Sup. Prob. 4 Key.pdf - Dickey Physics

CHAPTER 41. Bob walks 80 m and then he walks125 m.a. What is Bob’s displacement if he walkseast both times?80 m 125 m 205 mb. What is Bob’s displacement if he walkseast then west?80 m 125 m –45 mc. What distance does Bob walk in eachcase?80 m 125 m 205 m2. A cross-country runner runs 5.0 km eastalong the course, then turns around andruns 5.0 km west along the same path.She returns to the starting point in40 min. What is her average speed? heraverage velocity?average speeddistance traveled during time intervaltime 5.0 k m 5.0 km40min (0.25 km/min)(60 min/h) 15 km/haverage velocitydisplacement during time intervaltime+5.0 km 5.0 km 040 min3. Car A is traveling at 85 km/h while car Bis at 60 km/h. What is the relativevelocity of car A to car Ba. if they both are traveling in the samedirection?85 km/h 60 km/h 25 km/hb. if they are headed toward each other?85 km/h 60 km/h 145 km/h4. Find θ ifa. tan θ 9.5143.θ 84.000°b. sin θ 0.4540.θ 27.00°c. cos θ 0.8192θ 35.00°d. tan q 0.1405θ 7.998°e. sin q 0.7547.θ 49.00°f. cos θ 0.9781.θ 12.01°5. Find the value of:a. tan 28°.0.53b. sin 86°.1.0c. cos 2°.1d. tan 58°.1.6e. sin 40°.0.64f. cos 71°.0.336. You walk 30 m south and 30 m east.Draw and add vectors representing thesetwo displacements.30 mNttan θ 3 0 m 130mθ 45°So, the resultant is at45° 270° 315°.R 2 (30 m) 2 (30 m) 2R 42 m, 315°R30 mECopyright © by Glencoe/McGraw-Hill46 Supplemental Problems Manual Physics: Principles and Problems

Chapter 4 (continued)7. Solve for all sides and all angles for thefollowing right triangles.a.65˚6.82.925˚25˚6.26.2vwtNACopyright © by Glencoe/McGraw-Hillb.c.d.e.7.039˚58˚12.15.111.94.08.29.47.039˚54˚58˚58˚7.64.72.512.19.75.111.99.651˚32˚4.032˚8.236˚8. A plane flying at 90° at 1.00 10 2 m/s isblown toward 180° at 5.0 10 1 m/s by astrong wind. Find the plane’s resultantvelocity and direction.A 1.00 10 2 m/s, northw 5.0 10 1 m/s, west5.0 101m/stan θ 1 .00 102 0.5m/sθ 27°So the resultant is at 90° θ 117°.v 2 A 2 w 2v (1.00 10 2 m/s) 2 (5.0 10 1 m/s) 2 112 m/sThus, v 112 m/s, 117°9. A man hops a freight car 15.0 m long and3.0 m wide. The car is moving east at2.5 m/s. Exploring the surroundings, theman walks from corner A to corner B in20.0 s; then from corner B to corner C in5.0 s as shown. With the aid of a vectordiagram, compute the man’sdisplacement relative to the ground.CB3.0 mR15.0 m48 m63 md car vt (2.5 m/s)(25.0 s) 63 md E 63 m 15.0 m 48 mR 2 (48 m)2 (3.0 m)2R 48.1 m3.0mtan θ 0.06244 8.1mθ 3.6°Thus, R 48.1 m, 3.6° N of EE15 mAEast3.0 mPhysics: Principles and Problems Supplemental Problems Manual 47

Chapter 4 (continued)10. A plane travels on a heading of 40.0° fora distance of 3.00 10 2 km. How farnorth and how far east does the planetravel?40˚ 40.0˚d 3.0 10 2 km at 40.0°d N d sin θ (3.00 10 2 km)(sin 40.0°) 1.93 10 2 km, northd E d cos θd E (3.00 10 2 km)(cos 40.0°) 2.30 10 2 km, east11. What are the x and y components of avelocity vector of magnitude 100 km/hand direction of 240°?v xvv 100 km/h at 240°d60˚240˚d Nv x v cos θ (100 km/h) cos 240° –(100 km/h cos 60°) –50.0 km/hv y v sin θ (100 km/h) sin 240° –(100 km/h) sin 60° –86.6 km/hv x –50.0 km/h, v y –86.6 km/hv y12. You are a pilot on an aircraft carrier. Youmust fly to another aircraft carrier, now1450 km at 45° of your position, movingat 56 km/h due east. The wind isblowing from the south at 72 km/h.Calculate the heading and air speedneeded to reach the carrier 2.5 h after youtake off. Hint: Draw a displacementvector diagram.AN54.0˚45˚Aircraft carrier1450 kmt t 12Position A: plane leaving aircraftcarrier A, needing to reach carrier B2.5 hr after takeoffPosition B: aircraft carrier B, whichis 1450 km at 45° from carrier Amoving at v B 56 km/h, eastThe distance components (eastand north) of the aircraft carrierared E 1450 km cos 45° 1025 kmd N 1450 km sin 45° 1025 kmPosition C: The position of aircraftcarrier B in 2.5 hThe distance traveled by carrier B is(56 km/h)(2.5 h) 140 km, ESolving for the resultant by thecomponent method,R E 1025 km 140 km 1165 kmR N 1025 kmPlane140 km, ERBCD180 km, NECopyright © by Glencoe/McGraw-Hill48 Supplemental Problems Manual Physics: Principles and Problems

Copyright © by Glencoe/McGraw-HillChapter 4 (continued)12. (continued)tan θ 1 1 025km1165kmSo θ 1 41.3°R 1450 km, 45° 140 km, 0°So R 2 (1165 km) 2 (1025 km) 2R 1550 km, 41.3°The wind will carry the plane northfrom position D to position C, adistance of (72 km/h)(2.5 h) 180 kmduring the 2.5 h.Therefore, withd 1 distance the plane travelsd 1 180 km, N R 1550 km, 41.3°Comparing N and E components ofthe distance:d 1E 0 R E 1165 kmSo d 1E 1165 kmd 1N 180 km R N 1025 kmSo d 1N 845 km845kmtan θ 2 1 165kmSo θ 2 36°d 1 (1165 km) 2 (845 km) 2 1440 kmSo d 1 1440 km, 36.0°Heading 0.0° 36.0° 54.0° E of NAir speed 14 40km 580 km/h2.5 hThus, the airspeed needed to reachthe carrier 2.5 h after you take off is580 km/h with heading 54° E of N.13. An 80-N and a 60-N force actconcurrently on a point. Find themagnitude of the vector sum if the forcespulla. in the same direction.60 N80 N60 N 80 N 140 Nb. in opposite directions.80 N 60 N 20 Nc. at a right angle to each other.R (80 N) 2 +(60 N) 2 100 N14. One force of 60 N and a second of 30 Nact on an object at point P. Graphicallyadd the vectors and find the magnitudeof the resultant when the angle betweenthem is as follows.a. 0°60 N30 N 60 N 90 Nb. 30°x-component: 30 N 60 N cos 30° 82 Ny-component: 0 N 60 N sin 30° 30 NR (82 N) 2 (30 N) 2 87 Nc. 45°30 N30 N80 N 60 N60 N150˚135˚80 N60 N60 N30 Nx-component: 30 N 60 N cos 45° 72 Ny-component: 0 N 60 N sin 45° 42 NR (72 N) 2 (42 N) 2 83 NPhysics: Principles and Problems Supplemental Problems Manual 49

Chapter 4 (continued)14. (continued)d. 60°650 N500 NRtx-component: 30 N 60 N cos 60° 60 Ny-component: 0 N 60 N sin 60° 52 NR (60 N) 2 (52 N) 2 79 Ne. 90°R (60 N) 2 (30 N) 2 67 Nf. 180°30 N120˚30 N90˚60 N 30 N 30 N15. In tackling a running back from theopposing team, a defensive linemanexerts a force of 500 N at 180°, while alinebacker simultaneously applies a forceof 650 N at 270°. What is the resultantforce on the ball carrier?R 2 (650 N) 2 (500 N) 2R 820 Ntan θ 6 50N500Nθ 52°180° 52° 232°F 820 N, 232°60 N60 N60 N 30 N16. A water skier is towed by a speedboat.The skier moves to one side of the boat insuch a way that the tow rope forms anangle of 55° with the direction of theboat. The tension on the rope is 350 N.What would be the tension on the rope ifthe skier were directly behind the boat?F T (350 N)(cos 55°) 200 N17. Two 15-N forces act concurrently onpoint P. Find the magnitude of theirresultant when the angle between them isa. 0.0°15 N 15 N 30 Nb. 30.0°F h 15 N 15 N cos 30.0° 28 NF v 0 N 15 N sin 30.0° 7.5 NF (28 N) 2 (7.5 N) 2 29 Nc. 90.0°F (15 N) 2 (15 N) 2 21 Nd. 120.0°15 N120˚15 NF h 15 N (15 N) cos 120° 15 N 7.5 N 7.5 NF v (15 N) sin 120° 13 NF (7.5 N) 2 (13 N) 2 15 NCopyright © by Glencoe/McGraw-Hill50 Supplemental Problems Manual Physics: Principles and Problems

Chapter 4 (continued)17. (continued)e. 180.0°15 N 15 N 0 N18. Kim pushes a lawn spreader across a lawnby applying a force of 95 N along thehandle that makes an angle of 60.0° withthe horizontal.a. What are the horizontal and verticalcomponents of the force?F h F cos θ (95 N)(cos 60°) 48 NF v F sin θ (95 N)(sin 60°) 82 Nb. The handle is lowered so it makes anangle of 30.0° with the horizontal.Now what are the horizontal andvertical components of the force?F h F cos θ (95 N)(cos 30°) 82 NF v F sin θ (95 N)(sin 30°) 48 NCopyright © by Glencoe/McGraw-HillPhysics: Principles and Problems Supplemental Problems Manual 51