101_PartUniversity Physics Solution

101_PartUniversity Physics Solution - Motion in Two or...

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Motion in Two or Three Dimensions 3-31 0 (30.0 m/min)(4.00 min) 120.0 m xx −= = . You will land 120.0 m east of point B , which is 45.0 m east of point C . The distance you will have traveled is 22 (400.0 m) (120.0 m) 418 m += . (b) B/W v ! is directed at angle φ east of north, where 75.0 m tan 400.0 m = and 10.6 = ° . B/W- (100.0 m/min)sin10.6 18.4 m/min x v == ° and B/W- (100.0 m/min)cos10.6 98.3 m/min y v ° . B/E- B/W- W/E- 18.4 m/min 30.0 m/min 48.4 m/min xxx vvv =+= + = . B/E- B/W- W/E- 98.3 m/min yyy . 0 B/E- 400.0 m 4.07 min 98.3 m/min y yy t v = . 0 (48.4 m/min)(4.07 min) 197 m = . You will land 197 m downstream from B , so 122 m downstream from C . (c) (i) If you reach point C , then B/E v ! is directed at 10.6 ° east of north, which is 79.4 ° north of east. We don’t know the magnitude of B/E v ! and the direction of B/W v ! . In part (a) we found that if we aim the boat due north we will land east of C , so to land at C we must aim the boat west of north. Let B/W v ! be at an angle of north of west. The relative velocity addition diagram is sketched in Figure 3.76. The law of sines says W/E B/W sin sin79.4 vv θ = ° . 30.0 m/min sin sin79.4 100.0 m/min ⎛⎞ = ⎜⎟ ⎝⎠ ° and 17.15 = ° . Then 180 79.4 17.15 83.5 =− = °° . The boat will head 83.5 ° north of west, so 6.5 ° west of north. B/E- B/W- W/E- (100.0 m/min)cos83.5 30.0 m/min 18.7 m/min + = ° . B/E- B/W- W/E- (100.0 m/min)sin83.5 99.4 m/min = ° . Note that these two components do give the direction of B/E v ! to be 79.4 ° north of east, as required. (ii) The time to cross the river is 0 B/E- 400.0 m 4.02 min 99.4 m/min y t v = . (iii) You travel from A to C , a distance of (400.0 m) (75.0 m) 407 m . (iv) B/E B/E- B/E- ( ) ( ) 101 m/min xy v =+ = . Note that B/E 406 m vt = , the distance traveled (apart from a small difference due to rounding). EVALUATE: You cross the river in the shortest time when you head toward point B , as in part (a), even though you travel farther than in part (c). Figure 3.76 3.77. IDENTIFY: / x vd x d t = , / y y d t = , / ad t = and / t = . SET UP: (sin ) cos( ) dt t dt ω = and (cos ) sin( ) t dt . EXECUTE: (a) The path is sketched in Figure 3.77. (b) To find the velocity components, take the derivative of x and y with respect to time: (1 cos ), x vR ω t and sin . y ω t = To find the acceleration components, take the derivative of x v and y v with respect to time: 2 sin x aR t , = and 2 cos y t . = (c) The particle is at rest ( 0) yx every period, namely at 0 2 / 4 / .... t, πω , , = At that time, 0 2 4 . . . ; x, π R, π R, = and 0. y = The acceleration is 2 = in the - y + direction.
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3-32 Chapter 3 (d) No, since () ( ) 1/2 22 2 sin cos . aR t R t R ω ωω ⎡⎤ =+ = ⎢⎥ ⎣⎦ The magnitude of the acceleration is the same as for uniform circular motion. EVALUATE: The velocity is tangent to the path. 0 x v is always positive; y v changes sign during the motion. Figure 3.77 3.78. IDENTIFY: At the highest point in the trajectory the velocity of the projectile relative to the earth is horizontal. The velocity P/E v ! of the projectile relative to the earth, the velocity F/P v ! of a fragment relative to the projectile, and the velocity F/E v ! of a fragment relative to the earth are related by F/E F/P P/E v= v+ v !
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101_PartUniversity Physics Solution - Motion in Two or...

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