4-1 MOTION IN TWO AND THREE DIMENSIONS. 4-2 Position and Displacement WHAT IS PHYSICS? CHAPTER

Transcription

1 CHAPTER (5 m)k ˆ r 4 (2 m)j ˆ ( 3 m)i ˆ z Fig. 4-1 The position ector r for a particle is the ector sum of its ector components. To locate the particle, this is how far parallel to z. This is how far parallel to. This is how far parallel to. MTIN IN TW AND THREE DIMENSINS 4-1 WHAT IS PHYSICS? motion, but now the motion can be in two or three dimensions. For eample, medical researchers and aeronautical engineers might concentrate on the phsics of the two- and three-dimensional turns taken b fighter pilots in dogfights because a modern high-performance jet can take a tight turn so quickl that the pilot immediatel loses consciousness. A sports engineer might focus on the phsics of basketball. For eample, in a free throw (where a plaer gets an uncontested shot at the basket from about 4.3 m), a plaer might emplo the oerhand push shot, in which the ball is pushed awa from about shoulder height and then released. r the plaer might use an underhand loop shot, in which the ball is brought upward from about the belt-line leel and released. The first technique is the oerwhelming choice among professional plaers, but the legendar Rick Barr set the record for free-throw shooting with the underhand technique. Motion in three dimensions is not eas to understand. For eample, ou are probabl good at driing a car along a freewa (one-dimensional motion) but would probabl hae a difficult time in landing an airplane on a runwa (threedimensional motion) without a lot of training. In our stud of two- and three-dimensional motion, we start with position and displacement. In this chapter we continue looking at the aspect of phsics that analzes 4-2 Position and Displacement ne general wa of locating a particle (or particle-like object) is with a position ector r, which is a ector that etends from a reference point (usuall the origin) to the particle. In the unit-ector notation of Section 3-5, r can be written r î ĵ zkˆ, (4-1) where î, ĵ, and zkˆ are the ector components of r and the coefficients,, and z are its scalar components. The coefficients,, and z gie the particle s location along the coordinate aes and relatie to the origin; that is, the particle has the rectangular coordinates (,, z). For instance, Fig. 4-1 shows a particle with position ector r ( 3 m)î (2 m)ĵ (5 m)kˆ and rectangular coordinates ( 3 m, 2 m, 5 m). Along the ais the particle is 3 m from the origin, in the î direction. Along the ais it is 2 m from the origin, in the ĵ direction. Along the z ais it is 5 m from the origin, in the kˆ direction. 8

2 4-2 PSITIN AND DISPLACEMENT 59 PART 1 As a particle moes, its position ector changes in such a wa that the ector alwas etends to the particle from the reference point (the origin). If the position ector changes sa, from r to during a certain time interal then the particle s displacement 1 r 2 r during that time interal is r r 2 r 1. (4-2) Using the unit-ector notation of Eq. 4-1, we can rewrite this displacement as r ( 2 î 2 ĵ z 2 kˆ ) ( 1 î 1 ĵ z 1 kˆ ) or as r ( 2 1 )î ( 2 1 )ĵ (z 2 z 1 )kˆ, (4-3) where coordinates ( 1, 1, z 1 ) correspond to position ector and coordinates ( 2, 2, z 2 ) correspond to position ector r 2. We can also rewrite the displacement b substituting for ( 2 1 ), for ( 2 1 ), and z for (z 2 z 1 ) r î ĵ zkˆ. (4-4) r 1 Sample Problem A rabbit runs across a parking lot on which a set of coordinate aes has, strangel enough, been drawn. The coordinates (meters) of the rabbit s position as functions of time t (seconds) are gien b 0.31t 2 7.2t 28 (4-5) and 0.22t 2 9.1t 30. (4-6) (a) At t 15 s, what is the rabbit s position ector r in unitector notation and in magnitude-angle notation? Two-dimensional position ector, rabbit run (m) To locate the rabbit, this is the component. 41 (m) KEY IDEA 40 r The and coordinates of the rabbit s position, as gien b Eqs. 4-5 and 4-6, are the scalar components of the rabbit s position ector r. Calculations We can write r (t) (t)î (t)ĵ. (4-7) (We write r (t) rather than r because the components are functions of t, and thus r is also.) At t 15 s, the scalar components are ( 0.31)(15) 2 (7.2)(15) m and (0.22)(15) 2 (9.1)(15) m, so r (66 m)î (57 m)ĵ, (Answer) which is drawn in Fig. 4-2a. To get the magnitude and angle of r, we use Eq. 3-6 r (66 m) 2 ( 57 m) 2 87 m, tan 1 tan 1 57 m (Answer) and. (Answer) 66 m 41 Fig. 4-2 (a) A rabbit s position ector r at time t 15 s. The scalar components of r are shown along the aes. (b)the rabbit s path and its position at si alues of t. (a) (b) 60 (m) This is the component. 25 s t = 0 s s 20 s 10 s 15 s This is the path with arious times indicated. (m)

3 60 CHAPTER 4 MTIN IN TW AND THREE DIMENSINS Check Although u 139 has the same tangent as 41, the components of position ector r indicate that the desired angle is (b) Graph the rabbit s path for t 0 to t 25 s. Graphing We hae located the rabbit at one instant, but to see its path we need a graph. So we repeat part (a) for seeral alues of t and then plot the results. Figure 4-2b shows the plots for si alues of t and the path connecting them. We can also plot Eqs. 4-5 and 4-6 on a calculator. Additional eamples, ideo, and practice aailable at WilePLUS 4-3 Aerage Velocit and Instantaneous Velocit If a particle moes from one point to another, we might need to know how fast it moes. Just as in Chapter 2, we can define two quantities that deal with how fast aerage elocit and instantaneous elocit. Howeer, here we must consider these quantities as ectors and use ector notation. If a particle moes through a displacement r in a time interal t, then its aerage elocit ag is aerage elocit displacement time interal, or ag (4-8) t. r 1 r 1 r 2 r As the particle moes, the position ector must change. Tangent 2 This is the displacement. Path Fig. 4-3 The displacement r of a particle during a time interal t, from position 1 with position ector r 1 at time t 1 to position 2 with position ector r 2 at time t 2.The tangent to the particle s path at position 1 is shown. This tells us that the direction of (the ector on the left side of Eq. 4-8) must be the same as that of the displacement r (the ector on the right side). Using Eq. 4-4, we can write Eq. 4-8 in ector components as î ĵ zkˆ ag î ĵ z kˆ. (4-9) t t t t For eample, if a particle moes through displacement (12 m)î (3.0 m)kˆ in 2.0 s, then its aerage elocit during that moe is r (12 m)î (3.0 m)kˆ ag (6.0 m/s)î (1.5 m/s)kˆ. t 2.0 s That is, the aerage elocit (a ector quantit) has a component of 6.0 m/s along the ais and a component of 1.5 m/s along the z ais. When we speak of the elocit of a particle, we usuall mean the particle s instantaneous elocit at some instant. This is the alue that ag approaches in the limit as we shrink the time interal t to 0 about that instant. Using the language of calculus, we ma write as the deriatie (4-10) Figure 4-3 shows the path of a particle that is restricted to the plane. As the particle traels to the right along the cure, its position ector sweeps to the right. During time interal t, the position ector changes from r to and the particle s displacement is 1 r 2 r. To find the instantaneous elocit of the particle at, sa, instant t 1 (when the particle is at position 1), we shrink interal t to 0 about t 1. Three things happen as we do so. (1) Position ector r in Fig. 4-3 moes toward so that 2 r shrinks toward zero. (2) The direction of 1 r r / t (and thus of ag) approaches the direction of the line tangent to the particle s path at position 1. (3) The aerage elocit approaches the instantaneous elocit at t 1. ag ag dr dt.

4 4-3 AVERAGE VELCITY AND INSTANTANEUS VELCITY 61 PART 1 ag In the limit as t 0, we hae ag and, most important here, takes on the direction of the tangent line. Thus, has that direction as well The direction of the instantaneous elocit of a particle is alwas tangent to the particle s path at the particle s position. The result is the same in three dimensions is alwas tangent to the particle s path. To write Eq in unit-ector form, we substitute for r from Eq. 4-1 d dt (î ĵ zkˆ ) d dt î d dt This equation can be simplified somewhat b writing it as ĵ dz dt kˆ. î ĵ z kˆ, (4-11) where the scalar components of are d dt, d dt, and z dz dt. (4-12) For eample, d/dt is the scalar component of along the ais.thus, we can find the scalar components of b differentiating the scalar components of r. Figure 4-4 shows a elocit ector and its scalar and components. Note that is tangent to the particle s path at the particle s position. Caution When a position ector is drawn, as in Figs. 4-1 through 4-3, it is an arrow that etends from one point (a here ) to another point (a there ). Howeer, when a elocit ector is drawn, as in Fig. 4-4, it does not etend from one point to another. Rather, it shows the instantaneous direction of trael of a particle at the tail, and its length (representing the elocit magnitude) can be drawn to an scale. The elocit ector is alwas tangent to the path. Tangent Fig. 4-4 The elocit of a particle, along with the scalar components of. Path These are the and components of the ector at this instant. CHECKPINT 1 The figure shows a circular path taken b a particle. If the instantaneous elocit of the particle is (2 m/s)î (2 m/s)ĵ, through which quadrant is the particle moing at that instant if it is traeling (a) clockwise and (b) counterclockwise around the circle? For both cases, draw on the figure.

5 62 CHAPTER 4 MTIN IN TW AND THREE DIMENSINS Sample Problem Two-dimensional elocit, rabbit run For the rabbit in the preceding Sample Problem, find the elocit at time t 15 s. KEY IDEA We can find b taking deriaties of the components of the rabbit s position ector. Calculations Appling the part of Eq to Eq. 4-5, we find the component of to be d dt 0.62t 7.2. (4-13) At t 15 s, this gies 2.1 m/s. Similarl, appling the part of Eq to Eq. 4-6, we find d dt d dt ( 0.31t2 7.2t 28) d dt (0.22t2 9.1t 30) 0.44t 9.1. (4-14) At t 15 s, this gies 2.5 m/s. Equation 4-11 then ields ( 2.1 m/s)î ( 2.5 m/s)ĵ, (Answer) which is shown in Fig. 4-5, tangent to the rabbit s path and in the direction the rabbit is running at t 15 s. To get the magnitude and angle of, either we use a ector-capable calculator or we follow Eq. 3-6 to write and ( 2.1 m/s) 2 ( 2.5 m/s) m/s tan 1 tan tan m/s Check Is the angle 130 or ? (m) Fig. 4-5 The rabbit s elocit at t 15 s These are the and components of the ector at this instant (Answer) (Answer) (m) Additional eamples, ideo, and practice aailable at WilePLUS 4-4 Aerage Acceleration and Instantaneous Acceleration When a particle s elocit changes from 1 to 2 in a time interal t, its aerage acceleration a during t is ag aerage change in elocit, acceleration time interal or ag a 2 1 (4-15) t t. aag If we shrink t to zero about some instant, then in the limit approaches the instantaneous acceleration (or acceleration) a at that instant; that is, a d dt. (4-16) If the elocit changes in either magnitude or direction (or both), the particle must hae an acceleration.

6 4-4 AVERAGE ACCELERATIN AND INSTANTANEUS ACCELERATIN 63 PART 1 We can write Eq in unit-ector form b substituting Eq for to obtain a d dt ( î ĵ z kˆ ) These are the and components of the ector at this instant. We can rewrite this as d dt î d dt ĵ d z dt kˆ. a a a where the scalar components of a are a a î a ĵ a z kˆ, a d dt, a d dt, and a z d z dt. (4-17) (4-18) Path Fig. 4-6 The acceleration a of a particle and the scalar components of a. To find the scalar components of a, we differentiate the scalar components of. Figure 4-6 shows an acceleration ector a and its scalar components for a particle moing in two dimensions. Caution When an acceleration ector is drawn, as in Fig. 4-6, it does not etend from one position to another. Rather, it shows the direction of acceleration for a particle located at its tail, and its length (representing the acceleration magnitude) can be drawn to an scale. CHECKPINT 2 Here are four descriptions of the position (in meters) of a puck as it moes in an plane (1) 3t 2 4t 2 and 6t 2 4t (3) r 2t 2 î (4t 3)ĵ (2) 3t 3 4t and 5t 2 6 (4) r (4t 3 2t)î 3ĵ Are the and acceleration components constant? Is acceleration a constant? Sample Problem Two-dimensional acceleration, rabbit run For the rabbit in the preceding two Sample Problems, find the acceleration a at time t 15 s. a ( 0.62 m/s 2 )î (0.44 m/s 2 )ĵ, (Answer) which is superimposed on the rabbit s path in Fig We can find components. a KEY IDEA b taking deriaties of the rabbit s elocit Calculations Appling the a part of Eq to Eq. 4-13, we find the component of a to be a d dt Similarl, appling the a part of Eq to Eq ields the component as a d dt d dt ( 0.62t 7.2) 0.62 m/s2. d dt (0.44t 9.1) 0.44 m/s2. We see that the acceleration does not ar with time (it is a constant) because the time ariable t does not appear in the epression for either acceleration component. Equation 4-17 then ields Fig. 4-7 The acceleration a of the rabbit at t 15 s.the rabbit happens to hae this same acceleration at all points on its path. (m) a 145 These are the and components of the ector at this instant. (m)

7 64 CHAPTER 4 MTIN IN TW AND THREE DIMENSINS To get the magnitude and angle of a, either we use a ector-capable calculator or we follow Eq For the magnitude we hae a 2a 2 a 2 2( 0.62 m/s 2 ) 2 (0.44 m/s 2 ) m/s 2. (Answer) For the angle we hae tan 1 a a 0.44 tan 1 m/s m/s2 35. Howeer, this angle, which is the one displaed on a calcula- a tor, indicates that is directed to the right and downward in Fig Yet, we know from the components that a must be directed to the left and upward. To find the other angle that has the same tangent as 35 but is not displaed on a calculator, we add (Answer) This is consistent with the components of a because it gies a ector that is to the left and upward. Note that a has the same magnitude and direction throughout the rabbit s run because the acceleration is constant. Additional eamples, ideo, and practice aailable at WilePLUS 4-5 Projectile Motion We net consider a special case of two-dimensional motion A particle moes in a ertical plane with some initial elocit 0 but its acceleration is alwas the freefall acceleration g, which is downward. Such a particle is called a projectile (mean- ing that it is projected or launched), and its motion is called projectile motion. A projectile might be a tennis ball (Fig. 4-8) or baseball in flight, but it is not an airplane or a duck in flight. Man sports (from golf and football to lacrosse and racquetball) inole the projectile motion of a ball, and much effort is spent in tring to control that motion for an adantage. For eample, the racquetball plaer who discoered the Z-shot in the 1970s easil won his games because the ball s peculiar flight to the rear of the court alwas perpleed his opponents. ur goal here is to analze projectile motion using the tools for twodimensional motion described in Sections 4-2 through 4-4 and making the assumption that air has no effect on the projectile. Figure 4-9, which is analzed in the net section, shows the path followed b a projectile when the air has no effect. The projectile is launched with an initial elocit 0 that can be written as 0 0 î 0 ĵ. (4-19) The components 0 and 0 can then be found if we know the angle u 0 between 0 and the positie direction 0 0 cos u 0 and 0 0 sin u 0. (4-20) During its two-dimensional motion, the projectile s position ector r and elocit ector change continuousl, but its acceleration ector a is constant and alwas directed erticall downward. The projectile has no horizontal acceleration. Projectile motion, like that in Figs. 4-8 and 4-9, looks complicated, but we hae the following simplifing feature (known from eperiment) In projectile motion, the horizontal motion and the ertical motion are independent of each other; that is, neither motion affects the other. Fig. 4-8 A stroboscopic photograph of a ellow tennis ball bouncing off a hard surface. Between impacts, the ball has projectile motion. Source Richard Megna/ Fundamental Photographs. This feature allows us to break up a problem inoling two-dimensional motion into two separate and easier one-dimensional problems, one for the horizontal motion (with zero acceleration) and one for the ertical motion (with constant downward acceleration). Here are two eperiments that show that the horizontal motion and the ertical motion are independent.

8 4-5 PRJECTILE MTIN 65 PART 1 Fig. 4-9 The projectile motion of an object launched into the air at the origin of a coordinate sstem and with launch elocit 0 at angle u 0.The motion is a combination of ertical motion (constant acceleration) and horizontal motion (constant elocit), as shown b the elocit components. A Vertical motion + Horizontal motion Projectile motion 0 Vertical elocit Launch 0 This ertical motion plus this horizontal motion produces this projectile motion Launch Launch elocit Launch angle Speed decreasing Constant elocit = 0 Stopped at maimum height = 0 Constant elocit Speed increasing Constant elocit Constant elocit

9 66 CHAPTER 4 MTIN IN TW AND THREE DIMENSINS Two Golf Balls Figure 4-10 is a stroboscopic photograph of two golf balls, one simpl released and the other shot horizontall b a spring. The golf balls hae the same ertical motion, both falling through the same ertical distance in the same interal of time. The fact that one ball is moing horizontall while it is falling has no effect on its ertical motion; that is, the horizontal and ertical motions are independent of each other. Fig ne ball is released from rest at the same instant that another ball is shot horizontall to the right. Their ertical motions are identical. Source Richard Megna/ Fundamental Photographs. A Great Student Rouser Figure 4-11 shows a demonstration that has enliened man a phsics lecture. It inoles a blowgun G, using a ball as a projectile. The target is a can suspended from a magnet M, and the tube of the blowgun is aimed directl at the can. The eperiment is arranged so that the magnet releases the can just as the ball leaes the blowgun. If g (the magnitude of the free-fall acceleration) were zero, the ball would follow the straight-line path shown in Fig and the can would float in place after the magnet released it. The ball would certainl hit the can. Howeer, g is not zero, but the ball still hits the can! As Fig shows, during the time of flight of the ball, both ball and can fall the same distance h from their zero-g locations. The harder the demonstrator blows, the greater is the ball s initial speed, the shorter the flight time, and the smaller the alue of h. CHECKPINT 3 At a certain instant, a fl ball has elocit 25î 4.9ĵ (the ais is horizontal, the ais is upward, and is in meters per second). Has the ball passed its highest point? The ball and the can fall the same distance h. Zero-g path M Can h Fig The projectile ball alwas hits the falling can. Each falls a distance h from where it would be were there no free-fall acceleration. G 4-6 Projectile Motion Analzed Now we are read to analze projectile motion, horizontall and erticall. The Horizontal Motion Because there is no acceleration in the horizontal direction, the horizontal component of the projectile s elocit remains unchanged from its initial alue 0 throughout the motion, as demonstrated in Fig At an time t, the projec-

10 4-6 PRJECTILE MTIN ANALYZED 67 PART 1 Fig The ertical component of this skateboarder s elocit is changing but not the horizontal component, which matches the skateboard s elocit. As a result, the skateboard stas underneath him, allowing him to land on it. Source Jamie Budge/ Liaison/Gett Images, Inc. tile s horizontal displacement 0 from an initial position 0 is gien b Eq with a 0, which we write as 0 0 t. Because 0 0 cos u 0, this becomes 0 ( 0 cos u 0 )t. (4-21) The Vertical Motion The ertical motion is the motion we discussed in Section 2-9 for a particle in free fall. Most important is that the acceleration is constant. Thus, the equations of Table 2-1 appl, proided we substitute g for a and switch to notation. Then, for eample, Eq becomes 0 0 t 1 2 gt 2 ( 0 sin 0)t 1 2 gt 2, (4-22) where the initial ertical elocit component 0 is replaced with the equialent 0 sin u 0. Similarl, Eqs and 2-16 become 0 sin u 0 gt (4-23) and 2 ( 0 sin 0) 2 2g( 0 ). (4-24) As is illustrated in Fig. 4-9 and Eq. 4-23, the ertical elocit component behaes just as for a ball thrown erticall upward. It is directed upward initiall, and its magnitude steadil decreases to zero, which marks the maimum height of the path. The ertical elocit component then reerses direction, and its magnitude becomes larger with time. The Equation of the Path We can find the equation of the projectile s path (its trajector) b eliminating time t between Eqs and Soling Eq for t and substituting into Eq. 4-22, we obtain, after a little rearrangement, (tan 0) g 2 2( 0 cos 0) 2 (trajector). (4-25)

11 68 CHAPTER 4 MTIN IN TW AND THREE DIMENSINS 0 I II This is the equation of the path shown in Fig In deriing it, for simplicit we let 0 0 and 0 0 in Eqs and 4-22, respectiel. Because g, u 0, and 0 are constants, Eq is of the form a b 2, in which a and b are constants. This is the equation of a parabola, so the path is parabolic. 60 Fig (I) The path of a fl ball calculated b taking air resistance into account. (II) The path the ball would follow in a acuum, calculated b the methods of this chapter. See Table 4-1 for corresponding data. (Adapted from The Trajector of a Fl Ball, b Peter J. Brancazio, The Phsics Teacher, Januar 1985.) Two Fl Balls a TABLE 4-1 Path I (Air) Path II (Vacuum) Range 98.5 m 177 m Maimum height 53.0 m 76.8 m Time of flight 6.6 s 7.9 s a See Fig The launch angle is 60 and the launch speed is 44.7 m/s. The Horizontal Range The horizontal range R of the projectile is the horizontal distance the projectile has traeled when it returns to its initial height (the height at which it is launched).to find range R, let us put 0 R in Eq and 0 0 in Eq. 4-22, obtaining R ( 0 cos u 0 )t and 0 ( 0 sin 0)t 1 2 gt 2. Eliminating t between these two equations ields R g sin 0 cos 0. Using the identit sin 2u 0 2 sin u 0 cos u 0 (see Appendi E), we obtain R 0 2 g sin 2 0. (4-26) Caution This equation does not gie the horizontal distance traeled b a projectile when the final height is not the launch height. Note that R in Eq has its maimum alue when sin 2u 0 1, which corresponds to 2u 0 90 or u The horizontal range R is maimum for a launch angle of 45. Howeer, when the launch and landing heights differ, as in shot put, hammer throw, and basketball, a launch angle of 45 does not ield the maimum horizontal distance. The Effects of the Air We hae assumed that the air through which the projectile moes has no effect on its motion. Howeer, in man situations, the disagreement between our calculations and the actual motion of the projectile can be large because the air resists (opposes) the motion. Figure 4-13, for eample, shows two paths for a fl ball that leaes the bat at an angle of 60 with the horizontal and an initial speed of 44.7 m/s. Path I (the baseball plaer s fl ball) is a calculated path that approimates normal conditions of pla, in air. Path II (the phsics professor s fl ball) is the path the ball would follow in a acuum. CHECKPINT 4 A fl ball is hit to the outfield. During its flight (ignore the effects of the air), what happens to its (a) horizontal and (b) ertical components of elocit? What are the (c) horizontal and (d) ertical components of its acceleration during ascent, during descent, and at the topmost point of its flight?

12 4-6 PRJECTILE MTIN ANALYZED 69 PART 1 Sample Problem Projectile dropped from airplane In Fig. 4-14, a rescue plane flies at 198 km/h ( 55.0 m/s) and constant height h 500 m toward a point directl oer a ictim, where a rescue capsule is to land. (a) What should be the angle f of the pilot s line of sight to the ictim when the capsule release is made? KEY IDEAS nce released, the capsule is a projectile, so its horizontal and ertical motions can be considered separatel (we need not consider the actual cured path of the capsule). Calculations In Fig. 4-14, we see that f is gien b tan 1 h, (4-27) where is the horizontal coordinate of the ictim (and of the capsule when it hits the water) and h 500 m. We should be able to find with Eq ( 0 cos u 0 )t. (4-28) Here we know that 0 0 because the origin is placed at the point of release. Because the capsule is released and not shot from the plane, its initial elocit 0 is equal to the plane s elocit. Thus, we know also that the initial elocit has magnitude m/s and angle u 0 0 (measured relatie to the positie direction of the ais). Howeer, we do not know the time t the capsule takes to moe from the plane to the ictim. To find t, we net consider the ertical motion and specificall Eq ( 0 sin 0)t 1 2 gt 2. (4-29) Here the ertical displacement 0 of the capsule is 500 m (the negatie alue indicates that the capsule moes downward). So, 500 m (55.0 m/s)(sin 0 )t 1 2 (9.8 m/s2 )t 2. (4-30) Soling for t, we find t 10.1 s. Using that alue in Eq ields 0 (55.0 m/s)(cos 0 )(10.1 s), (4-31) or m. Then Eq gies us tan m m (Answer) (b) As the capsule reaches the water, what is its elocit unit-ector notation and in magnitude-angle notation? KEY IDEAS (1) The horizontal and ertical components of the capsule s elocit are independent. (2) Component does not change from its initial alue 0 0 cos u 0 because there is no horizontal acceleration. (3) Component changes from its initial alue 0 0 sin u 0 because there is a ertical acceleration. in Calculations When the capsule reaches the water, h φ 0 Trajector Line of sight Fig A plane drops a rescue capsule while moing at constant elocit in leel flight. While falling, the capsule remains under the plane. 0 cos u 0 (55.0 m/s)(cos 0 ) 55.0 m/s. Using Eq and the capsule s time of fall t 10.1 s, we also find that when the capsule reaches the water, 0 sin u 0 gt (4-32) (55.0 m/s)(sin 0 ) (9.8 m/s 2 )(10.1 s) 99.0 m/s. Thus, at the water (55.0 m/s)î (99.0 m/s)ĵ. (Answer) Using Eq. 3-6 as a guide, we find that the magnitude and the angle of are 113 m/s and u (Answer) Additional eamples, ideo, and practice aailable at WilePLUS

13 70 CHAPTER 4 MTIN IN TW AND THREE DIMENSINS Sample Problem Cannonball to pirate ship Figure 4-15 shows a pirate ship 560 m from a fort defending a harbor entrance. A defense cannon, located at sea leel, fires balls at initial speed 0 82 m/s. (a) At what angle u 0 from the horizontal must a ball be fired to hit the ship? KEY IDEAS (1) A fired cannonball is a projectile. We want an equation that relates the launch angle u 0 to the ball s horizontal displacement as it moes from cannon to ship. (2) Because the cannon and the ship are at the same height, the horizontal displacement is the range. Fig R = 560 m A pirate ship under fire. Either launch angle gies a hit. Calculations We can relate the launch angle u 0 to the range R with Eq which, after rearrangement, gies (4-33) ne solution of sin 1 (54.7 ) is displaed b a calculator; we subtract it from 180 to get the other solution (125.3 ). Thus, Eq gies us 0 27 and (Answer) (b) What is the maimum range of the cannonballs? Calculations We hae seen that maimum range corresponds to an eleation angle u 0 of 45.Thus, R 0 2 sin 1 gr sin g sin sin 1 (9.8 m/s2 )(560 m) (82 m/s) 2 (82 m/s)2 sin (2 45 ) m/s 686 m 690 m. (Answer) As the pirate ship sails awa, the two eleation angles at which the ship can be hit draw together, eentuall merging at u 0 45 when the ship is 690 m awa. Beond that distance the ship is safe. Howeer, the cannonballs could go farther if the cannon were higher. Additional eamples, ideo, and practice aailable at WilePLUS 4-7 Uniform Circular Motion A particle is in uniform circular motion if it traels around a circle or a circular arc at constant (uniform) speed. Although the speed does not ar, the particle is accelerating because the elocit changes in direction. Figure 4-16 shows the relationship between the elocit and acceleration ectors at arious stages during uniform circular motion. Both ectors hae constant magnitude, but their directions change continuousl. The elocit is alwas directed tangent to the circle in the direction of motion. The acceleration is alwas directed radiall inward. Because of this, the acceleration associated with uniform circular motion is called a centripetal (meaning center seeking ) acceleration. As we proe net, the magnitude of this acceleration a is a 2 r (centripetal acceleration), (4-34) where r is the radius of the circle and is the speed of the particle. In addition, during this acceleration at constant speed, the particle traels the circumference of the circle (a distance of 2pr) in time T 2 r (period). (4-35)

14 4-7 UNIFRM CIRCULAR MTIN 71 PART 1 The acceleration ector alwas points toward the center. a a a Fig Velocit and acceleration ectors for uniform circular motion. The elocit ector is alwas tangent to the path. T is called the period of reolution, or simpl the period, of the motion. It is, in general, the time for a particle to go around a closed path eactl once. Proof of Eq To find the magnitude and direction of the acceleration for uniform circular motion, we consider Fig In Fig. 4-17a, particle p moes at constant speed around a circle of radius r.at the instant shown, p has coordinates p and p. Recall from Section 4-3 that the elocit of a moing particle is alwas tangent to the particle s path at the particle s position. In Fig. 4-17a, that means is perpendicular to a radius r drawn to the particle s position.then the angle u that makes with a ertical at p equals the angle u that radius r makes with the ais. The scalar components of are shown in Fig. 4-17b. With them, we can write the elocit as î ĵ ( sin )î ( cos )ĵ. (4-36) Now, using the right triangle in Fig. 4-17a, we can replace sin u with p /r and cos u with p /r to write p r î p r ĵ. (4-37) To find the acceleration a of particle p, we must take the time deriatie of this equation. Noting that speed and radius r do not change with time, we obtain a d dt r d p dt î r d p dt ĵ. (4-38) r p p p a a φ a (a) (b) (c) Fig Particle p moes in counterclockwise uniform circular motion. (a) Its position and elocit at a certain instant. (b) Velocit. (c) Acceleration a.

15 72 CHAPTER 4 MTIN IN TW AND THREE DIMENSINS Now note that the rate d p /dt at which p changes is equal to the elocit component. Similarl, d p /dt, and, again from Fig. 4-17b, we see that sin u and cos u. Making these substitutions in Eq. 4-38, we find a 2 r cos î 2 r sin ĵ. (4-39) This ector and its components are shown in Fig. 4-17c. Following Eq. 3-6, we find a 2a 2 a 2 2 r 2(cos ) 2 (sin ) 2 2 r 11 2 r, as we wanted to proe.to orient a, we find the angle f shown in Fig. 4-17c tan a (2 /r) sin tan. a ( 2 /r) cos a Thus, f u, which means that is directed along the radius r of Fig. 4-17a, toward the circle s center, as we wanted to proe. CHECKPINT 5 An object moes at constant speed along a circular path in a horizontal plane, with the center at the origin. When the object is at 2 m, its elocit is (4 m/s) ĵ. Gie the object s (a) elocit and (b) acceleration at 2m. Sample Problem Top gun pilots in turns Top gun pilots hae long worried about taking a turn too tightl. As a pilot s bod undergoes centripetal acceleration, with the head toward the center of curature, the blood pressure in the brain decreases, leading to loss of brain function. There are seeral warning signs. When the centripetal acceleration is 2g or 3g, the pilot feels hea. At about 4g, the pilot s ision switches to black and white and narrows to tunnel ision. If that acceleration is sustained or increased, ision ceases and, soon after, the pilot is unconscious a condition known as g-lc for g-induced loss of consciousness. What is the magnitude of the acceleration, in g units, of a pilot whose aircraft enters a horizontal circular turn with a elocit of i (400î 500ĵ) m/s and 24.0 s later leaes the turn with a elocit of ( 400î 500 ĵ) m/s? f KEY IDEAS We assume the turn is made with uniform circular motion. Then the pilot s acceleration is centripetal and has magnitude a gien b Eq (a 2 /R), where R is the cir- cle s radius. Also, the time required to complete a full circle is the period gien b Eq (T 2pR/). Calculations Because we do not know radius R, let s sole Eq for R and substitute into Eq We find a 2 T. Speed here is the (constant) magnitude of the elocit during the turning. Let s substitute the components of the initial elocit into Eq (400 m/s) 2 (500 m/s) m/s. To find the period T of the motion, first note that the final elocit is the reerse of the initial elocit. This means the aircraft leaes on the opposite side of the circle from the initial point and must hae completed half a circle in the gien 24.0 s. Thus a full circle would hae taken T 48.0 s. Substituting these alues into our equation for a, we find a 2 ( m/s) m/s 2 8.6g. (Answer) 48.0 s Additional eamples, ideo, and practice aailable at WilePLUS

16 4-8 RELATIVE MTIN IN NE DIMENSIN 73 PART Relatie Motion in ne Dimension Suppose ou see a duck fling north at 30 km/h.to another duck fling alongside, the first duck seems to be stationar. In other words, the elocit of a particle depends on the reference frame of whoeer is obsering or measuring the elocit. For our purposes, a reference frame is the phsical object to which we attach our coordinate sstem. In eerda life, that object is the ground. For eample, the speed listed on a speeding ticket is alwas measured relatie to the ground. The speed relatie to the police officer would be different if the officer were moing while making the speed measurement. Suppose that Ale (at the origin of frame A in Fig. 4-18) is parked b the side of a highwa, watching car P (the particle ) speed past. Barbara (at the origin of frame B) is driing along the highwa at constant speed and is also watching car P. Suppose that the both measure the position of the car at a gien moment. From Fig we see that PA PB BA. (4-40) The equation is read The coordinate PA of P as measured b A is equal to the coordinate PB of P as measured b B plus the coordinate BA of B as measured b A. Note how this reading is supported b the sequence of the subscripts. Taking the time deriatie of Eq. 4-40, we obtain d dt ( PA) d dt ( PB) d dt ( BA). Thus, the elocit components are related b PA PB BA. (4-41) This equation is read The elocit PA of P as measured b A is equal to the elocit PB of P as measured b B plus the elocit BA of B as measured b A. The term BA is the elocit of frame B relatie to frame A. Here we consider onl frames that moe at constant elocit relatie to each other. In our eample, this means that Barbara (frame B) dries alwas at constant elocit BA relatie to Ale (frame A). Car P (the moing particle), howeer, can change speed and direction (that is, it can accelerate). To relate an acceleration of P as measured b Barbara and b Ale, we take the time deriatie of Eq Because BA is constant, the last term is zero and we hae In other words, d dt ( PA) d dt ( PB) d dt ( BA). a PA a PB. (4-42) bserers on different frames of reference that moe at constant elocit relatie to each other will measure the same acceleration for a moing particle. Fig Ale (frame A) and Barbara (frame B) watch car P, as both B and P moe at different elocities along the common ais of the two frames.at the instant shown, BA is the coordinate of B in the A frame.also, P is at coordinate PB in the B frame and coordinate PA PB BA in the A frame. Frame A Frame B BA Frame B moes past frame A while both obsere P. P BA PB PA = PB + BA

17 74 CHAPTER 4 MTIN IN TW AND THREE DIMENSINS Sample Problem Relatie motion, one dimensional, Ale and Barbara In Fig. 4-18, suppose that Barbara s elocit relatie to Ale is a constant BA 52 km/h and car P is moing in the negatie direction of the ais. (a) If Ale measures a constant PA 78 km/h for car P, what elocit PB will Barbara measure? KEY IDEAS We can attach a frame of reference A to Ale and a frame of reference B to Barbara. Because the frames moe at constant elocit relatie to each other along one ais, we can use Eq ( PA PB BA ) to relate PB to PA and BA. Calculation We find 78 km/h PB 52 km/h. Thus, PB 130 km/h. (Answer) Comment If car P were connected to Barbara s car b a cord wound on a spool, the cord would be unwinding at a speed of 130 km/h as the two cars separated. (b) If car P brakes to a stop relatie to Ale (and thus relatie to the ground) in time t 10 s at constant acceleration, what is its acceleration a PA relatie to Ale? KEY IDEAS To calculate the acceleration of car P relatie to Ale, we must use the car s elocities relatie to Ale. Because the acceleration is constant, we can use Eq ( 0 at) to relate the acceleration to the initial and final elocities of P. Calculation The initial elocit of P relatie to Ale is PA 78 km/h and the final elocit is 0.Thus, the acceleration relatie to Ale is a PA 0 t 2.2 m/s 2. (Answer) (c) What is the acceleration a PB of car P relatie to Barbara during the braking? KEY IDEA To calculate the acceleration of car P relatie to Barbara, we must use the car s elocities relatie to Barbara. Calculation We know the initial elocit of P relatie to Barbara from part (a) ( PB 130 km/h).the final elocit of P relatie to Barbara is 52 km/h (this is the elocit of the stopped car relatie to the moing Barbara).Thus, a PB 0 t 2.2 m/s 2. 0 ( 78 km/h) 10 s 1 m/s 3.6 km/h 52 km/h ( 130 km/h) 10 s 1 m/s 3.6 km/h (Answer) Comment We should hae foreseen this result Because Ale and Barbara hae a constant relatie elocit, the must measure the same acceleration for the car. Additional eamples, ideo, and practice aailable at WilePLUS Fig Frame B has the constant twodimensional elocit BA relatie to frame A. The position ector of B relatie to A is r BA.The position ectors of particle P are r relatie to A and r relatie to B. PA P r PB r PA BA Frame B r BA Frame A PB 4-9 Relatie Motion in Two Dimensions ur two obserers are again watching a moing particle P from the origins of reference frames A and B, while B moes at a constant elocit BA relatie to A.(The corresponding aes of these two frames remain parallel.) Figure 4-19 shows a certain instant during the motion. At that instant, the position ector of the origin of B relatie to the origin of A is r BA.Also, the position ectors of particle P are r PA relatie to the origin of A and r PB relatie to the origin of B. From the arrangement of heads and tails of those three position ectors, we can relate the ectors with B taking the time deriatie of this equation, we can relate the elocities and of particle P relatie to our obserers PB r PA r PB r BA. PA PB BA. (4-43) PA (4-44)

18 4-9 RELATIVE MTIN IN TW DIMENSINS 75 PART 1 B taking the time deriatie of this relation, we can relate the accelerations and apb of the particle P relatie to our obserers. Howeer, note that because is constant, its time deriatie is zero.thus, we get BA apa apa a PB. (4-45) As for one-dimensional motion, we hae the following rule bserers on different frames of reference that moe at constant elocit relatie to each other will measure the same acceleration for a moing particle. Sample Problem Relatie motion, two dimensional, airplanes In Fig. 4-20a, a plane moes due east while the pilot points the plane somewhat south of east, toward a stead wind that blows to the northeast. The plane has elocit PW relatie to the wind, with an airspeed (speed relatie to the wind) of 215 km/h, directed at angle u south of east. The wind has elocit WG relatie to the ground with speed 65.0 km/h, directed 20.0 east of north. What is the magnitude of the elocit PG of the plane relatie to the ground, and what is? Similarl, for the components we find PG, PW, WG,. Here, because PG is parallel to the ais, the component PG, is equal to the magnitude PG. Substituting this notation and the alue u 16.5, we find PG (215 km/h)(cos 16.5 ) (65.0 km/h)(sin 20.0 ) 228 km/h. (Answer) KEY IDEAS The situation is like the one in Fig Here the moing particle P is the plane, frame A is attached to the ground (call it G), and frame B is attached to the wind (call it W).We need a ector diagram like Fig but with three elocit ectors. Calculations First we construct a sentence that relates the three ectors shown in Fig. 4-20b elocit of plane elocit of plane elocit of wind relatie to ground relatie to wind relatie to ground. (PG) (PW) (WG) This relation is written in ector notation as PG PW WG. (4-46) We need to resole the ectors into components on the coordinate sstem of Fig. 4-20b and then sole Eq ais b ais. For the components, we find PG, PW, WG, or 0 (215 km/h) sin u (65.0 km/h)(cos 20.0 ). Soling for u gies us sin 1 (65.0 km/h)(cos 20.0 ) 215 km/h (Answer) N This is the plane's orientation. Fig This is the plane's actual direction of trael. PW PG (a) PW PG (b) A plane fling in a wind. N 20 WG WG E This is the wind direction. The actual direction is the ector sum of the other two ectors (head-to-tail arrangement). Additional eamples, ideo, and practice aailable at WilePLUS

19 76 CHAPTER 4 MTIN IN TW AND THREE DIMENSINS Position Vector The location of a particle relatie to the origin of a coordinate sstem is gien b a position ector r, which in unit-ector notation is r î ĵ zkˆ. (4-1) Here, î ĵ, and zkˆ are the ector components of position ector r, and,, and z are its scalar components (as well as the coordinates of the particle). A position ector is described either b a magnitude and one or two angles for orientation, or b its ector or scalar components. Displacement If a particle moes so that its position ector changes from r to r, the particle s displacement r is (4-2) The displacement can also be written as r ( 2 1 )î ( 2 1 )ĵ (z 2 z 1 )kˆ (4-3) î ĵ z kˆ. (4-4) Aerage Velocit and Instantaneous Velocit If a particle undergoes a displacement r in time interal t, its aerage elocit for that time interal is ag 1 2 r r 2 r 1. Projectile Motion Projectile motion is the motion of a particle that is launched with an initial elocit 0. During its flight, the particle s horizontal acceleration is zero and its ertical acceleration is the free-fall acceleration g. (Upward is taken to be a positie direction.) If 0 is epressed as a magnitude (the speed 0 ) and an angle u 0 (measured from the horizontal), the particle s equations of motion along the horizontal ais and ertical ais are 0 ( 0 cos u 0 )t, (4-21) 0 ( 0 sin 0)t 1 2 gt2, (4-22) 0 sin u 0 gt, (4-23) 2 ( 0 sin 0) 2 2g( 0 ). (4-24) The trajector (path) of a particle in projectile motion is parabolic and is gien b (tan 0) g 2, 2( 0 cos 0) 2 (4-25) if 0 and 0 of Eqs to 4-24 are zero. The particle s horizontal range R, which is the horizontal distance from the launch point to the point at which the particle returns to the launch height, is ag r t. (4-8) R 2 0 g sin 2 0. (4-26) As t in Eq. 4-8 is shrunk to 0, ag reaches a limit called either the elocit or the instantaneous elocit dr dt, which can be rewritten in unit-ector notation as (4-10) (4-11) where d/dt, d/dt, and z dz/dt. The instantaneous elocit of a particle is alwas directed along the tangent to the particle s path at the particle s position. Aerage Acceleration and Instantaneous Acceleration If a particle s elocit changes from 1 to 2 in time interal t, its aerage acceleration during t is (4-15) As t in Eq is shrunk to 0, aag reaches a limiting alue called either the acceleration or the instantaneous acceleration a In unit-ector notation, î ĵ z kˆ, ag a 2 1 t a d dt. a a î a ĵ a z kˆ, where a d /dt, a d /dt, and a z d z /dt. t. (4-16) (4-17) Uniform Circular Motion If a particle traels along a circle or circular arc of radius r at constant speed, it is said to be in uniform circular motion and has an acceleration a of constant magnitude a 2 (4-34) r. The direction of a is toward the center of the circle or circular arc, and a is said to be centripetal. The time for the particle to complete a circle is T 2 r. (4-35) T is called the period of reolution, or simpl the period, of the motion. Relatie Motion When two frames of reference A and B are moing relatie to each other at constant elocit, the elocit of a particle P as measured b an obserer in frame A usuall differs from that measured from frame B. The two measured elocities are related b BA PA PB BA, (4-44) where is the elocit of B with respect to A. Both obserers measure the same acceleration for the particle apa a PB. (4-45)

20 QUESTINS 77 PART 1 1 Figure 4-21 shows the path taken b a skunk foraging for trash food, from initial point i. The skunk took the same time T to go from each labeled point to the net along its path. Rank points a, b, and c according to the magnitude of the aerage elocit of the skunk to reach them from initial point a i b c i, greatest first. Fig Question 1. 2 Figure 4-22 shows the initial position i and the final position f of a particle. What are the (a) initial position ector r i and (b) final position ector r f, both in unit-ector notation? (c) What is the component of displacement r? z 4 m 3 m 3 m 5 m 4 m 1 m 2 m Fig Question 2. f 3 m i angle u between its elocit ector and its acceleration ector during flight. (a) Which of the lettered points on that 2 cure corresponds to the landing of the fruitcake on the ground? (b) Cure 2 is 1 a similar plot for the same launch speed but for a different launch angle. Does the fruitcake now land farther awa or A B closer to the launch point? Fig Question 6. 7 An airplane fling horizontall at a constant speed of 350 km/h oer leel ground releases a bundle of food supplies. Ignore the effect of the air on the bundle. What are the bundle s initial (a) ertical and (b) horizontal components of elocit? (c) What is its horizontal component of elocit just before hitting the ground? (d) If the airplane s speed were, instead, 450 km/h, would the time of fall be longer, shorter, or the same? 8 In Fig. 4-25, a cream tangerine is thrown up past windows 1, 2, and 3, which are identical in size and regularl spaced erticall. Rank those three windows according to (a) the time the cream tangerine takes to pass them and (b) the aerage speed of the cream tangerine during the passage, greatest first. The cream tangerine then moes down past windows 4, 5, and 6, which are identical in size and irregularl spaced horizontall. Rank those three windows according to (c) the time the cream tangerine takes to pass them and (d) the aerage speed of the cream tangerine during the passage, greatest first. 3 When Paris was shelled from 100 km awa with the WWI long-range artiller piece Big Bertha, the shells were fired at an angle greater than 45º to gie them a greater range, possibl een twice as long as at 45º. Does that result mean that the air densit at high altitudes increases with altitude or decreases? 4 You are to launch a rocket, from just aboe the ground, with one of the following initial elocit ectors (1) 0 20î 70ĵ, (2) 0 20î 70ĵ, (3) 0 20î 70ĵ, (4) 0 20î 70ĵ. In our coordinate sstem, runs along leel ground and increases upward. (a) Rank the ectors according to the launch speed of the projectile, greatest first. (b) Rank the ectors according to the time of flight of the projectile, greatest first. 5 Figure 4-23 shows three situations in which identical projectiles are launched (at the same leel) at identical initial speeds and angles. The projectiles do not land on the same terrain, howeer. Rank the situations according to the final speeds of the projectiles just before the land, greatest first. (a) (b) (c) Fig Question 5. 6 The onl good use of a fruitcake is in catapult practice. Cure 1 in Fig gies the height of a catapulted fruitcake ersus the Fig Question 8. 9 Figure 4-26 shows three paths for a football kicked from ground leel. Ignoring the effects of air, rank the paths according to (a) time of flight, (b) initial ertical elocit component, (c) initial horizontal elocit component, and (d) initial speed, greatest first. 10 A ball is shot from ground leel oer leel ground at a certain initial speed. Figure 4-27 gies the range R of the ball ersus its launch angle u 0. Rank the three lettered points on the plot according to (a) the total flight time of the ball and (b) the 4 R Fig Question 9. a b c 0 Fig Question 10.

21 78 CHAPTER 4 MTIN IN TW AND THREE DIMENSINS ball s speed at maimum height, greatest first. 11 Figure 4-28 shows four tracks (either half- or quarter-circles) that can be taken b a train, which moes at a constant speed. Rank the tracks according to the magnitude of a train s acceleration on the cured portion, greatest first. 12 In Fig. 4-29, particle P is in uniform circular motion, centered on the origin of an coordinate sstem. (a) Fig Question 11. At what alues of u is the ertical component r of the position ector greatest in magnitude? (b) At what alues of u is the ertical component of the particle s P elocit greatest in magnitude? (c) r At what alues of u is the ertical component a of the particle s acceleration greatest in magnitude? 13 (a) Is it possible to be accelerating while traeling at constant speed? Is it possible to round a cure with (b) zero Fig Question 12. acceleration and (c) a constant magnitude of acceleration? SSM Tutoring problem aailable (at instructor s discretion) in WilePLUS and WebAssign Worked-out solution aailable in Student Solutions Manual WWW Worked-out solution is at Number of dots indicates leel of problem difficult ILW Interactie solution is at Additional information aailable in The Fling Circus of Phsics and at flingcircusofphsics.com http// sec. 4-2 Position and Displacement 1 The position ector for an electron is r (5.0 m)î (3.0 m)ĵ (2.0 m)kˆ. (a) Find the magnitude of r. (b) Sketch the ector on a right-handed coordinate sstem. 2 A watermelon seed has the following coordinates 5.0 m, 8.0 m, and z 0 m. Find its position ector (a) in unit-ector notation and as (b) a magnitude and (c) an angle relatie to the positie direction of the ais. (d) Sketch the ector on a right-handed coordinate sstem. If the seed is moed to the z coordinates (3.00 m, 0 m, 0 m), what is its displacement (e) in unit-ector notation and as (f) a magnitude and (g) an angle relatie to the positie direction? 3 A positron undergoes a displacement r 2.0î 3.0ĵ 6.0kˆ, ending with the position ector r 3.0ĵ 4.0kˆ, in meters. What was the positron s initial position ector? 4 The minute hand of a wall clock measures 10 cm from its tip to the ais about which it rotates. The magnitude and angle of the displacement ector of the tip are to be determined for three time interals. What are the (a) magnitude and (b) angle from a quarter after the hour to half past, the (c) magnitude and (d) angle for the net half hour, and the (e) magnitude and (f) angle for the hour after that? sec. 4-3 Aerage Velocit and Instantaneous Velocit 5 SSM A train at a constant 60.0 km/h moes east for 40.0 min, then in a direction 50.0 east of due north for 20.0 min, and then west for 50.0 min. What are the (a) magnitude and (b) angle of its aerage elocit during this trip? 6 An electron s position is gien b r 3.00tî 4.00t 2 ĵ 2.00kˆ, with t in seconds and r in meters. (a) In unit-ector notation, what is the electron s elocit (t)? At t 2.00 s, what is (b) in unit-ector notation and as (c) a magnitude and (d) an angle relatie to the positie direction of the ais? 7 An ion s position ector is initiall r 5.0î 6.0ĵ 2.0kˆ, and 10 s later it is r 2.0î 8.0ĵ 2.0kˆ, all in meters. In unitector notation, what is its ag during the 10 s? 8 A plane flies 483 km east from cit A to cit B in 45.0 min and then 966 km south from cit B to cit C in 1.50 h. For the total trip, what are the (a) magnitude and (b) direction of the plane s displacement, the (c) magnitude and (d) direction of its aerage elocit, and (e) its aerage speed? 9 Figure 4-30 gies the path of a squirrel moing (m) 50 about on leel ground, from D point A (at time t 0), to points B (at t 5.00 min), C 25 (at t 10.0 min), and finall D (at t 15.0 min). Consider 0 (m) the aerage elocities of the squirrel from point A to each A C of the other three points. f them, what are the (a) magnitude 25 and (b) angle of the one with the least magnitude and 50 B the (c) magnitude and (d) angle of the one with the greatest magnitude? 10 The position ector Fig Problem 9. r 5.00tî (et ft 2 )ĵ locates 20 a particle as a function of time t. Vector r is in meters, t is in seconds, and factors 0 e and f are constants. Figure 4-31 gies the angle u of the particle s direction of 20 trael as a function of t (u is t (s) measured from the positie Fig direction). What are (a) e Problem 10. and (b) f, including units? sec. 4-4 Aerage Acceleration and Instantaneous Acceleration 11 The position r of a particle moing in an plane is gien b r (2.00t t)î ( t 4 )ĵ, with r in meters and t in seconds. In unit-ector notation, calculate (a) r, (b), and (c) a