Holt Physics Problem 3C

Transcription

1 NAME DATE CLASS Holt Physics Proble 3C ADDING ECTORS ALGEBRAICALLY PROBLEM SOLUTION 1. DEFINE The southernost point in the United States is called South Point, and is located at the southern tip of the large island of Hawaii. A plane designed to take off and land in water leaves South Point and flies to Honolulu, on the island of Oahu, in three separate stages. The plane first flies 83.0 at.0 west of north fro South Point to Kailua Kona, Hawaii. The plane then flies 146 at 1.0 west of north fro Kailua Kona to Kahului, on the island of Maui. Finally, the plane flies 15 at 17.5 north of west fro Kahului to Honolulu. What is the plane s resultant displaceent? Given: d 1 = 83.0 q 1 =.0 west of north d = 146 q = 1.0 west of north d 3 = 15 q 3 = 17.5 north of west Unknown: d =? q =? Diagra: d 3 = 15 N θ 3 = 17.5 d = 146. PLAN d θ = 1.0 θ d 1 = 83.0 θ1 =.0 Choose the equation(s) or situation: Express the coponents of each vector in ters of sine or cosine functions. x 1 = d 1 (sin q 1 ) y 1 = d 1 (cos q 1 ) x = d (sin q ) y = d (cos q ) x 3 = d 3 (cos q 3 ) y 3 = d 3 (sin q 3 ) Note that the angles q 1 and q are with respect to the y axis (north), and so the x coponents are in ters of sin q. Write the equations for x tot and y tot, the coponents of the total displaceent. Ch. 3 6 Holt Physics Proble Bank

2 NAME DATE CLASS x tot = x 1 + x + x 3 = d 1 (sin q 1 ) + d (sin q ) + d 3 (cos q 3 ) y tot = y 1 + y + y 3 = d 1 (cos q 1 ) + d (cos q ) + d 3 (sin q 3 ) 3. CALCULATE Use the coponents of the total displaceent, the Pythagorean theore, and the tangent function to calculate the total displaceent. d = ( x tot ) + ( y tot ) q = tan 1 y tot Substitute the values into the equation(s) and solve: x tot = (83.0 )(sin.0 ) + (146 )(sin 1.0 ) + (15 ) (cos 17.5 ) = = 8 y tot = (83.0 )(cos.0 ) + (146 )(cos 1.0 ) + (15 ) (sin 17.5 ) = 59 d = (8 ) + (59) = = = q = tan 1 = 48.6 north of west 4. EALUATE ADDITIONAL PRACTICE If the diagra is drawn to scale, copare the calculated results to the drawing. The length of the drawn resultant is fairly close to the scaled agnitude for d, while the angle appears to be slightly greater than U.S. Highway 1 extends 55 at 37 north of east between Newell and Mud Butte, South Dakota. It then continues for 66 nearly due east fro Mud Butte to Faith, South Dakota. If you drive along this part of U.S. Highway 1, what will be your total displaceent?. Wrigley Field is one of only three original ajor-league baseball fields that are still in use today. Suppose you want to drive to Wrigley Field fro the corner of 55th Street and Woodlawn Avenue, about 14 iles south of Wrigley Field. Although not the fastest or ost direct route, the ost straightforward way to reach Wrigley Field is to drive 4.1 west on 55th Street to Halsted Street, then turn north and drive 17.3 on Halsted until you reach Clark Street. Turning on Clark, you will reach Wrigley Field after traveling 1. at an angle of 4.6 west of north. What is your resultant displaceent? Proble 3C Ch. 3 7

3 NAME DATE CLASS 3. A bullet traveling 850 ricochets fro a rock. The bullet travels another 640, but at an angle of 36 fro its previous forward otion. What is the resultant displaceent of the bullet? 4. The cable car syste in San Francisco is the last of its kind that is still in use in the United States. It was originally designed to transport large nubers of people up the steep hills on which parts of the city are built. If you ride seven blocks on the Powell Street cable car fro the terinal at Market Street to Pine Street, you will travel on level ground, then at an incline of 3.0 to the horizontal, and finally at 8.8 to the horizontal. What will be your resultant displaceent? 5. An Arctic tern flying to Antarctica encounters a stor. The tern changes direction to fly around the stor. If the tern flies 46 at 15 south of east, at 13 east of south, and finally 14 at 14 west of south, what is the tern s resultant displaceent? 6. A technique used to change the direction of space probes, as well as to give the additional speed, is to use the gravitational pull of nearby planet. This technique was first used with the oyager probes. oyager had traveled about when it reached Jupiter. Jupiter s gravity changed oyager s direction by 68. The probe then traveled about when it reached Saturn, and its direction was changed by 94. oyager was now redirected; it encountered Uranus after traveling fro Saturn. Use this inforation to calculate the resultant displaceent of oyager as it traveled fro Earth to Uranus. 7. The city of Asterda, in the Netherlands, has several canals that connect different sections of the city. Suppose you take a barge trip to the harbor, starting at a point near the northwest corner of the ondelpark. You would sail at 58.5 north of east, 375 at 1.8 north of east, and 875 at 1.5 east of north. What would be your resultant displaceent? 8. The elevated train, or L, in Chicago is a ajor source for ass transit in that city. One of the lines extends fro Jefferson Park, in the northwest part of town, to the Clark Street station downtown. The route of this line runs 5.0 at 36.9 south of east, 1.5 due south, 8.5 at 4. south of east, and 0.8 due east. What is the resultant displaceent of an L train fro Jefferson Park to Clark Street? 9. A billiard table is positioned with its long side parallel to north. A cue ball is then shot so that it travels 1.41 at an angle of 45.0 west of north, is deflected by the table s left side, and continues to ove 1.98 east of north at an angle of The ball is then deflected by the table s right side, so that it oves 0.4 west of north at an angle of After a reflection on the north end of the table, the ball travels 1.56 at an angle of 45.0 south of west. Deterine the resultant displaceent of the cue ball. Ch. 3 8 Holt Physics Proble Bank

4 NAME DATE CLASS 10. Hurricane Iniki was the ost destructive cyclone to have crossed the Hawaiian Islands in the twentieth century. It s path was also unusual: it oved south of the islands for 790 at an angle of 18 north of west, then oved due west for 150, turned north and continued for 470, and finally turned back 15 east of north and oved 40 to cross the island of Kauai. What was the resultant displaceent of Hurricane Iniki? Proble 3C Ch. 3 9

5 8. v = 165. /s q = 3.7 v forward = v(cos q) = (165. /s)(cos 3.7 ) v forward = 139 /s, forward v side = v(sin q) = (165. /s)(sin 3.7 ) v side = 89. /s to the side 9. v = 55.0 /h q = 13.0 above horizontal v y = v(sin q) = (55.0 /h)(sin 13.0 ) v y = 1.4 /h, upward v x = v(cos q) = (55.0 /h)(cos 13.0 ) v x = 53.6 /h, forward 10. v = 13.9 /s q h = 4.0 east of north q v = 6.0 above the horizontal v z = v(sin q v ) = (13.9 /s)(sin 6.0 ) v z = 6.09 /s, upward horizontal velocity = v h = v(cos q v ) v y = v h (cos q h ) = v(cos q v )(cos q h ) = (13.9 /s)(cos 6.0 )(cos 4.0 ) v y = 11.4 /s, north v x = v h (sin q h ) = v(cos q v )(sin q h ) = (13.9 /s)(cos 6.0 )(sin 4.0 ) v x = 5.08 /s, east Additional Practice 3C 1. d 1 = 55 q 1 = 37 north of east d = 66 q = 0.0 (due east) x 1 = d 1 (cos q 1 ) = (55 )(cos 37 ) = 44 y 1 = d 1 (sin q 1 ) = (55 )(sin 37 ) = 33 x = d (cos q ) = (66 )(cos 0.0 ) = 66 y = d (sin q ) = (66 )(sin 0.0 ) = 0 x tot = x 1 + x = = 110 y tot = y 1 + y = = 33 d = ( x tot ) + ( y tot ) = (110 ) + (33 ) = = d = 115 q = tan 1 y t ot q = 17 north of east = tan Ch. 3 4 Holt Physics Solution Manual

6 . d 1 = 4.1 q 1 = 180 (due west) d = 17.3 q = 90.0 (due north) d 3 = 1. q 3 = 4.6 west of north = = x 1 = d 1 (cos 1 ) = (4.1 )(cos 180 ) = 4.1 y 1 = d 1 (sin q 1 ) = (4.1 )(sin 180 ) = 0 x = d (cos q ) = (17.3 )(cos 90.0 ) = 0 y = d (sin q ) = (17.3 )(sin 90.0 ) = 17.3 x 3 = d 3 (cos q 3 ) = (1. )(cos ) = 0.4 Dy 3 = d 3 (sin q 3 ) = (1. )(sin ) = 1.1 x tot = x 1 + x + x 3 = ( 0.4 ) = 4.5 y tot = y 1 + y + y 3 = = 18.4 d = ( x tot ) + ( y tot ) = ( 4. 5) + (18.4 ) = = 359 d = 18.9 q = tan 1 y t ot = tan = 76 = 76 north of west 3. d 1 = 850 q 1 = 0.0 d = 640 q = 36 x 1 = d 1 (cos q 1 ) = (850 )(cos 0.0 ) = 850 y 1 = d 1 (sin q 1 ) = (850 )(sin 0.0 ) = 0 x = d (cos q ) = (640 )(cos 36 ) = 50 y = d (sin q ) = (640 )(sin 36 ) = 380 x tot = x 1 + x = = 1370 y tot = y 1 + y = = 380 d = ( x tot ) + ( y tot ) = (1370) + (380) = = d = 1400 q = tan 1 y t ot = tan q = 16 to the side of the initial displaceent Section Five Proble Bank Ch. 3 5

7 4. d 1 = q 1 = 0.0 d = q = 3.0 d 3 = q 3 = 8.8 x 1 = d 1 (cos q 1 ) = ( )(cos 0.0 ) =.0 10 y 1 = d 1 (sin q 1 ) = ( )(sin 0.0 ) = 0 x = d (cos q ) = ( )(cos 3.0 ) = y = d (sin q ) = ( )(sin 3.0 ) = 16 x 3 = d 3 (cos q 3 ) = ( )(cos 8.8 ) =.0 10 y 3 = d 3 (sin q 3 ) = ( )(sin 8.8 ) = 31 x tot = x 1 + x + x 3 = = y tot = y 1 + y + y 3 = = 47 d = ( x tot ) + ( y tot ) = ( ) + (47 ) = = d = q = tan 1 y t ot = tan q = 3.8 above the horizontal d 1 = 46 q 1 = 15 south of east = 15 d = q = 13 east of south = 77 d 3 = 14 q 3 = 14 west of south = = 104 x 1 = d 1 (cos q 1 ) = (46 )[cos( 15 )] = 44 y 1 = d 1 (sin q 1 ) = (46 )[sin( 15 )] = 1 x = d (cos q ) = ( )[cos( 77 )] = 4.9 y = d (sin q ) = ( )[sin( 77 )] = 1 x 3 = d 3 (cos q 3 ) = (14 )[cos( 104 )] = 3.4 y 3 = d 3 (sin q 3 ) = (14 )[sin( 104 )] = 14 x tot = x 1 + x + x 3 = ( 3.4 ) = 46 y tot = y 1 + y + y 3 = 1 + ( 1 ) + ( 14 ) = 47 d = ( x tot ) + ( y tot ) = (46) + ( 47 ) = = d = 66 q = tan 1 y t ot q = 46 south of east = tan = 46 Ch. 3 6 Holt Physics Solution Manual

8 6. d 1 = q 1 = 0.0 d = q = 68 d 3 = q 3 = =16 x 1 = d 1 (cos q 1 ) = ( )(cos 0.0 ) = y 1 = d 1 (sin q 1 ) = ( )(sin 0.0 ) = 0 x = d (cos q ) = ( )(cos 68 ) = y = d (sin q ) = ( )(sin 68 ) = x 3 = d 3 (cos q 3 ) = ( )(cos 16 ) = y 3 = d 3 (sin q 3 ) = ( )(sin 16 ) = x tot = x 1 + x + x 3 = ( ) = y tot = y 1 + y + y 3 = = d = (x to t ) + (y tot ) = ( ) + ( ) = = d = q = tan 1 y t ot = tan = =138 q = 138 fro the probe s initial direction 7. d i = q 1 = 58.5 north of east d = 375 q = 1.8 north of east d 3 = 875 q 3 = 1.5 east of north x 1 = d 1 (cos q 1 ) = ( )(cos 58.5 ) = 1310 y 1 = d 1 (sin q 1 ) = ( )(sin 58.5 ) = 130 x = d (cos q ) = (375 )(cos 1.8 ) = 348 y = d (sin q ) = (375 )(sin 1.8 ) = 139 x 3 = d 3 (sin q 3 ) = (875 )(sin 1.5 ) = 31 y 3 = d 3 (cos q 3 ) = (875 )(cos 1.5 ) = 814 x tot = x 1 + x + x 3 = = y tot = y 1 + y + y 3 = = d = ( x tot ) + ( y tot ) = ( ) + ( ) = = d = q = tan 1 y t ot q = 57.3 north of east = tan Section Five Proble Bank Ch. 3 7

9 8. d 1 = 5.0 q 1 = 36.9 south of east = 36.9 d = 1.5 q = 90.0 due south = 90.0 d 3 = 8.5 q 3 = 4. south of east = 4. d 4 = 0.8 q 4 = 0 (due east) x 1 = d 1 (cos q 1 ) = (5.0 )[cos( 36.9 )] = 4.0 y 1 = d 1 (sin q 1 ) = (5.0 )[sin( 36.9 )] = 3.0 x = d (cos q ) = (1.5 )[cos( 90.0 )] = 0 y = d (sin q ) = (1.5 )[sin( 90.0 )] = 1.5 x 3 = d 3 (cos q 3 ) = (8.5 )[cos( 4. )] = 6.3 y 3 = d 3 (sin q 3 ) = (8.5 )[sin( 4. )] = 5.7 x 4 = d 4 (cos q 4 ) = (0.8 )(cos 0 ) = 0.8 y 4 = d 4 (sin q 4 ) = (0.8 )(sin 0 ) = 0 x tot = x 1 + x + x 3 + x 4 = = 11.1 y tot = y 1 + y + y 3 + y 4 = ( 3.0 ) + ( 1.5 ) + ( 5.7 ) + 0 = 10. d = ( x tot ) + ( y tot ) = (11. 1) + ( 10. ) = = 7 d = 15.1 q = tan 1 y t ot = tan 1 1 q = 4.6 south of east = d 1 = 1.41 q 1 = 45.0 west of north = = d = 1.98 q = 45.0 east of north = 45.0 d 3 = 0.4 q 3 = 45.0 west of north = d 4 = 1.56 q 4 = 45.0 south of west = = 5.0 x 1 = d 1 (cos q 1 ) = (1.41 )(cos ) = y 1 = d 1 (sin q 1 ) = (1.41 )(sin ) = x = d (cos q ) = (1.98 )(cos 45.0 ) = 1.40 y = d (sin q ) = (1.98 )(sin 45.0 ) = 1.40 x 3 = d 3 (cos q 3 ) = (0.4 )(cos ) = 0.30 y 3 = d 3 (sin q 3 ) = (0.4 )(sin ) = 0.30 x 4 = d 4 (cos q 4 ) = (1.56 )(cos 5.0 ) = 1.10 y 4 = d 4 (sin q 4 ) = (1.56 )(sin 5.0 ) = 1.10 x tot = x 1 + x + x 3 + x 4 = ( ) ( 0.30 ) + ( 1.10 ) = y tot = y 1 + y + y 3 + y 4 = ( 1.10 ) = 1.60 d = ( x tot ) + ( y tot = ( ) + (1. 60 ) = = d = 1.88 q = tan 1 y t ot = tan = 58.1 q = 58.1 north of west Ch. 3 8 Holt Physics Solution Manual

10 10. d 1 = 79 q 1 = 18 north of west =16 d = 150 q = due west d 3 = 470 q 3 = 90.0 due north d 4 = 40 q 4 = 15 east of north =75 x 1 = d 1 (cos q 1 ) = (790 )(cos 16 ) = 750 y 1 = d 1 (sin q 1 ) = (790 )(sin 16 ) = 4 x = d (cos q ) = (150 )(cos ) = 150 y = d (sin q ) = (150 )(sin ) = 0 x 3 = d 3 (cos q 3 ) = (470 )(cos 90.0 ) = 0 y 3 = d 3 (sin q 3 ) = (470 )(sin 90.0 ) = 470 x 4 = d 4 (cos q 4 ) = (40 )(cos 75 ) = 6 y 4 = d 4 (sin q 4 ) = (40 )(sin 75 ) = 30 x tot = x 1 + x + x 3 + x 4 = ( 750 ) + ( 150 ) = 840 y tot = y 1 + y + y 3 + y 4 = = 940 d = ( x tot ) + ( y tot ) = ( 840) + (940) = = d = 160 q = tan 1 y t ot = tan = 48 q = 48 north of west Additional Practice 3D 1. v x = 430 /s t = v x x x = 400 y = 1 g t g = 9.81 /s y = 1 g v x x height of ridge = 430 = 1 (9.81 /s 400 ) 4 30 /s = 430. x = 101 t = v x x v x = 14.5 /s y = 1 g t g = 9.81 /s y = 1 g v x x 3. v x = /h t = v x x x = 135 y = 1 g t g = 9.81 /s y = 1 g v x x = 1 (9.81 /s 101 ) /s = 46 height of building = 46 = 1 (9.81 /s 135 ) /h s/h 3 0 / = 68.6 airship s altitude = 68.6 Section Five Proble Bank Ch. 3 9