Chapter 3 MOTION IN A PLANE

Transcription

1 Chapter 3 MOTION IN A PLANE Conceptual Questions 1. No; to be equal the must also hae the same direction. If the magnitudes are different, the cannot be equal.. (a) Yes, since the direction matters. One of the ectors might be thirt degrees north of east and the other might be nearl equal in magnitude and thirt-fie degrees south of west. (b) No. The largest possible magnitude occurs when the two ectors point in the same direction. Then the magnitude of the sum equals the sum of the magnitudes. 3. A ector is a quantit that has both a magnitude and a direction associated with it. Velocit and displacement are both eamples of ector quantities. A scalar is a quantit that is defined b a single number (with unit) and no direction in space. Some scalars can be either positie or negatie, such as Celsius temperature, and others can onl be positie, such as speed and distance traeled. 4. Yes, it is possible for two different projectiles with identical initial speeds but different angles of eleation to land in the same spot. An object s range is proportional to its horizontal elocit and also its time of flight. With a small angle of eleation at firing the horizontal component of the elocit is greater than it is with a large angle of eleation, but the time of flight can be correspondingl less. The figure below shows the trajectories of two projectiles launched with an initial speed of 100 m s at angles 30 and The trajector is sometimes parabolic in another reference frame that moes with a constant elocit with respect to the first. The onl possibilit other than parabolic is a straight-line trajector. We on the ground see a ball released b a child on a coasting skateboard as moing in a parabola, but the child sees it as falling straight down. 6. After the first ball has reached its highest point and fallen back down to where it started, it will be moing downward with speed i just as it passes the height at which it was first thrown. From that point on, its trajector will be identical to the initial trajector of the second ball. The balls therefore reach the ground with the same speed, albeit at different times. 7. The two balls will cross at a height greater than one half of h. Prior to crossing, the bottom ball will be moing faster than the top ball, for the same time interal, and so will coer more distance. 8. Yes. A change in the elocit s direction is a change in the elocit. The path of an object can onl dierge from a straight line if the object accelerates. An object moing with constant elocit is not accelerating and therefore must trael in a straight line. 10

2 Phsics Chapter 3: Motion in a Plane 9. The aerage speed and the magnitude of the aerage elocit of an object are equal if and onl if the object traels along a straight-line path without changing direction. In all other cases, the aerage speed is greater than the magnitude of the aerage elocit because the total distance traeled must be greater than the straight-line distance between the starting and ending points. 10. An object traeling erticall downward in the graitational field of the Earth is accelerating in the same direction as its elocit. An object tossed erticall upward is accelerating in the direction opposite its elocit. A projectile traeling in a parabolic path under the influence of grait accelerates in a direction perpendicular to its elocit at the ape of its flight. 11. A car driing around a cure at a constant speed does not hae a constant elocit, because the direction is changing. 1. (a) A car driing around a cure at constant speed does not hae constant elocit because its elocit is changing in direction. (b) A car driing straight up an incline at constant speed has constant elocit, directed up the incline. (c) The Moon moes in a circular path around the Earth therefore has changing elocit. 13. (a) A ector of magnitude 1L ma be obtained b adding two ectors with lengths 3L and 4L aligned in opposite directions. (b) A ector of magnitude 7L ma be obtained b adding two ectors with lengths 3L and 4L aligned in the same direction. (c) A ector of magnitude 5L ma be obtained as the hpotenuse of a right triangle whose sides are ectors with magnitudes 3L and 4L. 14. The primar benefit of graphical ector addition is its use in proiding a isual understanding of the problem a feature that is often obscured in algebraic ector addition. Despite this benefit, adding ectors graphicall is a cumbersome and imprecise process, whereas the algebraic method is relatiel eas to perform and proides higher precision. These benefits make the two methods complement each other. A sketch is alwas good to epose sign errors associated with directions. 15. No single component of a ector can eer be greater than the magnitude of the ector. This is equialent to the statement that each side of a right triangle must be shorter than its hpotenuse a statement that can be erified using the Pthagorean theorem. 16. Neglecting air resistance, the trajector of a bullet that has eited the muzzle of a rifle is solel influenced b grait. The graitational force causes the bullet to accelerate downward toward the Earth but does not influence its horizontal motion. Thus, to hit a target, the muzzle must be aimed aboe the target b a distance equal to the amount that the bullet will fall in the course of its trael. If aimed at the target instead of aboe it, the bullet will miss low. 17. The demonstration works when the hunter is aiming either up or down at the coconut. In the absence of grait, either case will result in the arrial of a bullet at the position occupied b the coconut. Grait alters the ertical motion of the coconut and the bullet identicall in a gien time interal, both objects will fall an equal distance from the trajector the would hae followed in the absence of grait. Thus, either case concludes with the result that the bullet and coconut arrie at the same position. Multiple Choice Questions 1. (d). (b) 3. (b) 4. (d) 5. (a) 6. (c) 7. (e) 8. (c) 9. (a) 10. (d) 11. (b) 1. (a) 13. (b) 14. (c) 15. (c) 16. (e) 17. (d) 11

3 Chapter 3: Motion in a Plane Phsics Problems 1. Strateg Let east be the +-direction. Draw ector diagrams; then find the magnitudes and directions of the ectors. (a) (b) (.56 km 7.44 km) west km west (c) (.56 km 7.44 km) west 4.88 km west 4.88 km east (7.44 km.56 km) west 4.88 km west Discussion We will see more ariet when the ectors are not along the same line.. Strateg Let the +-direction be to the right. Draw ector diagrams; then find the magnitudes and directions of the ectors. (a) A B ( ) units in the -direction 0.73 units in the -direction (b) AB ( ) units in the -direction.73 units in the -direction (c) B A ( ) units in the -direction.73 units in the -direction 3. Strateg Use the properties of ectors to answer the questions. (a) The onl wa for the sum to hae a magnitude of 7.0 N is if the ectors are in the same direction. (b) Recognizing from 5 = that the three ectors form a right triangle, we know that the ectors are perpendicular. (c) The smallest magnitude sum can onl be obtained if the two ectors are in opposite directions; the magnitude of the smallest ector sum is

4 Phsics Chapter 3: Motion in a Plane 4. Strateg Use the definitions of distance and displacement, and the properties of circles. (a) Halfwa around the track is a distance equal to half of the circumference of the circular track. 300 m 150 m (b) The displacement is the straight-line location-change ector from the starting to the ending point of the run. Since the runner ran halfwa around the track, the magnitude of the displacement is equal to the diameter of the circular track. The runner ran from west to east, so the direction of the displacement is east. Find the diameter of the track. C d, so d C (300 m) 95 m. The runner s displacement is 95.4 m east. 5. Strateg Sketch the displacement ectors using graph paper, ruler, and protractor. Then find the ector sum b sketching a graphical addition of the displacement ectors. The combined sketches are shown. Each bo is 5 km on a side. The sum is 0 km in the + direction. Discussion If we look ahead, we can also use the component method. The first displacement is o ˆ o The ˆ second ector is o ˆ o (0 km)cos(60 ) (0 km)sin(60 ) is (0 km)cos(60 o ) ˆ0 ˆ 0 km in the + direction (0 km)cos(60 ) (0 km)sin(60 ) ˆ. Their sum 6. Strateg We interpret northeast to mean at 45 north of east. Draw a diagram of Orille s trip. Use it to find how far he walked during the second portion of the trip. The diagram is shown. Using a ruler and a protractor (and the scale of the diagram), we find that Orille s second leg was about 30 m at 40 north of west. 7. Strateg Look at how long each arrow is. Disregarding whether the point up or down, left or right, and how far erticall the go in comparison to how far horizontall, looking onl at the length in the scale picture, we see that B and C hae equal lengths and A is longer. In order of increasing magnitude, we hae B C, A. For confirmation we can use the theorem of Pthagoras to find the magnitude of each ector. A A 4 4 A B B ( 1) 3 B We see numericall the order B = C, A. C C 1 ( 3) C

5 Chapter 3: Motion in a Plane Phsics 8. Strateg Find graphicall the ectors D A B and E A C. Then, show graphicall that A BB A. Use graph paper, ruler, and protractor to find the magnitude and direction of the ector sum of the two forces in each case. (a) (b) 9. Strateg Graph the ectors and their sum. Use the scale of the graph to find the magnitude of the ector sum. The length of the ector sum is er nearl equal to seen times the edge length of a grid square, so the magnitude is 14 N. The ector points east, so the ector sum of the forces is 14 N to the east. (Note that F and the ector sum oerlap in the diagram. Their heads are in the same place.) Discussion You are proing how forces of magnitude 10 N, 4.5 N, and N can add to a force whose strength is 14 N. 10. Strateg Look at how long each arrow is. Disregarding whether the point up or down, left or right, and how far east the go in comparison to how far north or south, looking onl at the length in the scale picture, we see that D has the largest magnitude. Then E and then F. In order of increasing magnitude, we hae F, E, D. For confirmation we can use the Pthagorean theorem on the components of each ector. Their magnitudes are D D 3 4 D N 5 N E E ( 4) E N 5 N 4.5 N F F 0 F 4 N N In order of increasing magnitude, we see numericall F, E, D. 14

6 Phsics Chapter 3: Motion in a Plane 11. (a) Strateg Sketch the addition to find C. Use the fact that A B and smmetr to determine the direction of C. B smmetr, the horizontal components cancel when A and B are added, so C points downward;. The downward components of each ector hae the same magnitude, about 0.7 cm. So, the magnitude of C is about 1.4 cm. The sketch is shown: (b) Strateg Draw the subtraction of the second ector from the first. Use the fact that A B and smmetr to determine the direction of D. The ertical components cancel when B is subtracted from A. The direction of the horizontal component of B is reersed due to the subtraction, and so the ector resulting from the subtraction is in the direction of the horizontal component of A ; that is, to the left. The horizontal components of each ector hae the same magnitude, which is about 3.95 cm; so the magnitude of D is about 7.9 cm. The sketch is shown: The ector B is represented b the arrow pointing to the upper left. 1. Strateg Draw the displacement ectors. Use the diagram to answer the questions. (a) The magnitude of the displacement from Killarne to Cork is the hpotenuse of a right triangle with legs km and 39 km. r ( km) (39 km) 45 km (b) The distance along Michaela s chosen route is 39 km km 61 km, so the additional distance traeled is 61 km 45 km 16 km. 15

7 Chapter 3: Motion in a Plane Phsics 13. Strateg Use graph paper, ruler, and protractor to draw a diagram. scale: 1 km The scout troop must walk.0 km at 0 east of south to return to their starting point. Discussion The three ectors get the scouts back to their starting point for lunch, so the add to zero. We can sa that we hae figured out the negatie of the quantit 1. km east +.7 km at 45 west of north. 14. Strateg Sketch the displacement ectors using graph paper, ruler, and protractor. Then estimate the position b sketching a graphical addition of the displacement ectors. The combined sketches are shown. scale: 10 km The position is about 30 nautical miles at about 15 to 0 south of east. 15. Strateg Look just at how far to the right the head of each arrow is relatie to its tail. Going left ranks as less than zero. In order of increasing component, the ectors are B, C, A. We can do the ranking numericall b reading the -coordinate of the terminal point of each ector in bo-widths, starting from the initial point. We count A 4; B 1; C 1 So again the order of increasing -component is B, C, A. 16. Strateg Look just at how far up the head of each arrow is relatie to its tail. Going down ranks as less than zero. In order of increasing component, the ectors are E, F, D. We can do the ranking numericall b reading the -coordinate of the terminal point of each ector in bo-widths, starting from the initial point. We count D 4; E 4; F 0 So again the order of increasing -component is E, F, D. 17. Strateg Add the corresponding -components of each ector sum. Find the total -components. A B 4 ( 1) 3; B C 11 0; A C In order of increasing -component, we hae B C, A B, A C. Discussion. You can add to the picture to sketch in the additions of to, of to, and of to A B B C A C, using the boes to guide our counts. And then just counting how far each sum carries its arrow to the right gies ou the quantities that are to be ranked. 16

8 Phsics Chapter 3: Motion in a Plane 18. Strateg The ector makes an angle of 60.0 counterclockwise from the -ais. So, the angle counterclockwise from the positie -ais is = Find the components of the ector. -comp = (0.0 m) cos (150.0) 17.3 m and -comp = (0.0 m) sin (150.0) 10.0 m. 19. Strateg Let A be directed along the +-ais and let B be 60.0 CCW from A. Find the magnitude of A B. ( A B) A B cos and ( A B) A B 0 6.0sin , so A B units. 0. (a) Strateg Since each ector is directed along a different ais, each component of the ector sum is just equal to the ector that lies along that component s ais. Use the Pthagorean theorem. Find the magnitude of A B. A B [( A B) ] [( A B) ] ( 1.0) ( 3.0).0 units Find the direction of this ector in the second quadrant tan 60 CW from the ais, so 1.0 A B.0 units at 60 CW from the ais. (b) Strateg Subtract a ector b adding its opposite. Use the Pthagorean theorem. Find the magnitude. A B [( A B) ] [( ) ] 1.0 A B units Find the direction of this ector in the first quadrant tan 60 CCW from the ais, so A B.0 units at 60 CCW from the + ais. 1.0 (c) Strateg The ectors lie on the aes. Find the components of A, which lies in the third quadrant. B -comp B 1.0 unit and -comp A 3.0 units. 1. Strateg The components of a are gien. Since the -component is negatie and the -component is positie, the ector lies in the second quadrant. Gie the angle with respect to the ais to which it lies closest. Find the magnitude and direction of a. (a) a a ( 3.0 m s ) (4.0 m s ) 5.0 m s a (b) tan 53 CW from the ais or 37 CCW from the + ais 3.0 Discussion. Some people like to reduce routine tasks to routine. An arbitrar choice is to state all angles counterclockwise from the + ais. In this problem the direction angle would be 17. With its angle in this standard form, the component of a ector A is alwas gien b A cos and the component b A sin, with our calculator suppling just enough minus signs in the right places. If ou do it like this, leae the signs with the components wheneer ou compute tan 1 (A /A ). If the component is negatie, add 180 to the angle our calculator tells ou. And remember that turning b 360 gets ou back to our original direction, so 40 is the same as

9 Chapter 3: Motion in a Plane Phsics. Strateg Use the fact that A B, smmetr, and the component method to find the magnitude of C. B smmetr, C points downward, since the horizontal components cancel when A and B are added. The downward components of each ector are the same, A B (4.0 cm)sin cm. So, the magnitude of C is 1.4 cm. 3. Strateg Use the fact that A B, smmetr, and the component method to find the magnitude of D. The ertical components cancel when B is subtracted from A. The direction of the horizontal component of B is reersed due to the subtraction, and so the ector resulting from the subtraction is in the direction of the horizontal component of A; that is, to the left. The horizontal components of both of the ectors we combine are the same, A B (4.0 cm) cos cm. so the magnitude of D is 7.9 cm. Discussion If ou like just writing math, ou can write D A B 4 cm cos(190 ) ˆ 4 cm sin(190 ) ˆ [4 cm cos( 10 ) ˆ 4 cm sin( 10 ) ˆ ] = 3.94 cm ˆ cm ˆ [+3.94 cm ˆ cm ˆ ] = 3.94 cm ˆ cm ˆ 3.94 cm ˆ cm ˆ 7.9 cm ˆ 7.9 cm at Strateg We choose to determine and use the angle each ector makes with the positie -ais. Find the components of each ector. Vector A : A A (7.0 m)cos m (7.0 m)sin m Vector C : C C (7.0 m) cos m (7.0 m) sin m Vector B : B B (7.0 m s)cos( 0.0 ) 6.6 m s (7.0 m s)sin( 0.0 ).4 m s Vector D : D D (7.0 m s)cos(50 ).4 m s (7.0 m s)sin(50 ) 6.6 m s 5. Strateg and Multipling a ector b a scalar is equialent to multipling the ector s components b that scalar alue. Multipling a ector b a positie scalar other than 1 changes the magnitude of the ector and its components but not the direction. Therefore, doubling the magnitude of the ector doubles both components, without changing their signs. Discussion With B = B at = B ˆ B ˆ we can write B = (B) at = ( B ˆ B ˆ ) B ˆ B ˆ. The direction of the ector is unchanged while the arrow representing it gets twice as long. 6. Strateg Take the ector B in the tetbook diagram for problems 4 through 6. It is in the fourth quadrant. Reersing the sign of the -component results in both the - and the -component being negatie. The resulting ector lies in the third quadrant, 0.0 below the negatie -ais. The sketch is shown. B 18

10 Phsics Chapter 3: Motion in a Plane 7. Strateg The components of are gien. Since the -component is positie and the -component is negatie, the ector lies in the fourth quadrant. Gie the angle with respect to the aes. Find the magnitude and direction of. (a) (16.4 m s) ( 6.3 m s) 31.0 m s (b) The angle with the ais is = tan 1 ( / ) tan 58.1 with the + -ais and 31.9 with the -ais The 16.4 negatie angle with the ais b conention means 58.1 clockwise rather than counterclockwise. 8. Strateg Use the Pthagorean theorem to find the magnitude of each ector. Gie the angle with respect to the ais to which it lies closest. Find the magnitude and direction of each ector. (a) (b) (c) A ( 5.0 m s) (8.0 m s) 9.4 m s and tan 58 The component is negatie, so the angle is CCW from the + -ais. 5.0 B (10 m) ( 60.0 m) 130 m and tan 7 so the direction is 7 CW from the + -ais or 333 CCW from the + -ais 10 C ( 13.7 m s) ( 8.8 m s) 16.3 m s and tan 33. The component is negatie, so CCW from the -ais (d) D (.3 m s ) (6.510 m s ).3 m s and tan 1.6 CCW from the + -ais Strateg The ector makes an angle of 50.0 (because ) clockwise from the -ais. We can use this angle in determining the components. Find the components of the ector. A Acos (. cm) cos cm and A Asin (. cm)sin cm. Discussion Or we can use the gien angle, presuming it is measured counterclockwise from the positie ais: A A Acos (. cm) cos cm and Asin (. cm)sin cm. The diagram gies a useful check on the signs for this ector in the second quadrant. 19

11 Chapter 3: Motion in a Plane Phsics 30. (a) Strateg Since the angle is below the +-ais, it is negatie. Compute the components. B 7.1cos( 14 ) 6.9 and B 7.1sin( 14 ) 1.7. (b) Strateg The components of C are gien. Use the Pthagorean theorem. Compute the magnitude and direction of C. C C C ( 1.8) ( 6.7) tan 75 The component is negatie, so CCW from the -ais 1.8 (c) Strateg Add the components of the ectors to find the components of the ector sum. Use the Pthagorean theorem. Gie the angle with respect to a specified ais. Find the magnitude and direction of C B. C B ( C ) ( ) ( ) ( ) B C B 9.8 and tan CW from the + ais or 31 CCW from the -ais. 5.1 (d) Strateg Use the components of Cand B to find those of C B. Compute the magnitude and direction. C B ( C B ) ( C B ) ( ) [ 6.7 ( 1.7)] tan 30. The component is negatie, so CCW from the -ais 8.7 (e) Strateg Use the components of Cand B to find those of C B. Compute the components. -comp C B and -comp C B Strateg Margaret s total displacement is the ector sum of the three displacements along her path. Add the displacements mi west 0.00 mi north mi east mi west 0.00 mi north mi west 0.00 mi west 0.00 mi north Let north be along the +-ais and west be along the -ais. Then, the components of the total displacement are 0.00 mi and 0.00 mi. Find the displacement magnitude. r ( ) ( ) ( 0.00 mi) (0.00 mi) 0.83 mi Find the direction. tan The component is negatie, so CCW from the -ais 45.0 N of W 0

12 Phsics Chapter 3: Motion in a Plane 3. Strateg Add the displacement from Cind s apartment to Jerr s dorm to the displacement from Jerr s dorm to the fitness center to find the total displacement from Cind s apartment to the fitness center. Add the displacements mi east.00 mi north 3.00 mi east 4.50 mi east.00 mi north Let north be along the +-ais and east be along the +-ais. Then, the components of the total displacement are 4.50 mi and.00 mi. Find the magnitude. r ( ) ( ) (4.50 mi) (.00 mi) 4.9 mi Find the direction tan 4.0 north of east Strateg Let east be the +-direction and north be the +-direction. Compute the net displacement of the scouts. r1 1. km r1 0 r (.7 km) cos km r (.7 km) sin km r r1 r 1. km 1.9 km 0.7 km r r1 r 1.9 km r ( r ) ( ) ( 0.7 km) (1.9 km) r.0 km The direction of the return trip is opposite the displacement ector found. r 0.7 km and r 1.9 km tan CW from east = 70 south of east 0.7 So, the must trael.0 km at 70 south of east. Discussion The three displacements get the scouts back to their starting point for lunch, so the add to zero. We can sa that we hae soled the ector equation r1 r r ' 0 for r ' ( r1 r ). 34. Strateg Draw a diagram and use the component method. Let north be + and east be +. The diagram is shown (with the answer included). Find the position of the sailboat. 45 n.m. (0.0 n.m.) cos n.m. (10.0 n.m.) cos 60 6 n.m. 8 n.m. 0 (0.0 n.m.) sin 300 (10.0 n.m.) sin n.m. r ( ) ( ) (8 n.m.) ( 8.7 n.m.) 9 n.m tan 17 so 17 clockwise from the ais = 343 CCW from 17 south of east 8 So, r 9 nautical miles at 17 south of east. 1

13 Chapter 3: Motion in a Plane Phsics 35. Strateg Use the component method to find the displacement from the starting point. Compute the total displacement. Let north be + and east be mi cos 90 (. mi) cos 55 (1.1 mi) cos15.3 mi 1.6 mi sin 90 (. mi) sin 55 (1.1 mi) sin mi r ( ) ( ) (.3 mi) (3.7 mi) 4.4 mi tan 58 north of east.3 So, r 4.4 miles at 58 north of east. 36. Strateg Draw diagrams of the situation. Use the definitions of aerage speed and aerage elocit. (a) Find the runner s aerage speed mi 1609 m 1 min a t 4.00 min mi 60 s 5.03 m s (b) Find the location of the runner on the track , so the runner has gone around once plus times Find the angle shown in the diagram Find the radius of the track. C C r, so r. Find r. i f f r ( r r cos ) ( r sin ) ( r r cos ) ( r sin ) r (1 cos ) (sin ) C mi r 1 cos cos sin r 1 cos 1 (1 cos ) (1 cos 4.84 ) mi 1 r sin 1 sin 4.84 tan tan 1.4 r r cos 1 cos 4.84 Find the runner s aerage elocit. r mi 1609 m 1 min a m s, so a m s at 1.4 west of north. t 4.00 min mi 60 s 37. Strateg Use the definitions of aerage speed and aerage elocit. (a) Find the runner s aerage speed mi 1609 m 1 min a t 4.0 min mi 60 s 6.7 m s (b) Find the location of the runner on the track , so the runner has gone around.0 times The runner has gone around two complete times, so his displacement is zero; therefore, a 0. Discussion The aerage speed has no direction because it is defined to be a scalar. The aerage elocit has no unique direction because its magnitude happens to be zero.

14 Phsics Chapter 3: Motion in a Plane 38. (a) Strateg Draw the position ectors with respect to Illium. (b) Strateg Use the component method. r rf r i; f i rf cosf ri cos i; f i rf sinf ri sini Find the magnitude of the displacement. r ( ) ( ) [(7. km)cos195 (73.6 km)cos 45 ] [(7. km)sin195 (73.6 km)sin 45 ] 59.9 km Find the direction of the displacement. 1 (7. km) sin195 (73.6 km) sin 45 tan 85 north of east (7. km) cos195 (73.6 km) cos 45 So, r 59.9 km at 85 north of east. (c) Strateg Use the definition of aerage elocit. r 59.9 km at 85 north of east a 80 km h at 85 north of east t (45 min) 1 h 60 min 39. (a) Strateg Find the distance traeled during the first part of the trip; then compute the speed required to trael the remaining distance in the time required. There is 48.0 min = h left to complete the trip. The harpsichordist has traeled (55.0 mi h)(1.0 h) 66.0 mi, so he has 1 mi 66 mi 56 mi to go. To get to the concert on time, he must trael at a speed of (56 mi) (0.800 h) 70 mi h. (b) Strateg Use the definition of aerage elocit. Use components for the displacements. Let the +-direction be west and the +-direction be south. Then, 66.0 mi (56 mi) cos mi and (56 mi) sin mi. (114.5 mi) (8 mi) The magnitude of the aerage elocit is 59 mi h..00 h 1 8 The direction of the aerage elocit is tan So, the aerage elocit for the entire trip was 59 mi h at 14 south of west. 3

15 Chapter 3: Motion in a Plane Phsics 40. Strateg Use the definition of aerage elocit. Draw a diagram. Let east be in the +-direction and north be in the +-direction. Find the magnitude of r. r [3. km (4.8 km) cos km] [(4.8 km) sin 75.0 ] 8.9 km Find the direction of r km tan 31 north of east 7.6 km r 8.94 km r So, a 6 km h and a 6 km h at 31 north of east. t 0.10 h 0.15 h 0.10 h t 41. (a) Strateg Find the aerage speed b diiding the total distance traeled b the total time. Since the time traeled at each speed is the same, we can simpl add the speeds and diide b km h 18 km h 4 km h a 110 km h.0.0 (b) Strateg Use the definition of aerage elocit. Draw a diagram. Let east be the +-direction and north be the +-direction. Find the magnitude of r. r 96 km h (18 km h)cos60 (1.0 h) (18 km h)sin60 (1.0 h) 195 km Find the direction of r km tan 35 north of east 160 km r km r So, a 97 km h and a 97 km h at 35 north of east. t.0 h t Discussion The ector addition diagram shows displacements, not elocities. But addition for an kind of ector, including elocit, can be done using the same method as for displacements. 4

16 Phsics Chapter 3: Motion in a Plane 4. (a) Strateg Find the aerage speed b diiding the total distance traeled b the total time. Each distance is gien b the product of the speed and time. distance (108 km h)(0.0 min) (90.0 km h)(10.0 min) a 10 km h time 0.0 min 10.0 min (b) Strateg Use the definition of aerage elocit. Draw a diagram. Compute the distance of each leg of the trip, then draw the diagram. 1 h 1 h (108 km h)(0.0 min) 36.0 km and (90.0 km h)(10.0 min) 15.0 km. 60 min 60 min Find r. Let east be in the +-direction and north be in the +-direction. r [ 36.0 km (15.0 km) cos 40.0 ] [(15.0 km) sin 40.0 ] ( 43.5 km) ( 13.0 km) 45.4 km tan 16.6 so 16.6south of west CCW from east 43.5 r 45.4 km So, a 90.8 km h and t 1 h (10.0 min 0.0 min) 60 min r a 90.8 km h at 16.6 south of west. t 43. (a) Strateg From Problem 1, the distance Michaela traeled between Killarne to Cork ia Mallow was 61 km. Find Michaela s aerage speed. distance traeled 61 km 1000 m 1 min total time 48 min 1 km 60 s 1 m s (b) Strateg From Problem 4, the magnitude of Michaela s displacement was 45 km. Find the magnitude of Michaela s aerage elocit. r 45 km 1000 m 1 min a 16 m s t 48 min 1 km 60 s 5

17 Chapter 3: Motion in a Plane Phsics 44. Strateg Draw diagrams. Use the definitions of aerage speed and aerage elocit. (a) Find the displacement from his home to the second friend. r r 1+ r (90.0 km h)(80.0 min)[1 h (60 min)] east (76.0 km h)(45.0 min)[1 h (60 min)] south of west 10 km east 57.0 km at 30.0 south of west Find the distance. r [10 km (57.0 km)cos30.0 ] [ (57.0 km)sin 30.0 ] 76. km (b) A third displacement of r brings him home oer the same distance. Find the magnitude of the aerage elocit on this return trip. r 76. km 60 min a 10 km h t 45.0 min 1 h Find the direction. 1 (57.0 km) sin 30.0 tan.0 so CCW from east =.0 north of west 10 km (57.0 km)cos30.0 The aerage elocit on the third leg is 10 km h at.0 north of west. (c) Compute the time min 15.0 min 45.0 min min Find the magnitude of the aerage elocit. r km 60 min a 3.6 km h t min 1 h The direction is opposite that found in part (b), so the aerage elocit during the first two legs is 3.6 km h at.0 south of east. (d) Since the displacement is zero, the aerage elocit oer the entire trip is 0. (e) Compute the total time min 15.0 min 45.0 min 45.0 min 55.0 min 40.0 min, or h Compute the total distance. 10 km 57.0 km 76. km 53. km The aerage speed during the entire trip is 53. km 63.3 km h h 6

18 Phsics Chapter 3: Motion in a Plane 45. Strateg Use the definition of aerage acceleration. Take south as the positie direction. f a i t t 5.0 m s south.0 m s north 5.0 m s south.0 m s south 10.0 s 10.0 s 0.70 m s south Discussion Students are sometimes told that there is acceleration that shows up as speed increasing, acceleration that shows up as speed decreasing, and acceleration that shows up as direction changing. That can be instructie, but it is also true that all acceleration shows up as ector elocit changing. In the hawk s motion speed increase and direction change happen together. 46. Strateg Use the definition of aerage acceleration. Take up as the positie direction. f i 8.3 m s down 55 m s down a 13 m s down 13 m s up. t t 3.5 s 47. (a) Strateg Use the definition of aerage elocit. Draw a diagram. Let the center of the circle be the origin, then r rf r i 0.0 m east 0.0 m south. r (0.0 m) ( 0.0 m) 8.3 m Let east be the +-direction and north the +-direction tan 45.0 north of east 0.0 r 8.3 m r So, a 3.33 m s and a 3.33m s at 45 north of east. t 8.50 s t (b) Strateg Use the definition of aerage acceleration and the fact that C r. Find the aerage acceleration of the car. r(3 4) 3 r f i 3 r f i, so aa (south west). t t t t ( t) 3r 3r a a 1 ( 1) ( t) ( t) 1 1 tan 45 so 45 south of east 1 3 (0.0 m) a a at 45 south of east 1.84 m s at 45 south of east t (8.50 s) (c) Strateg Consider the definitions of acceleration and elocit. Although the magnitude of the elocit is constant, its direction must change continuousl for the car to trael in a circle; changing the direction of the elocit requires an acceleration. Discussion We drew one diagram. It might be clearer to draw that diagram separatel without the elocit ectors to illustrate the change in position and how the aerage elocit is to the northeast ; and a separate diagram showing the elocit ectors starting from the same origin i f to reeal that the change in elocit is to the southeast. 7

19 Chapter 3: Motion in a Plane Phsics 48. Strateg The magnitude of the elocit is constant, but the direction changes. Recall that the circumference of a circle is gien b r. (a) Find the car s speed. C 4 C r r (10.0 m) t 4t 4t t (1.60 s) 9.8 m s (b) Let east be the +-direction and north the +-direction. f i east north east + south ( ) tan 1 tan 1( 1) south of east = due southeast (9.8 m s) southeast 13.9 m s southeast (c) a a m s southeast t 1.60 s 8.68 m s southeast 49. (a) Strateg Let east be in the +-direction and north be in the +-direction. See the figure. (b) Strateg Use the component method to subtract the initial elocit ector from the final elocit ector. f i 40 km h NW 19 km h N, so (40 km h)cos135 0 (40 km h)sin13519 km h 170 km h km h (40 km h)sin135 tan 7 Both components are negatie, so 187 = 7 south of west (40 km h)cos135 So, 170 km h at 7 south of west. (c) Strateg Use the definition of aerage acceleration. 170 km h at 7 south of west a a 57 km h at 7 south of west t 3.0 h Discussion In the diagram the ector change runs from the head of i to the head of f. If ou isualize the plane as slowl swinging around an arc of a circle, turning to the left, the acceleration points toward the center of the circle. 8

20 Phsics Chapter 3: Motion in a Plane 50. (a) Strateg Draw a diagram showing the initial and final elocities. Use the component method to subtract the initial elocit ector from the final elocit ector. Let east be in the +-direction and north be in the +direction. f i 100 km h SE 90 km h W 100 km h SE 90 km h E, so (100 km h)cos km h (100 km h)sin km h. 1 (100 km h)sin315 tan 4 so 4 south of east, (100 km h)cos31590 km h and 180 km h at 4 south of east. (b) Strateg Find the total time of trael. Then use the definition of aerage acceleration. 16 km 8.0 km 34 km t 0.6 h, so 90 km h 80 km h 100 km h 176 km h at 4 south of east a a 80 km h at 4 south of east. t 0.6 h 51. Strateg Since the particle is moing to the east and is accelerated to the south, its elocit in 8.00 s will be between east and south. Use the component method. Let north be in the +-direction and east be in the +direction m s and (.50 m s at )(8.00 s) 0.0 m s. (40.0 m s) ( 0.0 m s) 44.7 m s m s tan tan 6.6 so 6.6 south of east 40.0 m s So, 44.7 m s at 6.6 south of east. Discussion The trajector of the particle is a parabola opening toward the south. As it goes around part of this cure, its acceleration ector is pointing toward the center of curature, which is the center of a circle that osculates with this section of the parabola. 9

21 Chapter 3: Motion in a Plane Phsics 5. Strateg Use the component method. Sole for the time. Let north be in the +-direction and east be in the +direction. 60 m s and a t. Use the Pthagorean theorem. (100 m s) (60 m s) a t t a 100 m s 53. Strateg Think about ertical component of elocit. ( ), so 0.80 s. The time in flight depends just on the initial ertical component of elocit, because the constant graitational acceleration takes a predictable time to reduce this i to zero, and then an equal time to bring the projectile down again. The projectiles all hae the same initial elocit magnitude, so the more steepl launched ones hae more ertical elocit and a longer time of flight. In order in increasing flight time the are 15, 30, 45, 60, 75. Discussion Can ou find an animated ersion of this diagram on the web or make one ourself? It is fun to throw two snowballs at an opponent, the first at a high angle to get his attention and, after the right dela time, the second at a low angle to hit him without warning at almost the same time. 54. Strateg Use equations of motion with constant acceleration to determine the ertical and horizontal positions of the baseball after 1.40 s hae elapsed. Find the position of the ball. f it (30.0 m s)(1.40 s) 4.0 m 1 1 f i i t g( t) 9.60 m 0 (9.80 m s )(1.40 s) 0.00 m After 1.40 s hae elapsed, the ball is on the ground at a horizontal distance of 4.0 m from the launch point. 55. Strateg Use equations of motion with constant acceleration to determine the ertical and horizontal positions of the cla after 1.50 s hae elapsed. Find the position of the cla assuming it is in free fall for the whole time interal. f it (0.0 m s)(1.50 s) 30.0 m 1 1 f i i t g( t) 8.50 m 0 (9.80 m s )(1.50 s).53 m The cla cannot pass through the ground, so it hit and stuck prior to 1.50 s. Find the time it took for the cla to land. 1 1 f 0 i i t g( t) i g( t), so i (8.50 m) t s, and f it (0.0 m s)(1.317 s) 6.3 m. g 9.80 m s The cla hits and it is on the ground after 1.3 s, so the horizontal distance along the ground, from straight under its starting point, is 6.3 m. Discussion Keep our assumptions in mind. In man projectile-motion problems we deliberatel take the initial point just after the object is fired and the final point just before it hits, so that the acceleration is constant and known eerwhere in between. 30

22 Phsics Chapter 3: Motion in a Plane 56. (a) Strateg Use equations of motion with constant acceleration to determine the ertical and horizontal positions of the tennis ball after 1.60 s hae elapsed. Find the position of the ball, assuming it is still in flight. f it (0.0 m s)(1.60 s) 3.0 m 1 1 f i i t g( t) 14.0 m 0 (9.80 m s )(1.60 s) 14.0 m 1.5 m 1.5 m After 1.60 s hae elapsed, the ball has fallen 1.5 m erticall and has traeled 3.0 m horizontall. (b) Strateg The ball is still in the air. Set the final position equal to zero in Eq. (3-1) to find the time when the ball hits the ground. Find the elapsed time. 1 1 i (14.0 m) f 0 i it g( t) i g( t), so t 1.69 s. g 9.80 m s Find the range to the landing position. f i t (0.0 m s)(1.69 s) 33.8 m The ball will land after another 0.09 s and will then be at a horizontal distance of 33.8 m. 57. Strateg Use Eqs. (3-) and (3-3). Set f 0, since the ertical component of the elocit is zero at the maimum height. (a) Find the maimum height. i f i i f i 0 g, so and g i f i sin (19.6 m s) sin m 5.9 m. g (9.80 m s ) (b) At the ball s highest point, f 0, so the speed equals. i i cos 19.6 m s cos m s Discussion The choice of initial and final points is up to us, so long as the acceleration is constant eerwhere between them and the final point is after the initial one. Here in both parts (a) and (b) we choose initial just after release and final at the top of the flight. 58. (a) Strateg Use Eqs. (3-3) and (3-19). Find the components of the elocit. i i cos 0.0 m s cos m s i sin gt (0.0 m s)sin60.0 (9.80 m s )(3.0 s) 1 m s (b) Strateg Use Eq. (3-0 and 3-1). Calculate the -component of the displacement. t 10.0 m s (3.0 s) 30 m Calculate the -component of the displacement. 1 1 it g( t) 0.0 m ssin60.0 (3.0 s) (9.80 m s )(3.0 s) 8 m 31

23 Chapter 3: Motion in a Plane Phsics 59. Strateg Use Equations 3-1 and 3-0. For the horizontal motion we hae t i cos(35 ) t 15.0 m. This equation contains two unknowns. We can find t, the one quantit in common between horizontal and ertical motions, b using for the whole jump from (just after) takeoff to (just before) landing 1 1 sin 35 diide b t to obtain t i Now b substitution g it g( t) : 0 isin(35 ) t (9.80 m s )( t) Now t is not zero for a real jump, so we can sin 35 i i i icos(35 ) t i cos(35 ) sin(35 ) cos(35 ) sin(70 ) 15.0 m g g g And g 15.0 m(9.80 m/s ) i sin( i ) sin m/s 60. Strateg Sole t for the time and substitute the result into Eq. (3-1). Then, sole for to find the required distance from the cannon. t ( icos ) t, so t ( icos ). Substitute. 1 f i i t ( ) a t 1 ( ) g( ) ( i sin ) g tan, so i cos i cos i cos g 0 ( ) (tan ). cos i Use the quadratic formula. g (9.80 m s )( 5.0 m) tan tan 4 tan 35.0 tan 35.0 i cos (18.0 m s) cos m or 6.0 m g 9.80 m s (18.0 m s) cos 35.0 i cos Since the cannon won t fire backward, 6.0 m is etraneous. So, ou tell the ringmaster to place the net such that its center is 37.1 m in front of the cannon, and to do a rehearsal with a dumm. 3

24 Phsics Chapter 3: Motion in a Plane 61. (a) Strateg At the maimum height of the cannonball s trajector, f 0. Use Eq. (3-). Find the maimum height reached b the cannonball. f i 0 ( i sin ) a g( f i ), so i f i sin (40 m s) sin m 37 m. g (9.80 m s ) (b) Strateg Sole t for the time and substitute the result into Eq. (3-19). Then, sole for to find the horizontal distance from the release point. When the cannonball reaches the ground, 7.0 m. t ( i cos ) t, so t. Substitute. i cos 1 1 ( ) g( ) f i it a t i g i cos i cos i cos ( ) ( sin ) tan, so g 0 ( ) (tan ). Use the quadratic formula. cos i 4g (9.80 m s )( 7.0 m) i cos (40 m s) cos 37 g 9.80 m s i cos (40 m s) cos37 tan tan tan 37 tan m or 9 m Since the catapult doesn t fire backward, release point. 9 m is unphsical. So, the cannonball lands 170 m from its (c) Strateg The -component is the same as the initial alue. Find the -component using Eq. (3-). The -component of the elocit is f i i cos (40 m s)cos37 3 m s. Find the -component of the elocit. f i f i sin a g, so f i g where the negatie sign was chosen because the cannonball is on its wa down. sin (40 m s) sin 37 (9.80 m s )( 7.0 m) 7 m s, Discussion For some students the hardest part is the first step, choosing the final point at the top of the flight rather than at reaching the ground, to find the maimum altitude. A trick is to do part (c) before part (b): After ou know the ertical impact elocit component, ou can find the time of flight from f i = gt. And then the horizontal range follows from t. You hae to watch the minus signs with an method. 33

25 Chapter 3: Motion in a Plane Phsics 6. Strateg The minimum speed occurs when the spit goes no higher than is necessar in other words, when the grasshopper is struck at the ape of the trajector. Find i using Eq. (3-). f i a, so i f a f ( g) 0 (9.8 m s )(0.55 m) 3.08 m s. The time interal is found using Eq. (3-0). The spit needs to trael 0.00 m horizontall while it rises to the ape, so use t. Find i. 1 ( f i) t, so t. Then, f i ( f i) (0.00 m)( m s) m s. t ( f i) (0.55 m) Compute the minimum initial speed of the spit, and the necessar angle aboe the horizontal. i i 1 i m s i (0.611m s) (3.08 m s) 3.7 m s ; tan tan 79. i m s 63. Strateg Sole t for the time and substitute the result into Eq. (3-1). Then, sole for i. t ( i cos ) t, so t. Substitute. i cos 1 1 ( ) g( ) it a( t) ( isin ) g tan, so i cos i cos i cos g( ) (9.80 m s )(50 m) i cos ( tan ) cos 53 [(50 m) tan 53 ( 1 m)] 64. (a) Strateg Use Equations 3-0 and m s. For the flight from pod to earth 1 i t ( ) gies 1.1 m = ( 9.8 m/s )( ) so. m s / 9.8 m s a t t t Now the horizontal range is t (1. m/s)(0.474 s) 0.57 m (b) Strateg If we used the same pair of equations we would hae to sole a quadratic. For ariet let us use Equations 3- and 3-19 along with 3-0. The ertical component of impact speed is gien b f i f i a, so a (1. m/s) sin (17 ) ( 9.8 m s )( 1.1 m) 4.68 m s. Now the time of flight follows from f i sini 4.68 m/s (1. m/s)sin17 f i at as t s g 9.8 m/s And the horizontal range is t (1. m/s)cos(17 )(0.513 s) 0.59 m (c) Strateg and solution The measured 0.44 m is significantl less than 0.59 m. Air resistance is not negligible, but has slowed the flight of the seed. The seed ma be shaped like a feather, potato chip, or parachute. 34

26 Phsics Chapter 3: Motion in a Plane 65. (a) Strateg Sole for the time when the projectile is back at ground leel, using Eq. (3-1) i i sin 0 i t a ( t) i t g( t) i gt, so t. g g (b) Strateg Use t i t and the result of part (a) to find the range. Use the trigonometric identit sin sin cos. isin i sin( ) it becomes R i cos. g g (c) Strateg and The maimum alue of sin occurs when 90 or 45. Therefore, R ma i sin 90 i g g. Discussion Launching at twice the speed gies a range four times larger this is reasonable because the faster projectile spends more time in flight and traels farther horizontall in each second of that flight. Launching at 45 gies a greater range than other angles, as the chapter tet eplains and diagrams. If ou go to the Moon, with about one-sith the surface graitational acceleration, the range will be si times larger. Alan Shepard hit golf balls on the Moon to demonstrate the effect. Do not tr to memorize the range formula, because it is too specialized compared to the set of standard equations for constant-acceleration motion. 66. Strateg Use the epression for the range. (a) The maimum alue of sin occurs when 90. Therefore, R ma i sin 90 i (1) i g g g (b) The maimum alue of sin occurs when 90 or (a) Strateg At a projectile s highest point, the ertical component of its elocit is zero. Using Eqs. (3-3), we hae i cos and 0. (b) Strateg Use Eq. (3-19). i g t isin g t 0, so t i sin g. (c) Strateg Use Eq. (3-1) and the result from part (b). Find H. 1 1 H it a ( t) it g( t), so i sin 1 i sin i sin i sin ( i sin ) H i sin g. g g g g g. 35

27 Chapter 3: Motion in a Plane Phsics 68. (a) Strateg At the maimum height, f 0. Use Eq. (3-19) to find the time it takes the ball to reach its maimum height. i sin f i 0 i sin gt, so t. g Use Eq. (3-1) to find how much the ball rises. 1 i sin 1 i sin i sin i sin ( isin ) t g( t) isin g g g g g i sin (.0 m s) sin m higher than where it was hit g (9.80 m s ) (b) Strateg The elapsed time is twice that found in part (a). i t sin g.0 m s sin60.0 (9.80 m s ) 3.89 s (c) Strateg Use Eq. (3-5). Find the horizontal displacement of the ball. i i t ( cos ) t.0 m s cos60.0 (3.89 s) 4.8 m 69. Strateg In each case, use Eq. (3-1) to find the time it takes for the stone to reach the base of the gorge. (a) ( m) it a( t) 0 g( t) g( t), so t 3.49 s. g 9.83 m s (b) (c) 1 1 it a ( t) it g( t), so g( t) it (9.83 m s )( t) 0.0 m st 60.0 m. Sole for t using the quadratic formula. 0.0 m s (0.0 m s) (9.83 m s )( 60.0 m) t.01 s or 6.08 s 9.83 m s Since t 0, t.01 s. 1 1 ( isin ) t g( t), so 0 g( t) ( isin ) t. Sole for t using the quadratic formula. 1 isin i sin 4 g t g 0.0 m s sin m s sin 30.0 (9.83 m s )( 60.0 m) 4.66 s or.6 s 9.83 m s Since t 0, t 4.66 s. Find the horizontal distance. i ( cos ) t 0.0 m s cos30.0 (4.656 s) 80.6 m Discussion When ou are using the quadratic formula, the square root is actuall the ertical component of final elocit, and the rest of the equation is Equation 3-19 soled for the time. Coincidence? Mabe not! 36

28 Phsics Chapter 3: Motion in a Plane 70. Strateg The circus performer is moing horizontall as he clears the net, so at that moment, his ertical component of elocit is zero. His horizontal component of elocit is constant. Use Eqs. (3-19) and (3-0). Find the time that it takes the performer to reach the net. t i cos t, so t. i cos Find the muzzle speed of the cannon. g g (9.80 m s )(6.0 m) f 0 i gt i sin, so i 11 m s. i cos sin cos sin 40cos 40 Find the height of the net h ( f i) t i sint i sin tan tan 40 (6.0 m).5 m i cos 71. Strateg f 0 at the maimum height. Use Eqs. (3-19) and (3-1). For the upward flight to maimum height i i f i sin gt i sin 45 gt gt 0. Thus, t. g Find H ma, the maimum height of the projectile s trajector. 1 i i 1 i i i H i ma ( i sin ) t g( t) g g g g 4g 4g Find R, the range of the projectile. We use 1 ( isin ) t g( t) and ( icos ) t. When the projectile returns to its original height, isin ( isin ) t g( t) 0 so isin gt 0, and t. g Substitute this alue for t into R ( cos ) t. isin i sin cos R ( icos ) t icos g g Substitute for. i i sin 45cos 45 R i Therefore, H g g ma 7. Strateg Consider the relatie motion of the two ehicles. 1 i R. 4 g 4 Let north be in the +-direction. JR the elocit of the Jeep relatie to the road 8 km h RF the elocit of the road relatie to the Ford FR 48 km h JF the elocit of the Jeep relatie to the (obserer in the) Ford JR RF 8 km h 48 km h 130 km h So, 130 km h north. JF 37

29 Chapter 3: Motion in a Plane Phsics 73. Strateg Consider the relatie motion of the two ehicles. We choose to find BV the elocit of the BMW relatie to the VW. Let north be in the +-direction. BR the elocit of the BMW relatie to the road km h RV the elocit of the road relatie to the VW VR 4 km h BV BR RV km h 4 km h Find t km 3600 s t 54 s BV km h 4 km h 1 h Discussion The elocit of the road relatie to a car is a perfectl obserable quantit: it is the elocit of the world outside seen b an obserer in the car. We choose to minimize the number of minus signs to worr about b using ersion a of Equation Strateg Consider the relatie motion of the two ehicles. Draw a diagram and use the component method. Let north be in the +-direction and east the +-direction. tc the elocit of the truck relatie to the car tr the elocit of the truck relatie to the road cr the elocit of the car relatie to the road Find the elocit of the truck relatie to the car. tc tr rc tr ( cr ) tr cr tc tr cr tr cos15 0 (85 km h) cos15 tc tr cr tr sin15 cr (85 km h)sin km h tc (85 km h)cos15 (85 km h)sin15110 km h 63 km h 1(85 km h)sin15110 km h o tan 40 both components are negatie so is 40 south of (85 km h)cos15 So, the relatie elocit is 63 km h at 40 south of west. 75. Strateg Consider the relatie motion of the ship and the water. The relatie speeds are: upstream ship water up s w downstream ship water d s w Find the speed of the current, w. uptup ( s w ) tup, so s w (1). dtd ( s w ) td, so s w (). tup td Subtract (1) from () km 1 1 w, so w 0.4 km h. td tup td t up 19. h 0.8 h 76. Strateg The minimum air elocit is in the same direction as the airplane s. 10 m s east 160 m s east 50 m s east 38

30 Phsics Chapter 3: Motion in a Plane 77. (a) Strateg The ground speed of the small plane will be the magnitude of the ector sum of the two elocities. Draw a diagram and use the component method. Let north be in the +-direction and east the +-direction m s 30.0 m s (10.0 m s)cos 10 (10.0 m s)sin 10 (b) Strateg The new directional heading relatie to the ground is in the direction of the elocit relatie to the ground. 1 (10.0 m s)sin 10 tan m s (10.0 m s)cos 10 Both components are negatie, so south of west Discussion The effects of preailing winds as headwinds or tailwinds are built into airline schedules. 78. (a) Strateg To compensate for the wind, the plane s new heading will be north of west, and the north component of the plane s elocit relatie to the air must be equal in magnitude to that of the south component of the elocit of the wind. Draw a diagram and use the component method. The north component of the elocit relatie to the air is equal to (30.0 m s)sin. The south component of the wind is equal to (10.0 m s)sin30. We set these equal and sole for. (30.0 m s)sin (10.0 m s)sin30, so 1 (10.0 m s)sin30 sin 9.6 north of west m s (b) Strateg The new ground speed is equal to the sum of the west components of the two elocities in part (a). (30.0 m s)cos9.6 (10.0 m s)cos30 38 m s 79. Strateg The upstream component of the elocit of the boat must be equal in magnitude to the speed of the current. Let +-direction be toward the opposite shore (4.0 km h)sin 1.8 km h, so sin The direction of the elocit of the boat relatie to the water is 7 upstream. 39