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### chapter3

Course: PHYS 121, Fall 2009
School: University of Montana
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to Solutions Chapter 3 problems: Again, these are book solutions P1. The resultant vector displacement of the car is given by r r r DR = Dwest + Dsouth- . The westward displacement is west r Dsouthwest 215 + 85 cos 45o = 275.1 km and the south displacement is 85 sin 45o = 60.1 km . The resultant displacement has a magnitude of r Dwest r DR 275.12 + 60.12 = 281.6 km 282 km . The direction is = tan 1 60.1...

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to Solutions Chapter 3 problems: Again, these are book solutions P1. The resultant vector displacement of the car is given by r r r DR = Dwest + Dsouth- . The westward displacement is west r Dsouthwest 215 + 85 cos 45o = 275.1 km and the south displacement is 85 sin 45o = 60.1 km . The resultant displacement has a magnitude of r Dwest r DR 275.12 + 60.12 = 281.6 km 282 km . The direction is = tan 1 60.1 275.1 = 12.3o 12o south of west . P2. The truck has a displacement of 18 + ( 16 ) = 2 blocks north and 10 blocks east. The resultant has a magnitude of of tan 2 10 = 11 north of east . o 1 2 + 10 = 10 blocks and a direction 2 2 r Dnorth r Deast r DR r Dsouth P8. (a) V1x = 6.6 units V1 y = 0 units V2 V1 + V2 V2 x = 8.5 cos 45o = 6.0 units V2 y = 8.5 sin 45o = 6.0 units r r (b) V1 + V2 = (V1 x + V2 x , V1 y + V2 y ) = ( 0.6, 6.0 ) r V1 r r 6.0 2 2 V1 + V2 = ( 0.6 ) + ( 6.0 ) = 6.0 units = tan 1 = 84 o 0.6 The sum has a magnitude of 6.0 units, and is 84o clockwise from the negative x-axis, or 96o counterclockwise from the positive x-axis. r r r P9. (a) vnorth = ( 735 km h ) ( cos 41.5o ) = 550 km h vwest = ( 735 km h ) sin 41.5o = 487 km h ( ) (b) d north = vnorth t = ( 550 km h )( 3.00 h ) = 1650 km d west = vwest t = ( 487 km h )( 3.00 h ) = 1460 km P10. Ax = 44.0 cos 28.0 o = 38.85 Bx = 26.5 cos 56.0 o = 14.82 Ay = 44.0 sin 28.0 o = 20.66 B y = 26.5 sin 56.0 o = 21.97 C x = 31.0 cos 270 o = 0.0 C y = 31.0 sin 270 o = 31.0 r r r (a) ( A + B + C ) x = 38.85 + ( 14.82 ) + 0.0 = 24.03 = 24.0 r r r ( A + B + C ) y = 20.66 + 21.97 + ( 31.0 ) = 11.63 = 11.6 (b) r r r A+B+C = ( 24.03 ) 2 + (11.63 ) = 26.7 2 = tan 1 11.63 24.03 = 25.8o P11. Ax = 44.0 cos 28.0 o = 38.85 C x = 31.0 cos 270 o = 0.0 r r ( A C ) x = 38.85 0.0 = 38.85 r r A C = Ay = 44.0 sin 28.0 o = 20.66 C y = 31.0 sin 270 o = 31.0 r r ( A C ) y = 20.66 ( 31.0 ) = 51.66 ( 38.85 ) 2 + ( 51.66 ) = 64.6 2 = tan 1 51.66 38.85 = 53.1o P14. Ax = 44.0 cos 28.0 o = 38.85 Bx = 26.5 cos 56.0 o = 14.82 Ay = 44.0 sin 28.0 o = 20.66 B y = 26.5 sin 56.0 o = 21.97 C x = 31.0 cos 270 o = 0.0 C y = 31.0 sin 270 o = 31.0 r r r r B 2 A y = 21.97 2 ( 20.66 ) = 19.35 (a) B 2 A x = 14.82 2 ( 38.85 ) = 92.52 ( ) ( ) Note that since both components are negative, the vector is in the 3rd quadrant. r r 19.35 2 2 = tan 1 B 2 A = ( 92.52 ) + ( 19.35 ) = 94.5 = 11.8o below x axis 92.52 r r r (b) ( 2 A 3B + 2C ) x = 2 ( 38.85 ) 3 ( 14.82 ) + 2 ( 0.0 ) = 122.16 r r r 2 A 3B + 2C y = 2 ( 20.66 ) 3 ( 21.97 ) + 2 ( 31.0 ) = 86.59 ( ) Note that since the x component is positive and the y component is negative, the vector is in the 4th quadrant. r r r 86.59 2 2 = tan 1 2 A 3B + 2C = (122.16 ) + ( 86.59 ) = 149.7 = 35.3o below + x axis 122.16 P17. Choose downward to be the positive y direction. The origin will be at the point where the tiger leaps from the rock. In the horizontal direction, v x 0 = 3.5 m s and a x = 0 . In the vertical direction, v y 0 = 0 , a y = 9.80 m s , y0 = 0 , and the final location y = 6.5 m . The time for the tiger to reach 2 the ground is found from applying Eq. 2-11b to the vertical motion. y = y0 + v y 0 t + 1 a y t 2 2 6.5m = 0 + 0 + 1 2 ( 9.8 m s ) t 2 2 t= 2 ( 6.5m ) 9.8 m s 2 = 1.15 sec The horizontal displacement is calculated from the constant horizontal velocity. x = v x t = ( 3.5 m s )(1.15 sec ) = 4.0 m P18. Choose downward to be the positive y direction. The origin will be at the point where the diver dives from the cliff. In the horizontal direction, v x 0 = 1.8 m s and a x = 0 . In the vertical direction, v y 0 = 0 , a y = 9.80 m s , y0 = 0 , and the time of flight is t = 3.0 s . The height of the cliff is found 2 from applying Eq. 2-11b to the vertical motion. y = y0 + v y 0 t + 1 a y t 2 2 y = 0+0+ 1 2 ( 9.80 m s ) ( 3.0 s ) 2 2 = 44 m The distance from the base of the cliff to where the diver hits the water found is from the horizontal motion at constant velocity: x = v x t = (1.8 m s )( 3 s ) = 5.4 m P21. Choose downward to be the positive y direction. The origin will be at the point where the ball is 2 thrown from the roof of the building. In the vertical direction, v y 0 = 0 , a y = 9.80 m s , y0 = 0 , and the displacement is 45.0 m. The time of flight is found from applying Eq. 2-11b to the vertical motion. y = y0 + v y 0 t + 1 a y t 2 2 45.0 m = 1 2 ( 9.80 m s ) t 2 2 t= 2 ( 45.0 m ) 9.80 m s 2 = 3.03 sec The horizontal speed (which is the initial speed) is found from the horizontal motion at constant velocity: x = v x t v x = x t = 24.0 m 3.03 s = 7.92 m s . P22.Choose the point at which the football is kicked the origin, and choose upward to be the positive y direction. When the football reaches the ground again, the y displacement is 0. For the football, 2 v y 0 = 18.0 sin 35.0 o m s , a y = 9.80 m s and the final y velocity will be the opposite of the ( ) starting y velocity (reference problem 3-28). Use Eq. 2-11a to find the time of flight. 18.0 sin 35.0 o m s 18.0 sin 35.0 o m s vy vy0 v y = v y 0 + at t = = = 2.11 s a 9.80 m s 2 ( ) ( ) P23. Choose downward to be the positive y direction. The origin is the point where the ball is thrown 2 from the roof of the building. In the vertical direction, v y 0 = 0 , y0 = 0 , and a y = 9.80 m s . The initial horizontal velocity is 22.2 m/s and the horizontal range is 36.0 m. The time of flight is found from the horizontal motion at constant velocity. x = v x t t = x v x = 36.0 m 22.2 m s = 1.62 s The vertical displacement, which is the height of the building, is found by applying Eq. 2-11b to the vertical motion. y = y0 + v y 0 t + 1 a y t 2 2 y = 0+0+ 1 2 ( 9.80 m s ) (1.62 s ) 2 2 = 12.9 m P27. Choose downward to be the positive y direction. The origin is the point where the supplies are 2 dropped. In the vertical direction, v y 0 = 0 , a y = 9.80 m s , y0 = 0 , and the final position is y = 160 m . The time of flight is found from applying Eq. 2-11b to the vertical motion. y = y0 + v y 0 t + 1 a y t 2 2 t= 2 (160 m ) 160 m = 0 + 0 + 1 2 ( 9.80 m s ) t 2 2 = 5.71 s 9.80 m s 2 Note that the speed of the airplane does not enter into this calculation. P31. Choose the origin to be at ground level, under the place where ...

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