(b) A sketch is shown. The coordinate values are in meters.

Transcription

1 Chapter 4. (a) The magnitude of r is r (5. m) ( 3. m) (. m) 6. m. (b) A sketch is shown. The coordinate alues are in meters.. (a) The position ector, according to Eq. 4-, is r = ( 5. m) ˆi + (8. m)j ˆ. (b) The magnitude is r x + y + z ( 5. m) (8. m) ( m) 9.4 m. (c) Many calculators hae polar rectangular conersion capabilities that make this computation more efficient than what is shown below. Noting that the ector lies in the xy plane and using Eq. 3-6, we obtain: 8. m tan 58 or 5. m where the latter possibility ( measured counterclockwise from the +x direction) is chosen since the signs of the components imply the ector is in the second quadrant. (d) The sketch is shown to the right. The ector is counterclockwise from the +x direction. (e) The displacement is r r r where r is gien in part (a) and r (3. m)i. ˆ Therefore, r (8. m)i ˆ (8. m)j ˆ. (f) The magnitude of the displacement is r (8. m) ( 8. m) m. (g) The angle for the displacement, using Eq. 3-6, is 8. m tan = 45 or m 3

2 4 CHAPTER 4 where we choose the former possibility (45, or 45 measured clockwise from +x) since the signs of the components imply the ector is in the fourth quadrant. A sketch of r is shown on the right. 3. The initial position ector r o satisfies r r r, which results in r r r (3.j ˆ 4.k)m ˆ (.i ˆ 3.j ˆ 6.k)m ˆ (. m) ˆi (6. m) ˆj ( m) kˆ. o o 4. We choose a coordinate system with origin at the clock center and +x rightward (toward the 3: position) and +y upward (toward : ). (a) In unit-ector notation, we hae r ˆ ˆ ( cm)i and r ( cm)j. Thus, Eq. 4- gies r r r ( cm)i ˆ ( cm)j. ˆ The magnitude is gien by r ( cm) ( cm) 4 cm. (b) Using Eq. 3-6, the angle is cm tan 45 or 35. cm We choose 35since the desired angle is in the third quadrant. In terms of the magnitude-angle notation, one may write r r r ( cm)i ˆ ( cm)j ˆ (4cm 35 ). (c) In this case, we hae r ˆ ˆ ˆ ( cm)j and r ( cm)j, and r ( cm)j. Thus, r cm. (d) Using Eq. 3-6, the angle is gien by cm tan 9. cm (e) In a full-hour sweep, the hand returns to its starting position, and the displacement is zero. (f) The corresponding angle for a full-hour sweep is also zero.

3 5 5. THINK This problem deals with the motion of a train in two dimensions. The entire trip consists of three parts, and we re interested in the oerall aerage elocity. EXPRESS The aerage elocity of the entire trip is gien by Eq. 4-8, ag r / t, where the total displacement r r r r3 is the sum of three displacements (each result of a constant elocity during a gien time), and t t t t3 is the total amount of time for the trip. We use a coordinate system with +x for East and +y for North. ANALYZE (a) In unit-ector notation, the first displacement is gien by km 4. min = 6. ˆ i = (4. km)i. ˆ h 6 min/h r The second displacement has a magnitude of direction is 4 north of east. Therefore, (6. )( ). km, and its km h. min 6 min/h r (. km) cos(4. )i ˆ (. km) sin(4. ) ˆj (5.3 km)i ˆ (.9 km) ˆj. Similarly, the third displacement is km 5. min r ˆ ˆ 3 6. i = ( 5. km)i. h 6 min/h Thus, the total displacement is r r r r (4. km)i ˆ (5.3 km) ˆi (.9 km) ˆj (5. km) ˆi 3 (5.3 km) ˆi (.9 km) ˆj. The time for the trip is t ( ) min = min, which is equialent to.83 h. Eq. 4-8 then yields ag (5.3 km) ˆi (.9 km) ˆj (.9 km/h) ˆi (7. km/h) ˆj..83 h The magnitude of ag is (b) The angle is gien by ag (.9 km/h) (7. km/h) 7.59 km/h. 7. km/h ag, x.9 km/h or.5 east of due north. ag, y tan tan 67.5 (north of east),

4 6 CHAPTER 4 LEARN The displacement of the train is depicted in the figure below: Note that the net displacement r is found by adding r, r and r ectorially To emphasize the fact that the elocity is a function of time, we adopt the notation (t) for dx / dt. (a) Equation 4- leads to ˆ ˆ ˆ ˆ ( t) d (3.ti 4.t j +.k) (3. m/s)i (8. m/s ) t ˆj dt (b) Ealuating this result at t =. s produces = (3.i ˆ 6.j) ˆ m/s. (c) The speed at t =. s is (3. m/s) ( 6. m/s) 6.3 m/s. (d) The angle of at that moment is 6. m/s tan 79.4 or 3. m/s where we choose the first possibility (79.4 measured clockwise from the +x direction, or 8 counterclockwise from +x) since the signs of the components imply the ector is in the fourth quadrant. 7. Using Eq. 4-3 and Eq. 4-8, we hae ag (.i ˆ + 8.j ˆ.k) ˆ m (5.i ˆ 6.j ˆ +.k) ˆ m (.7i ˆ +.4j ˆ.4k) ˆ m/s. s 8. Our coordinate system has i pointed east and j pointed north. The first displacement is r (483 km)i ˆ and the second is r ( 966 km)j. ˆ AB BC

5 7 (a) The net displacement is r (483 km)i ˆ (966 km)j ˆ AC rab rbc which yields 3 rac (483 km) ( 966 km).8 km. (b) The angle is gien by 966 km tan km We obsere that the angle can be alternatiely expressed as 63.4 south of east, or 6.6 east of south. (c) Diiding the magnitude of r AC by the total time (.5 h) gies ag (483 km)i ˆ(966 km)j ˆ (5 km/h)i ˆ (49 km/h) ˆj.5 h with a magnitude ag (5 km/h) ( 49 km/h) 48 km/h. (d) The direction of ag is 6.6 east of south, same as in part (b). In magnitude-angle notation, we would hae ag (48 km/h 63.4 ). (e) Assuming the AB trip was a straight one, and similarly for the BC trip, then r AB is the distance traeled during the AB trip, and r BC is the distance traeled during the BC trip. Since the aerage speed is the total distance diided by the total time, it equals 483 km 966 km 644 km/h..5 h 9. The (x,y) coordinates (in meters) of the points are A = (5, 5), B = (3, 45), C = (, 5), and D = (45, 45). The respectie times are t A =, t B = 3 s, t C = 6 s, and t D = 9 s. Aerage elocity is defined by Eq Each displacement r is understood to originate at point A. (a) The aerage elocity haing the least magnitude (5. m/6 s) is for the displacement ending at point C:.83 m/s. ag (b) The direction of ag is (measured counterclockwise from the +x axis).

6 8 CHAPTER 4 (c) The aerage elocity haing the greatest magnitude ( for the displacement ending at point B:. m/s. ag (5 m) (3 m) / 3 s ) is (d) The direction of ag is 97 (counterclockwise from +x) or 63 (which is equialent to measuring 63 clockwise from the +x axis).. We differentiate r t ˆ et ft 5. i ( )j. ˆ (a) The particle s motion is indicated by the deriatie of r : = 5. i^ + (e + ft) j^. The angle of its direction of motion is consequently = tan ( y / x ) = tan [(e + ft)/5.]. The graph indicates o = 35., which determines the parameter e: e = (5. m/s) tan(35.) = 3.5 m/s. (b) We note (from the graph) that = when t = 4. s. Thus, e + ft = at that time. This determines the parameter f : f e 3.5 m/s t (4. s).5 m/s.. In parts (b) and (c), we use Eq. 4- and Eq For part (d), we find the direction of the elocity computed in part (b), since that represents the asked-for tangent line. (a) Plugging into the gien expression, we obtain r [.(8) 5.()]i ˆ + [6. 7.(6)] ˆj (6. ˆi 6ˆ j) m t. (b) Taking the deriatie of the gien expression produces t t ˆ t 3 ( ) = (6. 5.) i 8. j ˆ where we hae written (t) to emphasize its dependence on time. This becomes, at t =. s, = (9. ˆi 4 ˆj) m/s. (c) Differentiating the ( t) found aboe, with respect to t produces which yields a =(4.i ˆ 336 ˆj) m/s at t =. s. ˆ ˆ.ti 84.t j, (d) The angle of, measured from +x, is either

7 9 4 m/s tan 85. or m/s where we settle on the first choice ( 85., which is equialent to 75 measured counterclockwise from the +x axis) since the signs of its components imply that it is in the fourth quadrant.. We adopt a coordinate system with i pointed east and j pointed north; the coordinate origin is the flagpole. We translate the gien information into unit-ector notation as follows: r (4. m)i ˆ and = (. m/s)j ˆ (a) Using Eq. 4-, the displacement r is o r (4. m) ˆj and (. m/s)i. ˆ r r r ( 4. m)i ˆ (4. m) ˆj o o with a magnitude r ( 4. m) (4. m) 56.6 m. (b) The direction of r is y 4. m tan tan 45. or 35. x 4. m Since the desired angle is in the second quadrant, we pick 35 ( 45 north of due west). r r r o in terms of the Note that the displacement can be written as magnitude-angle notation. (c) The magnitude of ag is simply the magnitude of the displacement diided by the time (t = 3. s). Thus, the aerage elocity has magnitude (56.6 m)/(3. s) =.89 m/s. (d) Equation 4-8 shows that ag points in the same direction as r, that is, 35 ( 45 north of due west). (e) Using Eq. 4-5, we hae a ag t o (.333 m/s )i (.333 m/s )j. ˆ ˆ The magnitude of the aerage acceleration ector is therefore equal to (.333 m/s ) (.333 m/s ).47 m/s aag.

8 3 CHAPTER 4 (f) The direction of a ag is.333 m/s tan 45 or m/s Since the desired angle is now in the first quadrant, we choose 45, and a ag points north of due east. 3. THINK Knowing the position of a particle as function of time allows us to calculate its corresponding elocity and acceleration by taking time deriaties. EXPRESS From the position ector rt, () the elocity and acceleration of the particle can be found by differentiating rt () with respect to time: dr d d r, a. dt dt dt ANALYZE (a) Taking the deriatie of the position ector respect to time, we hae, in SI units (m/s), ˆ ˆ ˆ r( t) i (4 t )j tk with d ˆ (i 4t ˆj t k) ˆ 8t ˆj k. ˆ dt (b) Taking another deriatie with respect to time leads to, in SI units (m/s ), d a (8tˆj k) ˆ 8ˆj. dt LEARN The particle undergoes constant acceleration in the +y-direction. This can be seen by noting that the y component of rt () is 4t, which is quadratic in t. 4. We use Eq. 4-5 with designating the initial elocity and designating the later one. (a) The aerage acceleration during the t = 4 s interal is a ag (.i ˆ. ˆj+5.k) ˆ m/s (4.i ˆˆj+3.k) ˆ m/s ˆ (.5 m/s )i (.5m/s )k. ˆ 4s (b) The magnitude of a ag is (.5 m/s ) (.5 m/s ).6m/s. (c) Its angle in the xz plane (measured from the +x axis) is one of these possibilities:

9 3.5 m/s.5 m/s tan 8 or 6 where we settle on the second choice since the signs of its components imply that it is in the second quadrant. 5. THINK Gien the initial elocity and acceleration of a particle, we re interested in finding its elocity and position at a later time. EXPRESS Since the acceleration, ˆ ˆ ˆ i j (. m/s )i (.5 m/s )j ˆ x y a a a, is constant in both x and y directions, we may use Table - for the motion along each direction. This can be handled indiidually (for x and y) or together with the unit-ector notation (for r ). Since the particle started at the origin, the coordinates of the particle at any time t are gien by r t at. The elocity of the particle at any time t is gien by at, where is the initial elocity and a is the (constant) acceleration. Along the x-direction, we hae x( t) xt axt, x( t) x axt Similarly, along the y-direction, we get y( t) yt ayt, y( t) y ayt. Known: 3. m/s,, a. m/s, a.5 m/s. x y x y ANALYZE (a) Substituting the alues gien, the components of the elocity are t a t t t a t t x( ) x x (3. m/s) (. m/s ) y ( ) y y (.5 m/s ) When the particle reaches its maximum x coordinate at t = t m, we must hae x =. Therefore, 3..t m = or t m = 3. s. The y component of the elocity at this time is Thus, t (.5 m/s)j ˆ. m y ( 3. s) (.5 m/s )(3.).5 m/s (b) At t = 3. s, the components of the position are

10 3 CHAPTER 4 x t t a t y t t a t ( 3. s) x x (3. m/s)(3. s) (. m/s )(3. s) 4.5 m ( 3. s) y y (.5 m/s )(3. s).5 m Using unit-ector notation, the results can be written as r (4.5 m) ˆi (.5 m) ˆj. LEARN The motion of the particle in this problem is two-dimensional, and the kinematics in the x- and y-directions can be analyzed separately. 6. We make use of Eq (a) The acceleration as a function of time is d d a 6.t 4.t ˆ i + 8. ˆ j 6. 8.t ˆ i dt dt m in SI units. Specifically, we find the acceleration ector at 6. 8.(3.) ˆi ( 8 m/s )i. ˆ t 3. s to be (b) The equation is a b tgi = ; we find t =.75 s. (c) Since the y component of the elocity, y = 8. m/s, is neer zero, the elocity cannot anish. (d) Since speed is the magnitude of the elocity, we hae 6.t 4.t 8. in SI units (m/s). To sole for t, we first square both sides of the aboe equation, followed by some rearrangement: t t t t Taking the square root of the new expression and making further simplification lead to 6.t 4.t 6. 4.t 6.t 6. Finally, using the quadratic formula, we obtain

11 t 8. where the requirement of a real positie result leads to the unique answer: t =. s. 7. We find t by applying Eq. - to motion along the y axis (with y = characterizing y = y max ): = ( m/s) + (. m/s )t t = 6. s. Then, Eq. - applies to motion along the x axis to determine the answer: x = (8. m/s) + (4. m/s )(6. s) = 3 m/s. Therefore, the elocity of the cart, when it reaches y = y max, is (3 m/s)i^. 8. We find t by soling x x xt axt :. m (4. m/s) t (5. m/s ) t where we hae used x =. m, x = 4. m/s, and a x = 5. m/s. We use the quadratic formula and find t =.53 s. Then, Eq. - (actually, its analog in two dimensions) applies with this alue of t. Therefore, its elocity (when x =. m) is ˆ ˆ ˆ at (4. m/s)i (5. m/s )(.53 s)i (7. m/s )(.53 s)j (.7 m/s) ˆi (.7 m/s) ˆj. Thus, the magnitude of is (b) The angle of, measured from +x, is 9. We make use of Eq. 4-6 and Eq. 4-. Using a 3t ˆi 4tˆj, we hae (in m/s) (.7 m/s) (.7 m/s) 5.8 m/s..7 m/s tan m/s t t ( t) a dt (5.i ˆ.j) ˆ (3tˆi 4tˆj) dt 5. 3 t / iˆ. t ˆj Integrating using Eq. 4- then yields (in meters)

12 34 CHAPTER 4 t t ˆ ˆ ˆ ˆ ˆ ˆ 3 ˆ 3 ˆ r( t) r dt (.i 4. j) [(5. 3 t / )i (. t )j] dt (.i 4.j) (5. t t /)i (.t t /3)j 3 ˆ 3 (.5. t t /)i (4..t t /3)j ˆ (a) At t 4. s, we hae rt ( 4. s) (7. m)i ˆ (9.7 m) ˆj. (b) t ( 4. s) (9. m/s)i ˆ (34. m/s) ˆj. Thus, the angle between the direction of trael and +x, measured counterclockwise, is tan [(34. m/s)/(9. m/s)] The acceleration is constant so that use of Table - (for both the x and y motions) is permitted. Where units are not shown, SI units are to be understood. Collision between particles A and B requires two things. First, the y motion of B must satisfy (using Eq. -5 and noting that is measured from the y axis) y ayt 3 m (.4 m/s ) cos t. Second, the x motions of A and B must coincide: t axt (3. m/s) t (.4 m/s ) sin t. We eliminate a factor of t in the last relationship and formally sole for time: (3. m/s) t a x (.4 m/s ) sin. This is then plugged into the preious equation to produce (3. m/s) 3 m (.4 m/s ) cos (.4 m/s ) sin which, with the use of sin = cos, simplifies to 9. cos 9.. cos. 3 3 cos cos. We use the quadratic formula (choosing the positie root) to sole for cos : cos

13 35 F which yields H G I K J 6 cos.. We adopt the positie direction choices used in the textbook so that equations such as Eq. 4- are directly applicable. The initial elocity is horizontal so that y and m s. x (a) With the origin at the initial point (where the dart leaes the thrower s hand), the y coordinate of the dart is gien by y gt, so that with y = PQ we hae PQ 9.8 m/s.9 s.8 m. (b) From x = t we obtain x = ( m/s)(.9 s) =.9 m.. We adopt the positie direction choices used in the textbook so that equations such as Eq. 4- are directly applicable. (a) With the origin at the initial point (edge of table), the y coordinate of the ball is gien by y gt. If t is the time of flight and y =. m indicates the leel at which the ball hits the floor, then t. m.495s. 9.8 m/s (b) The initial (horizontal) elocity of the ball is i. Since x =.5 m is the horizontal position of its impact point with the floor, we hae x = t. Thus, x.5 m 3.7 m/s. t.495 s 3. (a) From Eq. 4- (with = ), the time of flight is h (45. m) t 3.3 s. g 9.8 m/s (b) The horizontal distance traeled is gien by Eq. 4-: x t (5 m/s)(3.3 s) 758 m. (c) And from Eq. 4-3, we find y gt (9.8 m/s )(3.3 s) 9.7 m/s.

14 36 CHAPTER 4 4. We use Eq. 4-6 R 9.5m/s sin 9.9 m 9.m max g g 9.8m/s max to compare with Powell s long jump; the difference from R max is only R =(9.m 8.95m) =.59 m. 5. Using Eq. (4-6), the take-off speed of the jumper is gr (9.8 m/s )(77. m) 43. m/s sin sin (. ) 6. We adopt the positie direction choices used in the textbook so that equations such as Eq. 4- are directly applicable. The coordinate origin is the throwing point (the stone s initial position). The x component of its initial elocity is gien by cos and the y component is gien by x sin, where = m/s is the initial speed and = y 4. is the launch angle. (a) At t =. s, its x coordinate is (b) Its y coordinate at that instant is b gb g x t cos. m/ s. s cos m y t gt sin.m/s.s sin m/s.s 8. m. (c) At t' =.8 s, its x coordinate is x (d) Its y coordinate at t' is b gb g. m / s 8. s cos m. y.m/s.8s sin m/s.8s 7.6m. (e) The stone hits the ground earlier than t = 5. s. To find the time when it hits the ground sole y t sin gt for t. We find b g t g. m / s sin sin. 6 s.. m / s

15 37 Its x coordinate on landing is x t cos. m/s.6 s cos 4 4. m. (f) Assuming it stays where it lands, its ertical component at t = 5. s is y =. 7. We adopt the positie direction choices used in the textbook so that equations such as Eq. 4- are directly applicable. The coordinate origin is at ground leel directly below the release point. We write = 3. since the angle shown in the figure is measured clockwise from horizontal. We note that the initial speed of the decoy is the plane s speed at the moment of release: = 9 km/h, which we conert to SI units: (9)(/36) = 8.6 m/s. (a) We use Eq. 4- to sole for the time: 7 m x ( cos ) t t. s. (8.6 m/s) cos( 3. ) (b) And we use Eq. 4- to sole for the initial height y : y y ( sin ) t gt y ( 4.3 m/s)(. s) (9.8 m/s )(. s) which yields y = 897 m. 8. (a) Using the same coordinate system assumed in Eq. 4-, we sole for y = h: h y sint gt which yields h = 5.8 m for y =, = 4. m/s, = 6., and t = 5.5 s. (b) The horizontal motion is steady, so x = x = cos, but the ertical component of elocity aries according to Eq Thus, the speed at impact is cos sin gt 7.4 m/s. (c) We use Eq. 4-4 with y = and y = H: b sin g H m. g

16 38 CHAPTER 4 9. We adopt the positie direction choices used in the textbook so that equations such as Eq. 4- are directly applicable. The coordinate origin is at its initial position (where it is launched). At maximum height, we obsere y = and denote x = (which is also equal to x ). In this notation, we hae 5. Next, we obsere cos = x =, so that we arrie at an equation (where cancels) which can be soled for : 5 (5 )cos cos Although we could use Eq. 4-6 to find where it lands, we choose instead to work with Eq. 4- and Eq. 4- (for the soccer ball) since these will gie information about where and when and these are also considered more fundamental than Eq With y =, we hae (9.5 m/s)sin 45. y ( sin ) t gt t.8 s. (9.8 m/s ) / Then Eq. 4- yields x = ( cos )t = 38.7 m. Thus, using Eq. 4-8, the player must hae an aerage elocity of ag r (38.7 m) ˆi (55 m)i ˆ ( 5.8 m/s) ˆi t.8s which means his aerage speed (assuming he ran in only one direction) is 5.8 m/s. 3. We first find the time it takes for the olleyball to hit the ground. Using Eq. 4-, we hae y y ( sin ) t gt.3 m (. m/s)sin(8. ) t (9.8 m/s ) t which gies t.3 s. Thus, the range of the olleyball is R cos t (. m/s) cos8. (.3 s) 5.7 m On the other hand, when the angle is changed to 8., using the same procedure as shown aboe, we find y y ( sin ) t gt.3 m (. m/s)sin(8. ) t (9.8 m/s ) t which yields t.46 s, and the range is R cos t (. m/s) cos8. (.46 s) 9.6 m

17 39 Thus, the ball traels an extra distance of R R R 9.6 m5.7 m 3.35 m 3. We adopt the positie direction choices used in the textbook so that equations such as Eq. 4- are directly applicable. The coordinate origin is at the release point (the initial position for the ball as it begins projectile motion in the sense of 4-5), and we let be the angle of throw (shown in the figure). Since the horizontal component of the elocity of the ball is x = cos 4., the time it takes for the ball to hit the wall is (a) The ertical distance is x. m t.5 s. (5. m/s) cos 4. x y t gt ( sin ) (5. m/s)sin 4. (.5 s) (9.8 m/s )(.5 s). m. (b) The horizontal component of the elocity when it strikes the wall does not change from its initial alue: x = cos 4. = 9. m/s. (c) The ertical component becomes (using Eq. 4-3) gt y sin (5. m/s) sin 4. (9.8 m/s )(.5 s) 4.8 m/s. (d) Since y > when the ball hits the wall, it has not reached the highest point yet. 33. THINK This problem deals with projectile motion. We re interested in the horizontal displacement and elocity of the projectile before it strikes the ground. EXPRESS We adopt the positie direction choices used in the textbook so that equations such as Eq. 4- are directly applicable. The coordinate origin is at ground leel directly below the release point. We write = 37. for the angle measured from +x, since the angle 53.gien in the problem is measured from the y direction. The initial setup of the problem is shown in the figure to the right (not to scale). ANALYZE (a) The initial speed of the projectile is the plane s speed at the moment of release. Gien that y 73 m and y at t 5. s, we use Eq. 4- to find :

18 4 CHAPTER 4 y y ( sin ) t gt 73 m sin( 37. )(5. s) (9.8 m/s )(5. s) which yields = m/s. (b) The horizontal distance traeled is R t ( cos ) t [( m/s)cos( 37. )](5. s) 86 m. x (c) The x component of the elocity (just before impact) is cos ( m/s)cos( 37. ) 6 m/s. x (d) The y component of the elocity (just before impact) is gt. y sin ( m/s)sin( 37. ) (9.8 m/s )(5. s) 7 m/s LEARN In this projectile problem we analyzed the kinematics in the ertical and horizontal directions separately since they do not affect each other. The x-component of the elocity, x cos, remains unchanged throughout since there s no horizontal acceleration. 34. (a) Since the y-component of the elocity of the stone at the top of its path is zero, its speed is cos (8. m/s)cos4..4 m/s. x y x (b) Using the fact that y at the maximum height y max, the amount of time it takes for the stone to reach y max is gien by Eq. 4-3: sin sin g y gt t. Substituting the aboe expression into Eq. 4-, we find the maximum height to be sin sin sin max y ( sin ) t gt sin g. g g g To find the time the stone descends to y y / max, we sole the quadratic equation gien in Eq. 4-: sin ( ) sin y ymax ( sin ) t gt t. 4g g

19 4 Choosing t t (for descending), we hae x cos (8. m/s) cos 4..4 m/s ( ) sin y sin g sin (8. m/s)sin 4..7 m/s g Thus, the speed of the stone when y y / is max (.4 m/s) (.7 m/s) 4.9 m/s. x y (c) The percentage difference is 4.9 m/s.4 m/s %..4 m/s 35. THINK This problem deals with projectile motion of a bullet. We re interested in the firing angle that allows the bullet to strike a target at some distance away. EXPRESS We adopt the positie direction choices used in the textbook so that equations such as Eq. 4- are directly applicable. The coordinate origin is at the end of the rifle (the initial point for the bullet as it begins projectile motion in the sense of 4-5), and we let be the firing angle. If the target is a distance d away, then its coordinates are x = d, y =. The projectile motion equations lead to d ( cos ) t, t sin gt where is the firing angle. The setup of the problem is shown in the figure aboe (scale exaggerated). ANALYZE The time at which the bullet strikes the target is gien by t d /( cos ). Eliminating t leads to sin cos gd. Using sin cos sinb g, we obtain

20 4 CHAPTER 4 gd (9.8 m/s )(45.7 m) sin ( ) gd sin( ) (46 m/s) 3 which yields sin( )., or =.66. If the gun is aimed at a point a distance aboe the target, then tan d so that d tan (45.7 m) tan(.66 ).484 m 4.84 cm. LEARN Due to the downward graitational acceleration, in order for the bullet to strike the target, the gun must be aimed at a point slightly aboe the target. 36. We adopt the positie direction choices used in the textbook so that equations such as Eq. 4- are directly applicable. The coordinate origin is at ground leel directly below the point where the ball was hit by the racquet. (a) We want to know how high the ball is aboe the court when it is at x =. m. First, Eq. 4- tells us the time it is oer the fence: x. m t.58 s. cos 3.6 m/s cos At this moment, the ball is at a height (aboe the court) of y y sin t gt.m which implies it does indeed clear the.9-m-high fence. (b) At t =.58 s, the center of the ball is (. m.9 m) =. m aboe the net. (c) Repeating the computation in part (a) with = 5. results in t =.5 s and y.4 m, which clearly indicates that it cannot clear the net. (d) In the situation discussed in part (c), the distance between the top of the net and the center of the ball at t =.5 s is.9 m.4 m =.86 m. 37. THINK The trajectory of the dier is a projectile motion. We are interested in the displacement of the dier at a later time. EXPRESS The initial elocity has no ertical component ( ), but only an x component. Eqs. 4- and 4- can be simplified to

21 43 x x xt y y yt gt gt. where x, x. m/s and y = +. m (taking the water surface to be at y ). The setup of the problem is shown in the figure below. ANALYZE (a) At t.8 s, the horizontal distance of the dier from the edge is x x xt (. m/s)(.8 s).6 m. (b) Similarly, using the second equation for the ertical motion, we obtain y y gt. m (9.8 m/s )(.8 s) 6.86 m. (c) At the instant the dier strikes the water surface, y =. Soling for t using the equation y y gt leads to t y (. m) g 9.8 m/s.43 s. During this time, the x-displacement of the dier is R = x = (. m/s)(.43 s) =.86 m. LEARN Using Eq. 4-5 with, the trajectory of the dier can also be written as gx y y. Part (c) can also be soled by using this equation: gx y (. m/s) (. m) y y x R.86 m. g 9.8 m/s

22 44 CHAPTER In this projectile motion problem, we hae = x = constant, and what is plotted is We infer from the plot that at t =.5 s, the ball reaches its maximum height, x y. where y =. Therefore, we infer from the graph that x = 9 m/s. (a) During t = 5 s, the horizontal motion is x x = x t = 95 m. (b) Since (9 m/s) y 3 m/s (the first point on the graph), we find y 4.5 m/s. Thus, with t =.5 s, we can use ymax y yt gt or y g y y y b max g, or max y y y y t to sole. Here we will use the latter: ymax y ( y y) t ymax ( 4.5m/s)(.5 s) 3 m where we hae taken y = as the ground leel. 39. Following the hint, we hae the time-reersed problem with the ball thrown from the ground, toward the right, at 6 measured counterclockwise from a rightward axis. We see in this time-reersed situation that it is conenient to use the familiar coordinate system with +x as rightward and with positie angles measured counterclockwise. (a) The x-equation (with x = and x = 5. m) leads to 5. m = ( cos 6. )(.5 s), so that = 33.3 m/s. And with y =, and y = h > at t =.5 s, we hae y y t gt where y = sin 6.. This leads to h = 3.3 m. y (b) We hae x = x = (33.3 m/s)cos 6. = 6.7 m/s y = y gt = (33.3 m/s)sin 6. (9.8 m/s )(.5 s) = 4. m/s. The magnitude of is gien by x y (6.7 m/s) (4. m/s).9 m/s. (c) The angle is y 4. m/s tan tan 4.4. x 6.7 m/s (d) We interpret this result ( undoing the time reersal) as an initial elocity (from the edge of the building) of magnitude.9 m/s with angle (down from leftward) of 4.4.

23 45 4. (a) Soling the quadratic equation Eq. 4-: y y ( sin ) t gt.6 m (5. m/s)sin(45. ) t (9.8 m/s ) t the total trael time of the shot in the air is found to be horizontal distance traeled is t.35 s R cos t (5. m/s) cos 45. (.35 s) 4.95 m. (b) Using the procedure outlined in (a) but for 4., we hae. Therefore, the y y ( sin ) t gt.6 m (5. m/s)sin(4. ) t (9.8 m/s ) t and the total trael time is t.45 s. This gies R cos t (5. m/s) cos 4. (.45 s) 5. m. 4. With the Archer fish set to be at the origin, the position of the insect is gien by (x, y) where x R / sin / g, and y corresponds to the maximum height of the parabolic trajectory: y y sin / g. From the figure, we hae max y sin / g tan tan x sin / g Gien that 36., we find the launch angle to be. tan tan tan tan 36. tan Note that depends only on and is independent of d. 4. (a) Using the fact that the person (as the projectile) reaches the maximum height oer the middle wheel located at x 3 m (3/ ) m 34.5 m, we can deduce the initial launch speed from Eq. 4-6: R sin gx (9.8 m/s )(34.5 m) x 6.5 m/s sin sin(53 ). g Upon substituting the alue to Eq. 4-5, we obtain

24 46 CHAPTER 4 gx (9.8 m/s )(3 m) y y x tan 3. m (3 m) tan m. cos (6.5 m/s) (cos 53 ) Since the height of the wheel is hw 8 m, y y h w 3.3 m 8 m 5.3 m. the clearance oer the first wheel is (b) The height of the person when he is directly aboe the second wheel can be found by soling Eq With the second wheel located at x 3 m (3/ ) m 34.5 m, we hae gx (9.8 m/s )(34.5 m) y y x tan 3. m (34.5 m) tan 53 cos (6.5 m/s) (cos 53 ) 5.9 m. Therefore, the clearance oer the second wheel is y y h w 5.9 m 8 m 7.9 m. (c) The location of the center of the net is gien by gx sin (6.5 m/s) sin(53 ) y y x tan x 69 m. cos g 9.8 m/s 43. We designate the gien elocity (7.6 m/s)i ˆ (6. m/s) ˆj as, as opposed to the elocity when it reaches the max height or the elocity when it returns to the ground, and take 3 as the launch elocity, as usual. The origin is at its launch point on the ground. (a) Different approaches are aailable, but since it will be useful (for the rest of the problem) to first find the initial y elocity, that is how we will proceed. Using Eq. -6, we hae gy (6. m/s) (9.8 m/s )(9. m) y y y which yields y = 4.7 m/s. Knowing that y must equal, we use Eq. -6 again but now with y = h for the maximum height: yy gh (4.7 m/s) (9.8 m/s ) h which yields h = m. (b) Recalling the deriation of Eq. 4-6, but using y for sin and x for cos, we hae yt gt, R xt

25 47 which leads to R xy / g. Noting that x = x = 7.6 m/s, we plug in alues and obtain R = (7.6 m/s)(4.7 m/s)/(9.8 m/s ) = 3 m. (c) Since 3x = x = 7.6 m/s and 3y = y = 4.7 m/s, we hae 3 3 x 3 y (7.6 m/s) ( 4.7 m/s) 7 m/s. (d) The angle (measured from horizontal) for 3 is one of these possibilities: 4.7 m tan 63 or m where we settle on the first choice ( 63, which is equialent to 97 ) since the signs of its components imply that it is in the fourth quadrant. 44. We adopt the positie direction choices used in the textbook so that equations such as Eq. 4- are directly applicable. The initial elocity is horizontal so that y and 6 km h. Conerting to SI units, this is = 44.7 m/s. x (a) With the origin at the initial point (where the ball leaes the pitcher s hand), the y coordinate of the ball is gien by y gt, and the x coordinate is gien by x = t. From the latter equation, we hae a simple proportionality between horizontal distance and time, which means the time to trael half the total distance is half the total time. Specifically, if x = 8.3/ m, then t = (8.3/ m)/(44.7 m/s) =.5 s. (b) And the time to trael the next 8.3/ m must also be.5 s. It can be useful to write the horizontal equation as x = t in order that this result can be seen more clearly. (c) Using the equation y we see that the ball has reached the height of gt, 9.8 m/s.5 s.5 m at the moment the ball is halfway to the batter. (d) The ball s height when it reaches the batter is 9.8 m/s.49 s.8m, which, when subtracted from the preious result, implies it has fallen another.65 m. Since the alue of y is not simply proportional to t, we do not expect equal time-interals to correspond to equal height-changes; in a physical sense, this is due to the fact that the initial y-elocity for the first half of the motion is not the same as the initial y-elocity for the second half of the motion.

26 48 CHAPTER (a) Let m = d d =.6 be the slope of the ramp, so y = mx there. We choose our coordinate origin at the point of launch and use Eq Thus, (9.8 m/s ) x y tan(5. ) x.6x (. m/s) (cos5. ) which yields x = 4.99 m. This is less than d so the ball does land on the ramp. (b) Using the alue of x found in part (a), we obtain y = mx =.99 m. Thus, the Pythagorean theorem yields a displacement magnitude of x + y = 5.8 m. (c) The angle is, of course, the angle of the ramp: tan (m) = 3.º. 46. Using the fact that y when the player is at the maximum height y max, the amount of time it takes to reach y max can be soled by using Eq. 4-3: sin sin g y gt tmax. Substituting the aboe expression into Eq. 4-, we find the maximum height to be sin sin sin max max max y ( sin ) t gt sin g. g g g To find the time when the player is at y y / max, we sole the quadratic equation gien in Eq. 4-: sin ( ) sin y ymax ( sin ) t gt t. 4g g With t t (for ascending), the amount of time the player spends at a height y y / max is sin ( ) sin sin t t max t tmax t.77. g g g tmax Therefore, the player spends about 7.7% of the time in the upper half of the jump. Note that the ratio t/ tmax is independent of and, een though t and t max depend on these quantities. 47. THINK The baseball undergoes projectile motion after being hit by the batter. We d like to know if the ball clears a high fence at some distance away.

27 49 EXPRESS We adopt the positie direction choices used in the textbook so that equations such as Eq. 4- are directly applicable. The coordinate origin is at ground leel directly below impact point between bat and ball. In the absence of a fence, with 45, the horizontal range (same launch leel) is R 7 m. We want to know how high the ball is from the ground when it is at x 97.5 m, which requires knowing the initial elocity. The trajectory of the baseball can be described by Eq. 4-5: gx y y (tan ) x. ( cos ) The setup of the problem is shown in the figure below (not to scale). ANALYZE (a) We first sole for the initial speed. Using the range information y y when x R) and = 45, Eq. 4-5 gies ( gr 9.8 m/s 7 m 3.4 m/s. sin sin(45 ) Thus, the time at which the ball flies oer the fence is: x 97.5 m x ( cos ) t t 4.6 s. cos 3.4 m/s cos 45 At this moment, the ball is at a height (aboe the ground) of y y sin t gt. m [(3.4 m/s)sin 45 ](4.6 s) (9.8 m/s )(4.6 s) 9.88 m which implies it does indeed clear the 7.3 m high fence. (b) At t 4.6 s, the center of the ball is 9.88 m 7.3 m =.56 m aboe the fence.

28 5 CHAPTER 4 LEARN Using the trajectory equation aboe, one can show that the minimum initial elocity required to clear the fence is gien by gx y y (tan ) x, ( cos ) or about 3.9 m/s. 48. Following the hint, we hae the time-reersed problem with the ball thrown from the roof, toward the left, at 6 measured clockwise from a leftward axis. We see in this time-reersed situation that it is conenient to take +x as leftward with positie angles measured clockwise. Lengths are in meters and time is in seconds. (a) With y =. m, and y = at t = 4. s, we hae y y t gt y where y sin 6. This leads to = 6.9 m/s. This plugs into the x-equation x x xt (with x = and x = d) to produce d = (6.9 m/s)cos 6 (4. s) = 33.7 m. (b) We hae x x (6.9 m/s) cos m/s gt (6.9 m/s)sin 6. (9.8m/s )(4. s) 4.6 m/s. y y The magnitude of is x y (8.43 m/s) ( 4.6 m/s) 6. m/s. (c) The angle relatie to horizontal is y 4.6 m/s tan tan 7.. x 8.43 m/s We may conert the result from rectangular components to magnitude-angle representation: (8.43, 4.6) (6. 7. ) and we now interpret our result ( undoing the time reersal) as an initial elocity of magnitude 6. m/s with angle (up from rightward) of THINK In this problem a football is gien an initial speed and it undergoes projectile motion. We d like to know the smallest and greatest angles at which a field goal can be scored. EXPRESS We adopt the positie direction choices used in the textbook so that equations such as Eq. 4- are directly applicable. The coordinate origin is at the point where the ball is kicked. We use x and y to denote the coordinates of the ball at the goalpost, and try to find the kicking angle(s) so that y = 3.44 m when x = 5 m. Writing the kinematic equations for projectile motion:

29 5 x y t gt cos, sin, we see the first equation gies t = x/ cos, and when this is substituted into the second the result is gx y x tan. cos ANALYZE One may sole the aboe equation by trial and error: systematically trying alues of until you find the two that satisfy the equation. A little manipulation, howeer, will gie an algebraic solution: Using the trigonometric identity we obtain gx tan / cos = + tan, gx x tan y which is a second-order equation for tan. To simplify writing the solution, we denote c gx / 9.8 m/s 5 m / 5 m/s 9.6 m. Then the second-order equation becomes c tan x tan + y + c =. Using the quadratic formula, we obtain its solution(s). x x 4 y c c 5 m (5 m) m 9.6 m 9.6 m tan. c 9.6 m The two solutions are gien by tan =.95 and tan =.65. The corresponding (firstquadrant) angles are = 63 and = 3. Thus, (a) The smallest eleation angle is = 3, and (b) The greatest eleation angle is = 63. LEARN If kicked at any angle between 3 and 63, the ball will trael aboe the cross bar on the goalposts. 5. We apply Eq. 4-, Eq. 4-, and Eq (a) From x t x, we find x 4 m/ s m/s. (b) From y t gt y 53 m (9.8 m/s )( s) / 36 m/s., we find y

30 5 CHAPTER 4 (c) From y gt with y = as the condition for maximum height, we obtain y t (36 m/s) /(9.8 m/s ) 3.7 s. During that time the x-motion is constant, so x x ( m/s)(3.7 s) 74 m. 5. (a) The skier jumps up at an angle of.3 up from the horizontal and thus returns to the launch leel with his elocity ector.3 below the horizontal. With the snow surface making an angle of 9. (downward) with the horizontal, the angle between the slope and the elocity ector is (b) Suppose the skier lands at a distance d down the slope. Using Eq. 4-5 with x dcos and y dsin (the edge of the track being the origin), we hae Soling for d, we obtain gd ( cos ) dsin dcos tan. cos cos cos d gcos gcos cos sin( ). g cos Substituting the alues gien, we find cos tan sin cos sin cos sin ( m/s) cos(.3 ) sin(.3 9. ) 7.7 m. d (9.8 m/s )cos (9. ) which gies y d sin (7.7 m)sin(9. ). m. Therefore, at landing the skier is approximately. m below the launch leel. (c) The time it takes for the skier to land is x dcos (7.7 m) cos(9. ) t.7 s. cos ( m/s) cos(.3 ) x Using Eq. 4-3, the x-and y-components of the elocity at landing are x cos ( m/s) cos(.3 ) 9.8 m/s gt y sin ( m/s)sin(.3 ) (9.8 m/s )(.7 s) 5.7 m/ s

31 53 Thus, the direction of trael at landing is y 5.7 m/s tan tan 7.3. x 9.8 m/s or 7.3 below the horizontal. The result implies that the angle between the skier s path and the slope is , or approximately 8 to two significant figures. 5. From Eq. 4-, we find t x / x. Then Eq. 4-3 leads to gx y y gt y. x Since the slope of the graph is.5, we conclude g x ox = 9.6 m/s. And from the y intercept of the graph, we find oy = 5. m/s. Consequently, o = tan ( oy ox ) = Let y = h =. m at x = when the ball is hit. Let y = h (the height of the wall) and x describe the point where it first rises aboe the wall one second after being hit; similarly, y = h and x describe the point where it passes back down behind the wall four seconds later. And y f =. m at x f = R is where it is caught. Lengths are in meters and time is in seconds. (a) Keeping in mind that x is constant, we hae x x = 5. m = x (4. s), which leads to x =.5 m/s. Thus, applied to the full six seconds of motion: x f x = R = x (6. s) = 75. m. (b) We apply y y t gt y to the motion aboe the wall, y y y 4. s g 4. s and obtain y = 9.6 m/s. One second earlier, using y = y g(. s), we find y 9.4 m/s. Therefore, the elocity of the ball just after being hit is ˆi ˆj (.5 m/s) ˆi (9.4 m/s) ˆj x y

32 54 CHAPTER 4 Its magnitude is (.5 m/s) +(9.4 m/s) 3.9 m/s. (c) The angle is y 9.4 m/s tan tan 67.. x.5 m/s We interpret this result as a elocity of magnitude 3.9 m/s, with angle (up from rightward) of 67.. (d) During the first. s of motion, y y t gt yields y h. m 9.4 m/s. s 9.8 m/s. s 5.5 m. 54. For y =, Eq. 4- leads to t = o sin o /g, which immediately implies t max = o /g (which occurs for the straight up case: o = 9). Thus, t max = o /g = sin o. Therefore, the half-maximum-time flight is at angle o = 3.. Since the least speed occurs at the top of the trajectory, which is where the elocity is simply the x-component of the initial elocity ( o cos o = o cos3 for the half-maximum-time flight), then we need to refer to the graph in order to find o in order that we may complete the solution. In the graph, we note that the range is 4 m when o = 45.. Equation 4-6 then leads to o = 48.5 m/s. The answer is thus (48.5 m/s)cos3. = 4. m/s. 55. THINK In this problem a ball rolls off the top of a stairway with an initial speed, and we d like to know on which step it lands first. EXPRESS We denote h as the height of a step and w as the width. To hit step n, the ball must fall a distance nh and trael horizontally a distance between (n )w and nw. We take the origin of a coordinate system to be at the point where the ball leaes the top of the stairway, and we choose the y axis to be positie in the upward direction, as shown in the figure. The coordinates of the ball at time t are gien by x = x t and y gt (since y = ). ANALYZE We equate y to nh and sole for the time to reach the leel of step n:

33 55 The x coordinate then is t nh. g nh n(.3 m) x x (.5 m/s) (.39 m) n. g 9.8 m/s The method is to try alues of n until we find one for which x/w is less than n but greater than n. For n =, x =.39 m and x/w =.5, which is greater than n. For n =, x =.437 m and x/w =.5, which is also greater than n. For n = 3, x =.535 m and x/w =.64. Now, this is less than n and greater than n, so the ball hits the third step. LEARN To check the consistency of our calculation, we can substitute n = 3 into the aboe equations. The results are t =.353 s, y =.69 m and x =.535 m. This indeed corresponds to the third step. 56. We apply Eq to sole for speed and Eq to find acceleration a. (a) Since the radius of Earth is m, the radius of the satellite orbit is Therefore, the speed of the satellite is r = ( ) m = 7. 6 m. c h b gb g 6 r 7. m T 98. min 6 s / min m / s. (b) The magnitude of the acceleration is a r c7. 49 m / sh 6 7. m 3 8. m / s. 57. The magnitude of centripetal acceleration (a = /r) and its direction (toward the center of the circle) form the basis of this problem. (a) If a passenger at this location experiences a 83. m/ s east, then the center of the circle is east of this location. The distance is r = /a = (3.66 m/s) /(.83 m/s ) = 7.3 m. (b) Thus, relatie to the center, the passenger at that moment is located 7.3 m toward the west.

34 56 CHAPTER 4 (c) If the direction of a experienced by the passenger is now south indicating that the center of the merry-go-round is south of him, then relatie to the center, the passenger at that moment is located 7.3 m toward the north. 58. (a) The circumference is c = r = (.5 m) =.94 m. (b) With T = (6 s)/ =.5 s, the speed is = c/t = (.94 m)/(.5 s) = 9 m/s. This is equialent to using Eq (c) The magnitude of the acceleration is a = /r = (9 m/s) /(.5 m) =.4 3 m/s. (d) The period of reolution is ( re/min) = min, which becomes, in SI units, T =.5 s = 5 ms. 59. (a) Since the wheel completes 5 turns each minute, its period is one-fifth of a minute, or s. (b) The magnitude of the centripetal acceleration is gien by a = /R, where R is the radius of the wheel, and is the speed of the passenger. Since the passenger goes a distance R for each reolution, his speed is b g. 5 m s b7. 85 m / sg and his centripetal acceleration is a 5 m 7 85 m / s 4. m / s. (c) When the passenger is at the highest point, his centripetal acceleration is downward, toward the center of the orbit. (d) At the lowest point, the centripetal acceleration is a 4. m/s, same as part (b). (e) The direction is up, toward the center of the orbit. 6. (a) During constant-speed circular motion, the elocity ector is perpendicular to the acceleration ector at eery instant. Thus, a =. (b) The acceleration in this ector, at eery instant, points toward the center of the circle, whereas the position ector points from the center of the circle to the object in motion. Thus, the angle between r and a is 8º at eery instant, so r a =. 6. We apply Eq to sole for speed and Eq to find centripetal acceleration a. (a) = r/t = ( km)/. s = 6 km/s =.3 5 m/s.

35 57 (b) The magnitude of the acceleration is b g 5. m / s. 6 km / s a 7 9 r km (c) Clearly, both and a will increase if T is reduced. 6. The magnitude of the acceleration is b g.. m / s a 4 m / s r 5 m 63. We first note that a (the acceleration at t =. s) is perpendicular to a acceleration at t =5. s), by taking their scalar (dot) product: (the a a ˆ ˆ ˆ ˆ [(6. m/s )i+(4. m/s )j] [(4. m/s )i+( 6. m/s )j]=. Since the acceleration ectors are in the (negatie) radial directions, then the two positions (at t and t ) are a quarter-circle apart (or three-quarters of a circle, depending on whether one measures clockwise or counterclockwise). A quick sketch leads to the conclusion that if the particle is moing counterclockwise (as the problem states) then it traels three-quarters of a circumference in moing from the position at time t to the position at time t. Letting T stand for the period, then t t = 3. s = 3T/4. This gies T = 4. s. The magnitude of the acceleration is a a a (6. m/s ) (4. m/s) 7. m/s. x y Using Eqs and 4-35, we hae a 4 r / T, which yields at (7. m/s )(4. s) r.9 m When traeling in circular motion with constant speed, the instantaneous acceleration ector necessarily points toward the center. Thus, the center is straight up from the cited point. (a) Since the center is straight up from (4. m, 4. m), the x coordinate of the center is 4. m. (b) To find out how far up we need to know the radius. Using Eq we find

36 58 CHAPTER 4 5. m/s r. m. a.5 m/s Thus, the y coordinate of the center is. m + 4. m = 6. m. Thus, the center may be written as (x, y) = (4. m, 6. m). 65. Since the period of a uniform circular motion is T r /, where r is the radius and is the speed, the centripetal acceleration can be written as r 4 r a. r r T T Based on this expression, we compare the (magnitudes) of the wallet and purse accelerations, and find their ratio is the ratio of r alues. Therefore, a wallet =.5 a purse. Thus, the wallet acceleration ector is ˆ ˆ ˆ ˆ. a.5[(. m/s )i +(4. m/s )j]=(3. m/s )i +(6. m/s )j 66. The fact that the elocity is in the +y direction and the acceleration is in the +x direction at t = 4. s implies that the motion is clockwise. The position corresponds to the 9: position. On the other hand, the position at t =. s is in the 6: position since the elocity points in the x direction and the acceleration is in the +y direction. The time interal t. s 4. s 6. s is equal to 3/4 of a period: Equation 4-35 then yields 3 6. s 8. s. 4 T T T (3. m/s)(8. s) r 3.8 m. (a) The x coordinate of the center of the circular path is x 5. m3.8 m 8.8 m. (b) The y coordinate of the center of the circular path is y 6. m. In other words, the center of the circle is at (x,y) = (8.8 m, 6. m). 67. THINK In this problem we hae a stone whirled in a horizontal circle. After the string breaks, the stone undergoes projectile motion. EXPRESS The stone moes in a circular path (top iew shown below left) initially, but undergoes projectile motion after the string breaks (side iew shown below right). Since a / R, to calculate the centripetal acceleration of the stone, we need to know its

37 59 speed during its circular motion (this is also its initial speed when it flies off). We use the kinematic equations of projectile motion (discussed in 4-6) to find that speed. (top iew) (side iew) Taking the +y direction to be upward and placing the origin at the point where the stone leaes its circular orbit, then the coordinates of the stone during its motion as a projectile are gien by x = t and y gt (since y = ). It hits the ground at x = m and y. m. ANALYZE Formally soling the y-component equation for the time, we obtain t y / g, which we substitute into the first equation: b g b g g 9. 8 m / s x m 57. m / s. y. m Therefore, the magnitude of the centripetal acceleration is 5.7 m/s a R.5 m 6 m/s. gx LEARN The aboe equations can be combined to gie a. The equation implies yr that the greater the centripetal acceleration, the greater the initial speed of the projectile, and the greater the distance traeled by the stone. This is precisely what we expect. 68. We note that after three seconds hae elapsed (t t = 3. s) the elocity (for this object in circular motion of period T ) is reersed; we infer that it takes three seconds to reach the opposite side of the circle. Thus, T = (3. s) = 6. s. (a) Using Eq. 4-35, r = T/, where (3. m/s) (4. m/s) 5. m/s, we obtain r 4.77 m. The magnitude of the object s centripetal acceleration is therefore a = /r = 5.4 m/s.

38 6 CHAPTER 4 (b) The aerage acceleration is gien by Eq. 4-5: a ag ˆ ˆ ˆ ˆ t t 5. s. s ( 3.i 4.j) m/s (3.i 4.j) m/s ˆ (. m/s )i+(.67 m/s )j ˆ which implies aag (. m/s ) (.67 m/s ) 3.33 m/s. 69. We use Eq. 4-5 first using elocities relatie to the truck (subscript t) and then using elocities relatie to the ground (subscript g). We work with SI units, so km/ h 5.6 m/ s, 3 km/ h 8.3 m/ s, and 45 km/ h.5 m/ s. We choose east as the i direction. (a) The elocity of the cheetah (subscript c) at the end of the. s interal is (from Eq. 4-44) (.5 m/s) ˆi ( 5.6 m/s) ˆi (8. m/s) ˆi c t c g t g relatie to the truck. Since the elocity of the cheetah relatie to the truck at the beginning of the. s interal is ( 8.3 m/s)i ˆ, the (aerage) acceleration ector relatie to the cameraman (in the truck) is (8. m/s)i ˆ ( 8.3 m/s)i ˆ a ˆ ag (3 m/s )i,. s or a 3 m/s. ag (b) The direction of aag is +i ˆ, or eastward. (c) The elocity of the cheetah at the start of the. s interal is (from Eq. 4-44) ( 8.3 m/s)i ˆ ( 5.6 m/s)i ˆ ( 3.9 m/s)i ˆ cg ct tg relatie to the ground. The (aerage) acceleration ector relatie to the crew member (on the ground) is (.5 m/s)i ˆ ( 3.9 m/s)i ˆ a ˆ ag (3 m/s )i, aag 3 m/s. s identical to the result of part (a). (d) The direction of a ag is +i ˆ, or eastward. 7. We use Eq. 4-44, noting that the upstream corresponds to the ˆ +i direction. (a) The subscript b is for the boat, w is for the water, and g is for the ground.

39 6 (4 km/h) ˆi ( 9 km/h) ˆi (5 km/h) ˆi. bg bw wg Thus, the magnitude is (b) The direction of bg bg 5 km/h. is +x, or upstream. (c) We use the subscript c for the child, and obtain The magnitude is c g c b b g ( 6 km/ h) i ( 5 km/ h) i ( km/ h) i. cg km/h. (d) The direction of cg is x, or downstream. 7. While moing in the same direction as the sidewalk s motion (coering a distance d relatie to the ground in time t =.5 s), Eq leads to sidewalk + man running = d t. While he runs back (taking time t =. s) we hae sidewalk man running = d t. Diiding these equations and soling for the desired ratio, we get = 5 3 = We denote the elocity of the player with PF and the relatie elocity between the player and the ball be BP. Then the elocity BF of the ball relatie to the field is gien by BF PF BP. The smallest angle min corresponds to the case when BF PF. Hence, min 4. m/s BP 6. m/s PF 8 cos 8 cos We denote the police and the motorist with subscripts p and m, respectiely. The coordinate system is indicated in Fig (a) The elocity of the motorist with respect to the police car is ( 6 km/h) ˆj ( 8 km/h)i ˆ (8 km/h)i ˆ (6 km/h) ˆ m p m p j.