 S of E. 3 10 The x and y components of vector A are its projections. on lines parallel to the x and y axis respectively as shown. in the sketch The magnitude of these components can be. computed using the sine and cosine functions as shown below. Ax A cos 325 A cos 35 35 0 cos 35 28 7 units,Ay A sin 325 A sin 35 35 0 sin 35 20 1 units.
3 11 Using the vector diagram given at the right we find. R 6 00 m 2 5 40 m 2 8 07 m,tan 1 tan 1 0 900 42 0, Thus the required displacement is 8 07 m at 42 0 S of E. 3 12 a The skater s displacement vector d extends in a straight. line from her starting point A to the end point B When. she has coasted half way around a circular path as shown. in the sketch at the right the displacement vector d. coincides with the diameter of the circle and has,d 2r 2 5 00 m 10 0 m. b The actual distance skated s is one half the circumference of the circular path of radius r Thus. 2 r 5 00 m 15 7 m, c When the skater skates all the way around the circular path her end point B coincides with the start point A. Thus the displacement vector has zero length or d 0. 3 13 a Her net x east west displacement is 3 00 0 6 00 3 00 blocks while her net y north south. displacement is 0 4 00 0 4 00 bloc ks The magnitude of the resultant displacement is. R x 2 y 2 3 00 2 4 00 2 5 00 blocks, and the angle the resultant makes with the x axis eastward direction is. tan 1 tan 1 tan 1 1 33 53 1, The resultant displacement is then 5 00 blocks at 53 1 N of E.
b The total distance traveled is 3 00 4 00 6 00 13 0 blocks. ur ur ur ur ur ur ur, 3 14 a The resultant displacement is R A B C D where A 75 0 m due north B 250 m due east. C 125 m at 30 0 north of east and D 150 m due south Choosing east as the positive x direction. and north as the positive y direction we find the components of the resultant to be. Rx Ax Bx C x Dx 0 250 m 125 m cos 30 0 0 358 m, Ry Ay By C y Dy 75 0 m 0 125 m sin 30 0 150 m 12 5 m. The magnitude and direction of the resultant are then. R Rx2 Ry2 358 m 2 12 5 m 2 358 m,tan 1 tan 1 2 00,Thus R 358 m at 2 00 south of east. b Because of the commutative property of vector addition the net displacement is the same regardless of the. order in which the individual displacements are executed. 3 15 Ax 25 0 Ay 40 0,A Ax2 Ay2 25 0 2 40 0 2 47 2 units. From the triangle we find that,1 40 0 58 0,tan 1 tan so 180 122.
Thus A 47 2 units at 122 counterclockwise from x axis. 3 16 Let A be the vector corresponding to the 10 0 yd run B to the 15 0 yd run and C to the 50 0 yd pass Also we. choose a coordinate system with the y direction downfield and the x direction toward the sideline to which the. player runs,The components of the vectors are then. Ax 0 Ay 10 0 yds,Bx 15 0 yds By 0,Cx 0 C y 50 0 yds. From these Rx x 15 0 yds and Ry y 40 0 yds and,R Rx2 Ry2 15 0 yds 2 40 0 yds 2 42 7 yards. 3 17 After 3 00 h moving at 41 0 km h the hurricane is 123 km at 60 0 N of W from the island In the next 1 50 h it. travels 37 5 km due north The components of these two displacements are as follows. Displacement x component eastward y component northward. 123 km 61 5 km 107 km,37 5 km 0 37 5 km,Resultant 61 5 km 144 km. Therefore the eye of the hurricane is now,61 5 km 2 144 km.
R 157 km from the island, 3 19 The components of the displacements a b and c are. a x a cos 30 0 152 km,bx b cos 110 51 3 km,c x c cos 180 190 km. a y a sin 30 0 87 5 km,by b sin 110 141 km,c y c sin 180 0. Rx a x bx c x 89 7 km and Ry a y by c y 228 km,R Rx2 Ry2 245 km and tan 1 x tan 1 1 11 21 4. City C is 245 km at 21 4 W of N from the starting point. 3 20 a F1 120 N F1x 120 N cos 60 0 60 0 N F1y 120 N sin 60 0 104 N. F2 80 0 N F2 x 80 0 N cos 75 0 20 7 N F2 y 80 0 N sin 75 0 77 3 N. Fx 2 39 3 N 2 181 N 2,FR Fy 185 N,and tan 1 tan 1 4 61 77 8.
The resultant force is FR 185 N at 77 8 from the x axis. b To have zero net force on the mule the resultant above must be cancelled by a force equal in magnitude and. oppositely directed Thus the required force is,185 N at 258 from the x axis. 3 21 The single displacement required to sink the putt in one stroke is equal to the resultant of the three actual putts used. by the novice Taking east as the positive x direction and north as the positive y direction the components of the. three individual putts and their resultant are,Ax 0 Ay 4 00 m. Bx 2 00 m cos 45 0 1 41 m By 2 00 m sin 45 0 1 41 m. C x 1 00 m sin 30 0 0 500 m C y 1 00 m cos 30 0 0 866 m. Rx Ax Bx C x 0 914 m Ry Ay By C y 4 55 m, The magnitude and direction of the desired resultant is then. R Rx2 Ry2 4 64 m and tan 1 78 6,Thus R 4 64 m at 78 6 north of east. 3 23 a With the origin chosen at point O as shown in Figure P3 23 the coordinates of the original position of the. stone are x0 0 and y0 50 0 m, b The components of the initial velocity of the stone are v 0 x 18 0 m s and v 0 y 0.
c The components of the stone s velocity during its flight are given as functions of time by. v x v 0 x a x t 18 0 m s 0 t or v x 18 0 m s,v y v 0 y ay t 0 g t or. v x 9 80 m s2 t, d The coordinates of the stone during its flight are. a x t 2 0 18 0 m s t 0 t 2 x 18 0 m s t,x x0 v 0 x t or. y y0 v 0 y t,a t 2 50 0 m 0 t g t 2,y 50 0 m 4 90 m s2 t 2. e We find the time of fall from y v 0 y t a t 2 with v 0 y 0. 2 y 2 50 0 m,a 9 80 m s2, f At impact v x v 0 x 18 0 m s and the vertical component is.
v y v 0 y a y t 0 9 80 m s2 3 19 s 31 3 ms,v v 2x v 2y 18 0 m s 31 3 m s. tan 1 tan 18 0 60 1,or v 36 1 m s at 60 1 below the horizontal. 3 24 The constant horizontal speed of the falcon is. mi 0 447 m s,v x 200 89 4 m s,The time required to travel 100 m horizontally is. vx 89 4 m s,The vertical displacement during this time is. 9 80 m s2 1 12 s 2 6 13 m,or the falcon has a vertical fall of 6 13 m.
3 25 At the maximum height v y 0 and the time to reach this height is found from. v y v0 y 0 v0 y v0 y,v y v 0 y a y t as t, The vertical displacement that has occurred during this time is. v y v0 y 0 v0 y v0 y,y max v y t t 2 g,Thus if y max 12 ft 1 m 3 281 ft 3 7 m then. v0 y 2 g y max,2 9 80 m s2 3 7 m 8 5 m s, and if the angle of projection is 45 the launch speed is. v0 y 8 5 m s,sin sin 45, 3 26 a When a projectile is launched with speed v 0 at angle 0 above the horizontal the initial velocity components. are v 0 x v 0 cos 0 and v 0 y v 0 sin 0 Neglecting air resistance the vertical velocity when the projectile. returns to the level from which it was launched in this case the ground will be v y v 0 y From this in. formation the total time of flight is found from v y v 0 y a y t to be. v yf v 0 y v 0 y v 0 y 2v 0 y 2v 0 sin 0,t total or ttotal.
Since the horizontal velocity of a projectile with no air resistance is constant the horizontal distance it will. travel in this time i e its range is given by,2v s in 0 v 02 v 02 sin 2 0. R v 0 x t total v 0 cos 0 0 2 sin cos, Thus if the projectile is to have a range of R 81 1 when launched at an angle of 0 45 0 the required. initial speed is,Rg 81 1 m 9 80 m s2,v0 28 2 m s,sin 2 0 sin 90 0. b With v 0 28 2 m s and 0 45 0 the total time of flight as found above will be. 2v 0 sin 0 2 28 2 m s sin 45 0,t total 4 07 s,g 9 80 m s2. c Note that at 0 45 0 sin 2 0 1 and that sin 2 0 will decrease as 0 is increased above this opti. mum launch angle Thus if the range is to be kept constant while the launch angle is increased above 45 0. we see from v 0 Rg sin 2 0 that,the required initial velocity will increase.
Observe that for 0 90 the function sin 0 increases as 0 is increased Thus increasing the launch an. gle above 45 0 while keeping the range constant means that both v 0 and sin 0 will increase Considering. the expression for ttotal given above we see that the total time of flight will increase. 3 27 When y y max v y 0,Thus v y v 0 y a y t yields 0 v 0 sin 3 00 gt or. v 0 sin 3 00, The vertical displacement is y v 0 y t a t 2 At the maximum height this becomes. v sin 3 00 1 v sin 3 00 v 02 sin 2 3 00,y max v 0 sin 3 00 0 g 0. If y max 0 330 m the initial speed is,2 9 80 m s2 0 330 m. v0 48 6 m s,sin 3 00 sin 3 00, Note that it was unnecessary to use the horizontal distance of 12 6 m in this solution.
3 28 a With the origin at ground level directly below the window the original coordinates of the. ball are x y 0 y0,b v 0 x v 0 cos 0 8 00 m s cos 20 0 7 52 m s. v 0 y v 0 sin 0 8 00 m s sin 20 0 2 74 m s,a x t 2 0 7 52 m s t 0 t 2 x 7 52 m s t. c x x0 v 0 x t or,y y0 v 0 y t,a y t 2 y0 2 74 m s t. 9 80 m s2 t 2,y y0 2 74 m s t 4 90 m s2 t 2, d Since the ball hits the ground at t 3 00 s the x coordinate at the landing site is. xlanding x t 3 00 s 7 52 m s 3 00 s 22 6 m, e Since y 0 when the ball reaches the ground at t 3 00 s the result of c above gives.
3 00 s 4 90 s2 3 00 s,y0 y 2 74 t 4 90 2 t 2 0 2 74. s s t 3 00 s s,Or y0 52 3 m, f When the ball has a vertical displacement of y 10 0 it will be moving downward with a velocity given. by v 2y v 02 y 2 a y y as,v y v 02 y 2 a y y 2 74,m s 2 9 80 m s2. 10 0 m 14 3 m s,The elapsed time at this point is then. v y v0 y 14 3 m s 2 74 m s,ay 9 80 m s2, 3 29 We choose our origin at the initial position of the projectile After 3 00 s it is at ground level so the vertical dis.
placement is y H,To find H we use y v 0 y t a t 2 which becomes. H 15 m s sin 25 3 0 s,9 80 m s2 3 0 s 2 or H 25 m, 3 30 The initial velocity components of the projectile are. v 0 x 300 m s cos 55 0 172 m s and v 0 y 300 m s sin 55 0 246 m s. while the constant acceleration components are,ax 0 and a y g 9 80 m s2. The coordinates of where the shell strikes the mountain at t 42 0 s are. a t 2 172 m s 42 0 s 0 7 22 103 m 7 22 km,y v0 y t ay t 2. 246 m s 42 0 s,9 80 m s2 42 0 s 2 1 69 103 m 1 69 km.
3 31 The speed of the car when it reaches the edge of the cliff is. v v 02 2a x,0 2 4 00 m s2 50 0 m 20 0 m s, Now consider the projectile phase of the car s motion The vertical velocity of the car as it reaches the water is. v y v 02 y 2a y y,20 0 m s sin 24 0 2 9 80 m s2 30 0 m. v y 25 6 m s,b The time of flight is,v y v0 y 25 6 m s 20 0 m s sin 24 0. ay 9 80 m s2, a The horizontal displacement of the car during this time is. x v 0 x t 20 0 m s cos 24 0 1 78 s 32 5 m, 3 33 a At the highest point of the trajectory the projectile is moving horizontally with velocity components of and.
v x v 0 x v 0 cos 60 0 m s cos 30 0 52 0 m s, b The horizontal displacement is x v 0 x t 52 0 m s 4 00 s 208 m and from. y v 0 sin t 1,a y t 2 the vertical displacement is. y 60 0 m s sin 30 0 4 00 s,9 80 m s2 4 00 s 2 41 6 m. The straight line distance is,d x 2 y 2 208 m 2 41 6 m 2 212 m. 3 34 The horizontal kick gives zero initial vertical velocity to the rock Then from y v 0 y t 1. a y t 2 the time of flight,2 y 2 40 0 m,ay 9 80 m s2.
The extra time t 3 00 s 8 16 s 0 143 s is the time required for the sound to travel in a straight line back. to the player The distance the sound travels is,d x 2 y 2 v sound t 343 m s 0 143 s 49 0 m. where x represents the horizontal displacement of the rock when it hits the water Thus. d 2 y 49 0 m 2 40 0 m 2, The initial velocity given the ball must have been. v0 v0 x 9 91 m s, 3 35 a The jet moves at 3 00 102 mi h due east relative to the air Choosing a coordinate system with the positive. x direction eastward and the positive y direction northward the components of this velocity are. vr JA x 3 00 102 mi h and vr JA y 0, b The velocity of the air relative to Earth is 1 00 102 mi h at 30 0 north of east Using the coordinate sys. tem adopted in a above the components of this velocity are.

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