Problem 3A


1a) 24 m b) 8 m 2) 0.14 m, 40˚ below the horizontal 3) 171 m, 15˚ below the horizontal 4) 16 m, 55˚ below the horizontal 5) 57.9 m, 49.4˚ north of east, 17.1˚ above the horizontal 6) 592 m, east, 8.5˚ north of east 7) 4.8x102 km, west, 5.0˚ south of west 8) 6700 km, west


Problem 3B


1) 770 m 2) 1.0 m 3) 14.9 m, south 4) 33 km/h, downward 5)  ∆x=2.6 m along the table's length,  ∆y=0.61 m along the table's width 6) vx=0.290 m/s, east, vy=1.16 m/s, north 7)  ∆x=15.6 km, west,  ∆y=27.0 km, south 8) vforward=139 km/s, forward, vside=89.2 km/s to the side 9) vy=12.4 km/h, upward, vx=53.6 km/h, forward 10) vz=6.09 m/s, upward, vy=11.4 m/s, north, vx=5.08 m/s, east


Problem 3C


1) 115 km, 17˚ N of E 2) 18.9 km, 76˚ N of W


Problem 3D


1) 430 m 2) 246 m 3) 68.6 m 4) 404 m 5) 1230 m 6) 17.0 m


Problem 3E


1) 45.8 m/s 2) 68.2 m/s 3) 16.6 m/s 4a) 20.8 m/s b) The brick's maximum height is 11.0 m. c) The brick's maximum height is 22.1 m. 5) 4.07 m 6) 60˚


Problem 3F


1) 0.47 h = 28 min 2) 53 s 3) 79 s 4) 1800 s = 3.0x101 min 5)  ∆t=182 s,  ∆t'=33 s, 6)  ∆x=11m, west,  ∆y=18 m, north


Section 3-1 Review, page 87


1a) vector b) scalar c) scalar d) vector e) scalar 2) 126 m at 10˚ above the horizontal 3) 204 km/h at 75˚ north of east 4) 89 km/h at 54˚ north of east 5) zero


Section 3-2 Review, page 97


1a) x-axis: forward and backward on the sidewalk, y-axis: left and right on sidewalk b) x-axis: forward and backward on rope, y-axis: up and down c) x-axis: horizontal at water level, y-axis: up and down 2a) 5.8 m/s at 59˚ downriver from its intended path b) 6.1 m/s at 9.5˚ from the direction the wave is traveling 3a) 7.07 km, 7.07 km b) 1.6 m/s2, 1.1 m/s2 4) 13.5 m at 37˚ north of west


Section 3-3 Review, page 105


1) b, f 2) a, b, d, e, g 3) 319 m 4) 104.8 m/s at 17.4˚ below the horizontal 5) 5.05 s; 454 m 6) 3.6 m/s 7) 1.0 m


Section 3-4 Review, page 109


1a) south with a speed equal to the train's speed b) moves north c) appears to fall straight down d) moves in a parabola 2) 10 m/s away in the opposite direction 3) 1.51 m/s at 5.7˚ north of east


Section 3-1 Worksheet


1) {A,C,E,H,I}; {D,G}, {B,F,J} 2) {A,D,H}, {B,C,G}, {I,J} 3) {A,H} 4) Both diagrams should show a vector A that is twice as long as the original vector A, but still pointing up. The first diagram should have the tip of 2A next to the tail of B. The second diagram should have the tip of B next to the tail of 2A. The resultant vectors should have the same magnitude and direction, slanting towards the upper right. 5) Both diagrams should show a vector B thst is half as long as the original vector B. The first diagram should have the tip of A next to the tail of -B/2, and -B/2 should be pointing to the left. The second diagram should have the tip of B/2 next to the tail of -A, and -A should be pointing down. The resultant vectors should have the same magnitude but opposite directions. The first will slant towards the upper left. The second will slant towards the lower right.


Section 3-2 Worksheet


1) graph 2) Shot 1: 45 m; 45 m, Shot 2: 110 m; 64 m, Shot 3: 65 m; 33 m, Shot 4: 0 m; 14.89 m 3) 220 m


Section 3-3 Worksheet


1)  ∆t=visinø/g 2) h=vi2(sinø)2/2g 3) x=vi(cosø)(∆t) 4) R=(2vi2sinøcosø)/g 5) 15˚-8.5 m-130 m, 30˚-32 m-220 m, 45˚-65 m-250 m, 60˚-85 m-220 m, 75˚-120 m-130 m


Section 3-4 Worksheet


1) vBL=vBW+vWL 2) Diagrams should show vBW twice as long as vWL but both are in the same direction as vBL, which is long as both together 3) Diagrams should show vWL and vBW, longer and opposite in direction. The vector vBL should be as the difference between the two, and in the same direction and in the same direction as vBW. 4) Diagrams should show vWL and vBW at a right angle with vBL forming the hypotenuse of a right triangle. 5a) 6.0 km/h, due east b) 2.0 km/h, due west c) 4.5 km/h, 26.6˚