Physics Solution Manual for 1100 and 2101

Solution the first two rows of the following table

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Unformatted text preview: due north, and thus has no x component. Therefore the x component of vTG is equal to the x component of vTC: 143 North vCG vTG vTC 52.0° East vTG,x = vTC,x + vCG,x = vTC cos 52.0o + 0 m/s = ( 24.0 m/s ) cos 52.0o = 14.8 m/s Next, we find the y component of the truck’s velocity relative to the ground. Noting that the vector vCG points due north, so that vCG,y = vCG, we have vTG,y = vTC,y + vCG,y = vTC sin 52.0o + vCG = ( 24.0 m/s ) sin 52.0o + 16.0 m/s = 34.9 m/s The magnitude of vTG is found from the Pythagorean theorem: 2 2 vTG = vTG,x + vTG,y = (14.8 m/s )2 + ( 34.9 m/s )2 = 37.9 m/s 63. SSM REASONING The velocity v PM of the puck relative to Mario is the vector sum of the velocity v PI of the puck relative to the ice and the velocity v IM of the ice relative to Mario as indicated by Equation 3.7: v PM = v PI + v IM . The values of v MI and v PI are given in the statement of the problem. In order to use the data, we must make use of the fact that v IM = − v MI , with the result that v PM = v PI − v MI . SOLUTION The first two rows of the following table give the east/west and north/south components of the vectors v PI and – v MI . The third row gives the components of their resultant v PM = v PI − v MI . Due east and due north have been taken as positive. 144 KINEMATICS IN TWO DIMENSIONS Vector East/West Component North/South Component v PI –(11.0 m/s) sin 22° = –4.1 m/s –(11.0 m/s) cos 22° = –10.2 m/s – v MI 0 +7.0 m/s v PM = v PI – v MI –4.1 m/s –3.2 m/s Now that the components of v PM are known, the Pythagorean theorem can be used to find the magnitude. v PM = (–4.1 m / s) 2 + (–3.2 m / s) 2 = 5.2 m / s v PM φ 3.2 m/s 4.1 m/s The direction of v PM is found from φ = tan −1 4.1 F m / sI = G m / sJ HK 3.2 64. REASONING AND SOLUTION 52 ° west of south While flying west, the airplane has a ground speed of 2 vPG = 2.40 × 10 m/s – 57.8 m/s = 182 m/s and requires time tW = x/(182 m/s) to reach the turn-around point. While flying east the airplane has a ground speed of 2 vPG = 2.40 × 10 m/s + 57.8 m/s = 298 m/s and requires tE = x/(298 m/s) to return home. Now the total time for the trip is 4 t = tW + tE = 6.00 h = 2.16 × 10 s, so x/182 + x/298 = 2.16 × 10 4 or x = 2.44 × 106 m = 2440 km Chapter 3 Problems 65. SSM WWW REASONING 145 The velocity v OW of the object relative to the water is the vector sum of the velocity v OS of the object relative to the ship and the velocity vSW of the ship relative to the water, as indicated by Equation 3.7: v OW = v OS + vSW . The value of vSW is given in the statement of the problem. We can find the value of v OS from the fact that we know the position of the object relative to the ship at two different times. The initial position is rOS1, and the final position is rOS2. Since the object moves with constant velocity, ∆r r −r v OS = OS = OS2 OS1 (1) ∆t ∆t SOLUTION The first two rows of the following table give the east/west and north/south components of the vectors rOS2 and −rOS1 . The third row of the table gives the components of ∆rOS = rOS2 − rOS1 . Due east and due north have been taken as positive. Vector East/West Component North/South Component rOS2 –(1120 m) cos 57.0° = −6.10 × 10 2 m –(1120 m) sin 57.0° = −9.39 × 10 2 m −rOS1 –(2310 m) cos 32.0° = – 1.96 × 103 m +(2310 m) sin 32.0° = 1.22 × 103 m −2.57 × 103 m 2.81× 10 2 m ∆rOS = rOS2 − rOS1 Now that the components of ∆rOS are known, the Pythagorean theorem can be used to find the magnitude. ∆r OS 2.81 × 10 m 2 ∆rOS = (–2.57 × 10 m ) + ( 2 .81 × 10 m ) = 2.59 × 10 m 3 2 2 2 3 The direction of ∆rOS is found from φ = tan −1 Therefore, from Equation (1), F × 10 m I = 6.24 ° 2.81 G × 10 m J H K 2.57 2 3 φ 2.57 × 103 m 146 KINEMATICS IN TWO DIMENSIONS v OS = ∆rOS ∆t = rOS 2 − rOS1 ∆t = 2.59 × 10 3 m = 7 .19 m / s, 6.24 ° north of west 360 s Now that v OS is known, we can find v OW , as indicated by Equation 3.7: v OW = v OS + vSW . The following table summarizes the vector addition: Vector East/West Component North/South Component –(7.19 m/s) cos 6.24° = –7.15 m/s (7.19 m/s) sin 6.24° = 0.782 m/s v OS v SW +4.20 m/s 0 m/s v OW = v OS + vSW –2.95 m/s 0.782 m/s Now that the components of v OW are known, the Pythagorean theorem can be used to find the magnitude. v OW 0.782 m/s v OW = (–2.95 m / s) 2 + ( 0.782 m / s) 2 = 3.05 m / s φ 2.95 m/s The direction of v OW is found from φ = tan −1 0 F.782 m / s I = G.95 m / s J H K 2 14.8 ° north of west ____________________________________________________________________________________________ 66. REASONING The data given in the problem are summarized as follows: x-Direction Data x ax 4.11 × 106 m ? vx y-Direction Data v0x t y ay 4370 m/s 684 s 6.07 × 106 m ? vy v0y t 6280 m/s 684 s With these data, we can use Equations 3.5a and 3.5b to determine the acceleration component ax and ay. Chapter 3 Problems 147 SOLUTION Using Equations 3.5a and 3.5b, we can solve for the acceleration co...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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