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Unformatted text preview: CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS 77 CHAPTER 4 Answers to Checkpoint Questions 1. 2. 3. 4. 5. 6. 7. (a) (8 i 6 j) m; (b) yes, the xy plane (a) rst; (b) third 1 and 3: ax and ay are both constant and thus a is constant; 2 and 4: ay is constant but ax is not, thus a is not 4 m/s3 , 2 m/s, 3 m (a) vx constant; (b) vy initially positive, decreases to zero, and then becomes progressively more negative; (c) ax = 0 throughout; (d) ay = g throughout (a) (4 m/s) i; (b) (8 m/s2 ) j (1) 0, distance not changing; (2) +70 km/h, distance increasing; (3) +80 km/h, distance decreasing Answers to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 1 and 3: ay is constant but ax is not and thus a is not; 2: ax is constant but ay is not and thus a is not; 4: ax and ay are both constant and thus a is constant; (e) 2 m/s2 , 3 m/s 1, 2, 3 (a) highest point; (b) lowest point yes (the vertical component of v is downward) (a) all tie; (b) 1 and 2 tie (the rocket is shot upward), then 3 and 4 tie (it is shot into the ground!) (a) decreases; (b) increases (2 i 4j) m/s (a) 0; (b) 350 km/h; (c) 350 km/h; (d) same nothing changed about the vertical motion (a) all tie; (b) all tie; (c) c, b, a; (d) c, b, a 78 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS 10. 11. 12. 13. 14. 15. 16. 17. (a) 2; (b) 3; (c) 1; (d) 2; (e) 3; (f) 1 (a) no; (b) same (a) yes; (b) no; (c) yes (a) in your hands; (b) behind you; (c) in front of you (a) equal; (b) equal (a) straight down; (b) curved; (c) more curved decreases (a) 3; (b) 4 Solutions to Exercises & Problems 1E (a) The position vector of the watermelon is r = xi + yj + z k = ( 5:0i + 8:0j) m. (b) The magnitude of the vector is r = x2 + y2 + z2 = ( 5:0 m)2 + (8:0 m)2 + (0 m)2 = 9:4 m ;
and the angle it makes with the positive x axis is p p = tan
(c) The vector is shown below. 1 y = tan 8:0 m = 122 : x 5:0 m y r
=122 o O x CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS 79 (a) r = x2 + y2 + z 2 = (5:0 m)2 + ( 3:0 m)2 + (2:0 m)2 = 6:2 m : (b)
z 2E p p 2m 3 m O r
5m y x 3E (a) The displacement vector is r = rf ri = ( 2:0i + 6:0j + 2:0k) (5:0i 6:0j + 2:0k) = ( 7:0i + 12j) m : (b) Since r has no z component, it is parallel to the xy plane (i.e., the z = 0 plane). The former position vector r0 satises r r0 = r, which gives
r0 = r 4E r = (3:0j 4:0k) (2:0i 3:0j + 6:0k) = 2:0i + 6:0j 10k ; where the suppressed unit is meters.
5E (a) The magnitude of the displacement vector r is given by r = (300 mi)2 + (600 mi)2 = 671 mi ;
p while the angle which it makes with the east direction is = tan 1 600 mi = 63:4 ; 300 mi 80 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS i.e., it is pointed at 63.4 south of east. (b) The average velocity is 671 mi r = v= t (45:0 min)(1 h=60 min) + 1:50 h r0 = (298 mi/h) r0 ; where r0 is the unit vector in the direction of r. (c) Denote the total distance traveled as D. The average speed v is given by 300 v = t = (45:0 min)(1mi=+ 600 mi+ 1:50 h = 400 mi/h : D h 60 min)
6E The average velocity is the total displacement divided by the time interval. The total displacement r is the sum of 3 displacements, each calculated as the product of a velocity and a time interval. The rst has a magnitude of (60:0 km/h)(40:0 min)=(60:0 min/h) = 40:0 km. Its direction is east. If we take the x axis to be toward the east and the y axis to be toward the north, then this displacement is r1 = (40:0 i) km. The second has a magnitude of (60:0 km/h)(20:0 min)=(60:0 min/h) = 20:0 km. Its direction is 50:0 east of north, so it may be written r2 = (20:0 sin 50:0 i + 20:0 cos 50:0 j) km = (15:3 i + 12:9 j) km. The third has a magnitude of (60:0 km/h)(50:0 min)=(60:0 min/h) = 50:0 km. Its direction is west, so the displacement may be written r3 = ( 50 i) km. The total displacement, in km, is r = r1 + r2 + r3 = (40:0 i) km + (15:3 i + 12:9 j) km (50 i) km = (5:3 i + 12:9 j) km : The total time for the trip is 40:0 min + 20:0 min + 50:0 min = 110 min = 1:83 h. Divide r by this interval to obtain an average velocity of v = 2:89 i + 7:04 j, in km/h. The average velocity has a magnitude of 7:61 km/h and is directed 67:7 north of east.
7E (a) The magnitude of the dispalcement vector r is given by r = (21:5 km)2 + (9:70 km)2 + (2:88 km)2 = 23:8 km :
p Thus v = r=t = (23:8 km=3:50 h) = 6:79 km/h: (b) The angle in question is given by = tan 1 p 2:88 km 2 + (9:70 km)2 = 6:96 : (21:5 km) CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS 81 The displacement vector of the ion is r = ( 2:0 i + 8:0 j 2:0 k) m (5:0 i 6:0 j + 2:0 k) m = ( 7:0 i + 14:0 j 4:0 k) m : The average velocity in t = 10 s is thus v=
9E v(t) = 8E r = 7:0 i + 14:0 j 4:0 k = ( 0:70 i + 1:4 j 0:40 k) m/s: t 10 s (a) dr = d (3:0ti 4:0t2 j + 2:0 k) m/s dt dt = (3:0 i 8:0t j) m/s :
(8:0)(2:0) j] m/s = (3:0 i 16 j) m/s :
p (b) (c) v(t = 2:0 s) = [3:0i
q 2 2 v = vx + vy = (3:0 m/s)2 + (16 m/s)2 = 16 m/s : The angle between the velocity vector v(t = 2:0 s) and the positive x axis is given by = tan 1 16 m/s = 79 : 3:0 m/s (a) The change of velocity in t = 4:0 s is given by v = ( 2:0 i 2:0 j + 5:0 k) m/s (4:0 i 2:0 j + 3:0 k) m/s = ( 6:0 i + 2:0 k) m/s : v = ( 6:0 i + 2:0 k) m/s = ( 1:5 i + 0:50 k) m/s2 : t 4:0 s (b) The magnitude of a is a= Thus 10E a = ( 1:5 m/s2 )2 + (0:50 m/s2 )2 = 1:6 m/s2 : p 82 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS Since it has no y component, a lies in the xz plane, making an angle with the positive x axis, where 2 1 az = tan 1 0:50 m/s = tan a 2 = 162 : 1:5 m/s
x (a) The velocity is the derivative of the position vector with respect to time:
v= 11E (b) The acceleration is the derivative of the velocity with respect to time:
a= d i + 4t2 j + t k = 8t j + k : dt d (8t j + k) = 8 j : dt The velocity is in m/s and the acceleration is in m/s2 .
12E The values of the various kinematic quantities at t = 2:00 s are given by (a)
r = [(2:00)(2:00)3 (5:00)(2:00)] i + [6:00 = (6:00 i 106 j) m ; (7:00)(2:00)4 ] j (b) dr = d [(2:00t3 5:00t) i + (6:00 7:00t4 ) j] v= dt dt t=2:00 s 2 5:00] i (4)(7:00)(2:00)3 j = [(3)(2:00)(2:00) = (19:0 i 224 j) m/s ;
(c)
2 2 dr d a = 2 = 2 [(2:00t3 5:00t) i + (6:00 7:00t4 ) j] dt dt t=2:00 s 2j = (6)(2:00)(2:00) i (12)(7:00)(2:00) = (24:0 i 336 j) m/s2 : (d) From the expression for the vector v above, we know that the it makes an angle with respect to the positive x axis, with 224 m/s = tan 1 vy = 85:2 o : 1 = tan 19:0 m/s vx CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS 83 13E The average acceleration is a=
14P v = 0 (6:30 i 8:42 j) m/s = ( 2:10 i + 2:81 j) m/s2 : t 3s (a) The position vector of the particle measured from the origin is given by
r(t) = r0 +
Z t = v0 t + 1 at2 2 = (8:0 j)t + 1 (4:0 i + 2:0 j)t2 = (2:0t2 ) i + (8:0t + 1:0t2 ) j : 2 For x(t) = 2:0t2 = 29 m, we nd t = 3:8 s, which is used to nd y(t): 0 v0 + Z t 0 a(t0 ) dt0 dt y(t) = 8:0(3:8) m + 1:0(3:8)2 m = 45 m :
(b) The velocity of the particle is
v(t) = dr = d [(8:0 j)t + 1 (4:0 i + 2:0 j)t2 ] dt dt 2 = 8:0 j + (4:0 i + 2:0 j)t = (8:0 m/s) j + (4:0 i + 2:0 j)(3:8 m/s) = (15 i + 16 j) m/s :
q p The speed of the particle is then
2 2 v(t) = vx + vy = (15 m/s)2 + (16 m/s)2 = 22 m/s : (a) The velocity of the particle at any time t is given by v = v0 + at, where v0 is the initial velocity and a is the acceleration. The x component is vx = v0x + ax t = 3:00 1:00t and the y component is vy = v0y + ay t = 0:500t. When the particle reaches its maximum x coordinate vx = 0. This means 3:00 1:00t = 0 or t = 3:00 s. The y component of the velocity at this time is vy = ( 0:500)(3:00) = 1:50 m/s. (b) The coordinates of the particle at any time t are x = v0x t + 1 ax t2 and y = v0y t + 1 ay t2 . 2 2 At t = 3:00 s their values are x = (3:00)(3:00) 1 (1:00)(3:00)2 = 4:50 m 2 15P 84 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS and 1 y = 2 (0:50)(3:00)2 = 2:25 m : dv = d [(6:0t 4:0t2 ) i + 8:0 j] m/s = (6:0 8:0t) j m/s2 : dt dt (a) At t = 3:0 s, a = [6:0 8:0(3:0)] i = 18 i m/s2 : (b) Let a = (6:0 8:0t) i = 0, we nd t = 0:75 s. = (c) Since vy p 8:0 m/s 6= 0, the velocity will never be zero. (d) Let v = (6:0t 4:0t2 )2 + (8:0)2 m/s = 10 m/s, solve for t to obtain t = 2:2 s.
a= Use 16P Suppose that a collision takes place at point C at time t. From the gure we see that 17P y A v C AC sin = BC = 1vt 2 ; 2 at tan = vt : h
Eliminate t to obtain sin2 = 1 cos 2 = 2v2 = 2(3:0)2 = 1:5 ; cos cos ha 30(0:4) which gives a B x = cos
18E 1 1:5 + 1:52 + 4 = 60 o : 2 p (a) The distance PQ is equal to the vertical distance through which the dart falls in t = 0:19 s. Thus 1 1 PQ = 2 gt2 = 2 (9:8 m/s2 )(0:19 s)2 = 0:18 m : (b) The distance is given by D = vi t = (10 m/s)(0:19 s) = 1:9 m. CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS 85 Take the positive y axis to be downward and place the origin at the ring point. Then the y coordinate of the bullet is given by y = 1 gt2 . If t is the time of ight and y is the 2 distance the bullet hits below the target, then s p 2(0:75 in.) 2 t = 2y=g = 2 ) = 6:3 10 s : (12 in./ft)(32 ft/s Notice that 0:75 in. was converted to (0:75=12) ft. (b) The muzzle velocity is the initial velocity of the bullet. It is horizontal. If d is the horizontal distance to the target, then d = v0 t and ft v0 = d = 6:2510010 2 s = 1:6 103 ft/s : t (a) Since the time t it takes for the electron to y a horizontal distance of 1:0 m is given by t = 1:0 m=(3:0 106 m/s) = 3:3 10 7 s, it falls through a distance 1 1 H = 2 gt2 = 2 (9:8 m/s2 )(3:3 10 7 s)2 = 5:4 10 13 m : (b) As the initial speed increases, t decreases, and so does H . (a) The horizontal component of the velocity is constant. If ` is the length of a plate and t is the time an electron is between the plates, then ` = v0 t, where v0 is the initial speed. Thus 20 t = v` = 1:0 :10cm = 2:0 10 9 s = 2:0 ns : 9 cm/s 0 (b) The vertical displacement of the electron is 1 y = 1 at2 = 2 (1:0 1017 cm/s2 )(2:0 10 9 s)2 = 0:20 cm ; 2 where down was taken to be positive. (c) The x component of velocity is vx = v0 = 1:0 109 cm/s and the y component is vy = ay t = (1:0 1017 cm/s2 )(2:0 10 9 s) = 2:0 108 cm/s.
22E 21E 20E 19E (a) The time of ight is (b) H t = 2g = (2)(4:0 ft) = 0:50 s : 32 ft/s2 v0 = D = 5::0 ft = 10 ft/s : t 0 5s s s 86 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS (a) 23E
s t= 2h = s g 2(45:0 m) = 3:03 s : 9:80 m/s2 (b) D = vt = (250 m/s)(3:03 s) = 758 m: (c) v? = gt = (9:80 m/s2 )(3:03 s) = 29:7 m/s: (a) Since the horizontal component vh of the velocity of the ball is a constant, we have
24E D 30 ft t = v = (100 mi/h)(5280 ft/mi)(1 h=3600 s) = 0:20 s h
for both the rst and the second 30 ft of its horizontal travel. (b) The distance the ball fell during the rst 0:20 s is 1 1 d1 = 2 gt2 = 2 (32:2 ft/s2 )(0:205 s)2 = 0:67 ft = 8:1 in. . (c) During the second 0:20 s, the ball fell by 1 d2 = gt2 + 2 gt2 = (32:2 ft/s2 )(0:205 s)2 + 0:67 ft = 2:0 ft :
25E (a) At t = 2:0 s, the vertical component of the velocity is v? = v? (0) gt = v0 sin 60 o gt = (30 m/s)(sin 60 o ) (9:8 m/s2 )(2:0 s) = 6:4 m/s ;
while the horizontal component is vk = vk (0) = (30 m/s)(cos 60 o ) = 15 m/s :
Thus the magnitude of the velocity is
2 2 v = v? + vk = (6:4 m/s)2 + (15 m/s)2 = 16 m/s :
q p The velocity is directed at an angle above the horizontal, with = tan 1 v? = tan vk 1 6:4 m/s = 23 o : 15 m/s CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS 87 (b) Similary, at t = 5:0 s v? = (30 m/s)(sin 60 o ) (9:8 m/s2 )(5:0 s) = 23 m/s
and vk = 15 m/s: Thus v = ( 23 m/s)2 + (15 m/s)2 = 27 m/s :
The velocity is now directed at an angle 0 below the horizontal, with 23 m/s 0 = tan 1 v? = tan 1 15 m/s = 23 : vk
26E p Take the y axis to be upward and the x axis to be horizontal and to the right. Place the origin at the throwing point and suppose the stone is thrown at time t = 0. The x component of the initial velocity is given by v0x = v0 cos 0 and the y component is given by v0y = v0 sin 0 , where v0 is the initial speed and 0 is the launch angle. (a) At t = 1:10 s its x coordinate is x = v0 t cos 0 = (20:0 m/s)(1:10 s)(cos 40:0 ) = 16:9 m and the y component is 1 1 y = v0 t sin 0 2 gt2 = (20:0 m/s)(1:10 s)(sin 40 ) 2 (9:80 m/s2 )(1:10 s)2 = 8:21 m : (b) At t = 1:80 s, x = (20:0 m/s)(1:80 s)(cos 40:0 ) = 27:6 m and 1 y = (20:0 m/s)(1:80 s) sin 40 2 (9:80 m/s2 )(1:80 s)2 = 7:26 m : (c) The stone hits the ground earlier than t = 5:0 s. To nd the time when it hits the ground solve y = v0 t sin 0 1 gt2 = 0 for t. You should get 2 v 0 m/s) t = 2g 0 sin 0 = 2(20:m/s2 (sin 40 ) = 2:62 s : 9:8
Its x coordinate on landing is x = v0 t cos 0 = (20:0 m/s)(2:62 s)(cos 40 ) = 40:1 m :
Assuming it stays where it lands its coordinates at t = 5:00 s are x = 40:1 m and y = 0. 88 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS (a) Choose the positive y axis to be upward. The horizontal displacement is 27E x = (v0 cos )t = (15:0 m/s)(cos 20:0 )(2:30 s) = 32:4 m
and the vertical displacement is y = ( v0 sin )t 1 gt2 2 = ( 15:0 m/s)(sin 20:0 )(2:30 s) 1 (9:80 m/s2 )(2:30 s)2 = 37:7 m : 2
28E (a) Since the horizontal component of the velocity of the ball is vx = v0 cos 40:0 , the time it takes for the ball to hit the wall is m)(cos 40 D t = v = (22:025:0 m/s :0 ) = 1:15 s : x (b) The vertical distance is d = (v0 sin 40:0 )t 1 gt2 2 = (25:0 m/s)(sin 40:0 )(1:15 s) 1 (9:80 m/s2 )(1:15 s)2 = 12:0 m : 2 (c) The horizontal component of the velocity as it hits the wall remains at vx = v0 cos 40:0 = (25:0 m/s)(cos 40:0 ) = 19:2 m/s ;
while the vertical component becomes vy = v0 sin 40:0 gt = (25:0 m/s)(sin 40:0 )(1:15 s) (9:80 m/s2 )(1:15 s) = 4:80 m/s :
(d) Since vy > 0 when the ball hits the wall, it has not reached the highest point yet. (a) The maximum height H is given by
29E H = (v0 sin 0 ) ; 2g
while the range R is given by Eq. 420. Thus
2 sin2 0 1 H = v0 sin2 0 = sin2 0 = 2 =g ) sin 20 2 sin 20 2(2 sin 0 cos 0 ) = 4 tan 0 : R 2g(v0 2 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS 89 (b) Let we get H = 1 tan = 1 ; 0 R 4 0 = tan 1 4 = 76 : 30E (a) Use the result in 29E to obtain 1 H tan = R=2 = 2 4 tan 0 = 1 tan 0 : 2 (b) = tan 1 1 tan 45 = 27 : 2 (a) Let the time of ight of the stone be t, then its vertical displacement is 31E h = (v0 sin 60 )t 1 gt2 2 1 = (42:0 m/s)(sin 60:0 )(5:50 s) 2 (9:8 m/s2 )(5:50 s)2 = 51:8 m : (b) The horizontal component of the velocity of the stone just before impact is vk = v0 cos 60:0 ; while the vertical component is v? = v0 sin 60:0 gt: Thus
2 2 v = v? + vk p = (v0 cos 60 )2 + (v0 sin 60 gt)2 p = [(42:0 m/s)(cos 60:0 )]2 + [(42:0 m/s)(sin 60:0 ) (9:80 m/s2 )(5:50 s)]2 = 27:4 m/s :
q (c) 2 2 m/s)2 H = (v0 sin 0 ) = (42:02(9:80 (sin 60:0 ) = 67:5 m : 2g m/s2 ) 90 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS At the maximum height, the velocity has only a horizontal component. Thus the condition given implies that v0 = 5v0x , which gives 32P 0 = cos
33P 1 vx = 1 = 78:5 : v0 5 (a) The horizontal component of the velocity of the capsule is vx = v0 = (430 km/h)(1000 m/km)(1 h=3600 s) = 119 m/s ;
while the vertical component is 1 vy = v0 sin 0 + 2 gt2 = 0 + (9:80 m/s2 )(15:65 s) = 153 m/s : The speed of the capsule is then
2 2 v = vx + vy = (119 m/s)2 + (153 m/s)2 = 194 m/s :
q p (b) The angle of impact is given by = tan
34P 1 vx = tan vy 1 119 m/s = 37:9 : 153 m/s Since (see Eq. 420) Rmax = g sin 20 2 v 0 2 2 5 = vg0 = (9::80m/s)2 = 9:21 m ; 9 m/s max the dierence between Powell's long jump and Rmax is only R = 9:21 m 8:95 m = 0:26 m :
35P Take the y axis to be upward and the x axis to the horizontal. Place the origin at the ring point, let the time be 0 at ring, and let 0 be the ring angle. If the target is a distance d away then its coordinates are x = d, y = 0. The kinematic equations are d = v0 t cos 0 and 0 = v0 t sin 0 1 gt2 . Eliminate t and solve for 0 . The rst equation gives t = d=v0 cos 0 . 2 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS 91 2 This expression is substituted into the second equation to obtain 2v0 sin 0 cos 0 gd = 0. 1 sin(20 ) to obtain v 2 sin(20 ) = gd or Use the trigonometric identity sin 0 cos 0 = 2 0 gd = (32 ft/s2 )(150 ft) = 2:13 10 3 : sin(20 ) = v2 (1500 ft/s)2 0 The ring angle is 0:0611 . If the gun is aimed at a point a distance ` above the target then tan 0 = `=d or ` = d tan 0 = (150 ft) tan 0:0611 = 0:16 ft = 1:9 in.
Consider the vertical displacement of the projectile. The vertical component of its initial velocity is v0y = v0 sin 0 , while that of the nal velocity as the projectile reaches the top 2 2 of its path is vy = 0. Thus from vy = v0y + 2a(y y0 ) we have
2 2 v2 v2 v ymax = y y0 = y 2a 0y = 0 (2(0 sin 0 ) = (v0 sin)0 ) : g) 2g 36P (a) Let the time it takes for the ball to reach the given velocity be t. Then 37P vy (t) = 6:1 m/s = v0y gt ; y(t) = 9:1 m = v0y t 1 gt2 : 2
Solve for v0y to obtain v0y = 15 m/s. Thus the ball will reach a maximum height of v2 (15 m/s)2 ym = 20gy = 2(9:8 m/s) = 11 m :
(c) Since the horizontal component of the velocity is unchanged, we have : R = v0x 2v0y = 2(7:6 m/s)(1427 m/s) = 23 m : g 9:8 m/s
(d) The speed is
2 2 v = v0x + v0y = (15 m/s)2 + (7:6 m/s)2 = 17 m/s :
q p The angle v makes with the horizontal direction is = tan 1 v0y = tan v0x 1 15 m/s = 63 : 7:6 m/s 92 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS i.e., it is oriented 63 below the horizontal. The horizontal component of the initial velocity is = 6:7 ft/s ; v0x = D = p D = p 15 ft t 2H=g 2(80 ft)=32 ft/s2 which is suspiciously fast (about 20% of worldclass sprint speed). Thus the death is not likely accidental. From Eq. 420
39P 38P R(90 0 ) = vg0 sin[2(90 0 )] = vg0 sin 20 = R(0 ) ;
i.e., R(0 ) = R(90 0 ): Denote the physical quantities in Berlin and Melbourne with the subscripts b and m, respectively. Then from Eq. 420
2 Rb = (v0 =2gb ) sin 20 = gm : 2 Rm (v0 =2gm ) sin 20 gb 2 2 40P Thus 2 g Rm = g b Rb = 9::8128 m/s2 (8:09 m) = 8:10 m ; 9 7999 m/s m which means that his record would have been improved by 8:10 m 8:09 m = 0:01 m = 1 cm: Since the time of ight of the ball is 2(3 ft) t = 32:2 ft/s2 = 0:43 s ; the range of the ball is
s 41P D = v0 t = (85 mi/h)(5280 ft/mi)(1 h=3600 s)(0:43 s) = 54 ft :
Therefore the ball will land at 127 ft 54 ft = 73 ft from rst base. CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS 93 (b) Use Eq. 420:
2 v0 sin 20 = (124:7 ft/s2 )(sin 20 ) ; R = 127 ft = g 32:2 ft/s2 which gives 0 = 7:6 . (c) The time of ight in that case is R 127 ft t = v cos = (124:7 ft/s)(cos 7:6 ) = 1:0 s : 0
42P Since at point B in the path xB = 9:40 km and yB = 3:30 km, 3300 = (tan 35 )(9400) (a) Put the origin of a coordinate system at A, place the x axis horizontally to the right and the y axis vertically upward. The equation for the path of the bomb is then given by Eq. 419: g y = (tan 0 )x 2(v cos )2 x2 :
0 0 9:8 2 2(v0 cos 35 )2 (9400) ; which gives v0 = 260 m/s. (b) Let the time of ght be t, then in the horizontal direction (v0 cos 35 )t = R; so 9400 m R t = v cos 35 = (260 m/s)(cos 35 ) = 45 s : 0 (c) The presence of the air slows down the bomb. Thus the actual value of v0 has to be greater than the answer given in (a). Choose a coordinate system with its origin at the point where the player res the ball, and with the x axis pointing horizontally to the right and the y axis vertically upward. The trajectory of the ball must pass through the net, located at x = 13 ft, y = 3:0 ft. Using the trajectory equation, we get
2 y = x tan 0 2(v gx )2 0 cos 0 43P = (tan 55 )(13 ft) 2(v32 ft/s )2 0 cos 55 = 3:0 ft ; 2 94 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS which yields v0 = 23 ft/s: Take the y axis to be upward and the x axis to be horizontal. Place the origin at point where the ball is kicked and take the time to be 0 when it is kicked. x (= 50 yd = 150 ft) and y (= 5:0 ft) are coordinates of the landing point. The ball lands at time t (= 4:5 s). Since x = v0x t, v0x = x = 150 ft = 33 ft/s : t 4:5 s
1 gt2 , 2 44P Since y = v0y t The magnitude of the initial velocity is
q 1 5:0 ft + 2 (32 ft/s2 )(4:5 s)2 y + 1 gt2 = 71 ft/s : v0y = t2 = 4:5 s
2 2 v0 = v0x + v0y = (33 ft/s)2 + (71 ft/s)2 = 78 ft/s :
p The ring angle satises tan 0 = v0y =v0x = (71 ft/s)=(33 ft/s) = 2:1 : The angle is 0 = 65 . (a) Take the y axis to be upward and the x axis to be horizontal. Place the origin at the point where the ball is initially struck, and take the time to be 0 when it is struck. Then y = v0y t gt2 =2 and x = v0x t: As the ball hits the fairway, we have x = 180 m, which gives 180 m t = vx = (43 m/s)(cos 30 ) = 4:8 s : 0x Thus 1 y = (43 m/s)(4:8 s) 2 (9:8 m/s2 )(4:8 s)2 = 11 m ;
45P i.e., the rise is 11 m above the fairway. (b) As the ball strikes the fairway, we have vx = v0x = (43 m/s)(cos 30 ) = 37 m/s and 1 vy = v0y gt = (43 m/s)(sin 30 ) 2 (9:8 m/s2 )(4:83)2 = 26 m/s : Thus the speed of the ball is
2 2 v(t) = vx + vy = (37)2 + ( 26)2 m/s = 45 m/s :
q p CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS 95 2 Let the projection angle of the low trajectory be 1 . Then R = [v0 sin(21 )]=g, which gives 46P 1 1 = 2 sin 1 Rg = sin 2 v0 1 (20:0 m)(9:80 m/s2 ) = 6:29 : (30:0 m/s)2 The projection angle for the high trajectory is 2 = 90 1 = 83:7 (see 39P): The maximum range is given by
2 2 Rmax = vg0 = 9:8 v0 2 = 60 m ; m/s 47P which gives v0 = 24:3 m/s. Thus the maximum vertical height reached above the releasing point is 2 v0 (24 3 m/s)2 ym = 2g = 2(9::8 m/s2 ) = 30 m : Let the time of ight of the radar decoy be t. Then the range is R = 2300 ft = v0 t cos 30:0 = (180 mi/h)(t cos 30:0 ); which gives t = 10:1 s. The height of the plane was then
48P h = (v0 sin 30:0 )t + 1 gt2 2 = (180 mi/s)(5280 ft/mi)(1 h=3600 s)(sin 30:0 )(10:1 s) + 1 (32:2 ft/s2 )(21:6 s) 2 = 2900 ft: Let the range of the football be R and its time of ight be t, then the receiver has to travel a distance jR 60 ydj within time t. His average speed should then be
2 60 60 v = jR t ydj = (v0 =g)(sin 90 ) 45 yd (2v0 =g) sin 2 )=(32 ft/s2 ) 180 ftj (64 ft/s = j[2(64 ft/s) sin 45 ]=(32 ft/s2 ) = 19 ft/s : 49P 96 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS Choose a coordinate system with its origin at the point where the ball rolls o, and with the x axis horizontal and the y axis vertically upward. Then the trajectory equation of the ball is gx2 y = 3:0 ft = 2(v )2 : If the ball hits the nth step, the coordinates (xn ; yn ) of the point of impact must satisfy (n 1)(8:0 in.) < x < n(8:0 in.) and yn = n(8:0 in.). Plugging in g = 32:2 ft/s2 and v0 = 5:0 ft/s into the trajectory equation above, you can test this condition for various values of n. It is straightforward to nd that the smallest n that satises the condition is n = 3. Thus the ball will rst hit the third step.
51P
0 50P (a) Since the projectile is released its initial velocity is the same as the velocity of the plane at the time of release. Take the y axis to be upward and the x axis to be horizontal. Place the origin at the point of release and take the time to be 0 at release. Let x and y (= 730 m) be the coordinates of the point on the ground where the projectile hits and let t be the time when it hits. Then 1 y = v0 t cos 0 2 gt2 ; where 0 = 53:0 . This equation gives v0 = y + 1 gt2 ( 730 m) + 1 (9:80 m/s2 )(5:00 s)2 2 2 = = 202 m/s : t cos 0 (5:00 s)[cos( 53:0 )] x = v0 t sin 0 = (202 m/s)(5:00 s)[sin( 53:0 )] = 806 m : (b) The horizontal distance traveled is (c) The x component of the velocity is vx = v0 sin 0 = (202 m/s)(sin 53:0 ) = 161 m/s and the y component is given by vy = v0 cos 0 gt = (202 m/s) cos(53 ) (9:80 m/s2 )(5:00 s) = 171 m/s : 2 2 2 Using Eq. 2.16: v2 = v0 + 2a(x x0 ); which in our case reads (3v0 )2 = v0 + 2gh: Solve for v0 : r r gh = (9:8 m/s2 )(20 m) = 7:0 m/s : v0 = 4 4 52P CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS 97 (a) Refer to the gure to the right. The vertical coordinate y0 of the ball as it passes just above the net is given by the trajectory equation (with 0 = 0):
2 8 m/s2 )(12 2 0 y0 = gx2 = (9:2(23:6 m/s)m) 2 2v0 = 1:27 m : 53P y O y0 h H x0 x net ground level The clearance over the net is thus h0 = H h jy0 j = 2:37 m 0:90 m 1:27 m = 0:20 m :
(b) In this case
2 y0 = (tan 0 )(x0 ) 2(v gx )2 0 cos 0 2 (9: m/s2 = ( tan 5:00 )(12 m) 2(23:68m/s)2)(122m)00 ) (cos 5: = 2:3 m ; and the clearance is h0 = H h jy0 j = 2:37 m 0:90 m 2:3 m < 0 ;
which means that the ball will not be able to clear the net. Choose a coordinate system with its origin at the point where the second cannon res and with the x axis pointing horizontally to the right and the y axis vertically upward. Then the time of ight t of the cannon ball is given by
54P y = 30 m = v0 t 1 gt2 = (82 m/s)(sin 45 )t 1 (9:8 m/s2 )t2 ; 2 2
which gives t = 12:3 s: The range of the second cannon is then R2 = (v0 cos 0 )t = (82 m/s)(cos 45 )(12:3 s) = 715 m ;
which is longer by (715 m 690 m) = 25 m than that of the rst cannon. 98 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS You want to know how high the ball is from the ground when it is 320 ft from home plate. To calculate this quantity you need to know the initial speed v0 of the ball. Use the range information to obtain v0 . Put the origin at the point where the ball is hit, take the y axis to be upward and the x axis to be horizontal. If x (= 350 ft) and y (= 4:0 ft) are the coordinates of the ball when it lands, then x = v0 t cos 0 and y = v0 t sin 0 1 gt2 , where 2 t is the time of ight of the ball. The rst equation gives t = x=v0 cos 0 and when this is substituted into the second, the result is 1 gx2 y = x tan 0 2 v2 cos2 : 0 0 Solve for v0 . You should obtain (32 ft/s )(350 ft) v0 = 2(x tan y) cos2 = 2(354 ft)(cos2 45 ) = 105 ft/s : 0 0 Now take x and y to be the coordinates when the ball is at the fence. Again x = v0 t cos 0 and y = v0 t sin 0 1 gt2 . The time to reach the fence is given by t = x=v0 cos 0 . When 2
s 55P gx2 s 2 2 this is substituted into the second equation the result is 2 )(320 2 (32 = (320 ft))(tan 45 ) 2(105 ft/s 2 (cos2ft) ) = 24:1 ft : ft/s) 45 Since the ball started 4:0 ft above the ground it is 28:1 ft above the ground when it gets to the fence and it is 4:1 ft above the top of the fence. It goes over the fence. y = x tan 0 2v2 gx 2 0 cos 0 2 Take the y axis to be upward and the x axis to be horizontal. Place the origin at the point where the ball is kicked, on the ground, and take the time to be 0 at the instant it is kicked. x and y are the coordinates of ball at the goal post. You want to nd the kicking angle 0 so that y = 3:44 m when x = 50 m. Write the kinematic equations for projectile motion: x = v0 t cos 0 and y = v0 t sin 0 1 gt2 . The rst equation gives t = x=v0 cos 0 and when 2 this is substituted into the second the result is gx2 y = x tan 0 1 (v cos )2 : 2 0 0 You may solve this by trial and error: systematically try values of 0 until you nd the two that satisfy the equation. A little manipulation, however, will give you an algebraic solution. Use the trigonometric identity 1= cos2 0 = 1 + tan2 0 to obtain 1 gx2 tan2 x tan + y + 1 gx2 = 0 : 0 0 2 2 2 v0 2 v0 56P* CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS 99 2 This is a quadratic equation for tan 0 . To simplify writing the solution, let c = 1 gx2 =v0 = 2 2 )(50 m)2 =(25 m/s)2 = 19:6 m. Then the quadratic equation becomes c tan2 1 (9:80 m/s 0 2 x tan 0 + y + c = 0. It has the solution 2+ tan 0 = x x 2c 4(y + c)c p
p 2 4(3 m = 50 m (50 m) 2(19::44m) + 19:6 m)(19:6 m) : 6 The two solutions are tan 0 = 1:95 and tan 0 = 0:605. The corresponding angles are 0 = 63 and 0 = 31 . If kicked at any angle between these two, the ball will travel above the cross bar on the goalposts. (a) The centripetal acceleration a points toward the center of the circular path of the electron, with a magnitude
57E v2 = (2:18 106 m/s)2 = 9:00 1022 m/s2 : a= r 5:28 10 11 m
(b) The period of the motion is 2r = 2(5:28 10 11 m) = 1:52 10 T= v 2:18 106 m/s
58E
16 s : (a) The magnitude of the acceleration is a = vr = (10 m/s) = 4:0 m/s2 : 25 m
(b) a points toward the center of the circle. (a) The speed of the electron is
59E 2 2 v = ar = (3:0 1014 m/s2 )(15 cm) = 6:7 106 m/s :
(b) The period of the motion is p p T = 2r = 6:2(15 cm) = 1:4 10 7 s : v 7 106 m/s 100 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS (a) The radius of the track is given by
2 2 m/s)2 r = va = (9::8 m/s2 = 22 m : 3 60E (b) The period of the motion is T = 2r = 29:(22 m) = 15 s : v 2 m/s
61E (a) The circumference of the circular orbit is given by 2R, where R is the radius. Since the radius of the earth is 6:37 106 m, the radius of the satellite orbit is 6:37 106 m + 640 103 m = 7:01 106 m. The satellite travels a distance equal to one circumference in a time interval equal to one period., so its speed is 2 :01 106 m) v = 2R = (98:(7min)(60 s/min) = 7:49 103 m/s : T 0 (b) The magnitude of the acceleration is given by
2 2 49 103 a = v = (7:7:01 10m/s) = 8:00 m/s2 : 6m R (a) The minimum turning radius of the space probe is given by 62E v2 = [0:1(3:0 108 m/s)]2 = 4:6 1012 m : rmin = a 20(9:8 m/s2 ) max
(b) The time it takes to complete a 90 turn is m) T = 1 2r = 2(40:6 10 m/s) = 2:4 105 s ; 7 4 v 4(3: 10
12 which is equivalent to 2.8 days. (a) D = 2r = 2(0:15 m) = 0:94 m: (b) v = (1200D)=(60 s) = (1200)(0:94 m)=(60 s) = 19 m/s:
63E CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS 101 (c) a = v2 =r = (19 m/s)2 =(0:15 m) = 2:4 103 m/s2 : The direction of a is towards the center of the fan. (d) T = D=v = 0:94 m=(19 m/s) = 0:049 s: (a) The minimum turning radius of the train is given by 2 2 rmin = av = (0:(216 km/h) 2 ) = 7:3 103 m : 050)(9:8 m/s max (b) The speed of the train must be reduced to no more than p p v = ar = 0:050(9:8 m/s2 )(1:00 103 ) = 22 m/s ; which is about 79 km/h. (a) v = 2r=T = 2(20 km)=1:0 s = 1:3 105 km/s: (b) a = v2 = (126 km/s)2 = 8:0 104 ; g rg (20 km)(9:8 m/s2 )
65E 64E i.e., a = (8:0 104 )(9:80 m/s2 ) = 7:9 105 m/s2 . The direction of a is towards the center of the star. (c) Obviously, both v and a will increase.
66E (a) Let R be the radius of the orbit and solve a = v2 =R for v: p p v = Ra = (5:0 m)(7:0)(9:8 m/s2 ) = 19 m/s : (b) If the astronaut goes around N times in time t then his speed is v = 2NR=t. Solve for N=t and substitute v = 19 m/s, and R = 5:0 m: N = v = (19 m/s)(60 s/min) = 35 rev/min: t 2R 2(5:0 m) (c) The period of motion is :0 m) T = 2R = 2(5m/s = 1:7 s : v 19 (a) The speed of a person at the Earth's equator is v = 2R=T , where R is the radius of the Earth (6:37 106 m) and T is the length of a day (8:64 104 s): v = 2(6:37 106 m)=(8:64 104 s) = 463 m/s. The magnitude of the acceleration is given by a = v2 =R = (463 m/s)2 =(6:37 106 m) = 0:034 m/s2 .
67P 102 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS (b) If T is the period, then v = 2R=T is the speed and a = v2 =R = 42 R2 =T 2 R = 42 R=T 2
is the magnitude of the acceleration. Thus
6 T = 2 R = 2 6:37 10 2 m = 5:1 103 s = 84 min : a 9:8 m/s
r s (a) The period of motion is T = 1:0 min=5 = 60 s=5 = 12 s. (b) and (c) The magnitude of the acceleration is given by a = v2 =R, where R is the radius of the wheel, and v is the speed of the passenger. Since the passenger goes a distance 2R for each revolution, his speed is 5(2)(15 m)=(60 s) = 7:85 m/s and his acceleration is 68P a = (7:85 m/s) = 4:1 m/s2 : 15 m
At the highest point the acceleration is downward and at the lowest point it is upward, always toward the center of the circle. It has the same magnitude at every position. The person moves in a circle of radius r = R cos , where R is the radious of the Earth and is the lattitude. The period T of the circular motion is 1 day (i.e., 86; 400 s). Thus the acceleration is
69P 2 a = !2 r = 2 2 R cos = 2 2 (6:37 106 m)(cos 40 ) = 0:026 m/s2 : T 86400 s (a) The x and y components of the position vector r(t) as a function of time t are given by 70P t x(t) = r cos[(t)] = r sin 2T
and t y(t) = r + r sin[(t)] = r cos 2T ; CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS 103 respectively, where T is the period of the motion. Thus at t = 5:00 s (5 00 x(t) = (3:00 m) sin 220::0 s s) = 3:00 m and (5 00 y(t) = 3:00 m + (3:00 m) cos 220::0 s s) = 3:00 m : The magnitude of r(t) is P r (t) O x r(t) = x(t)2 + y(t)2 = (3:00 m)2 + (3:00 m)2 = 4:24 m ;
and the angle (t) that r(t) makes with the positive x axis is given by p p (t) = tan
Similarly, at t = 7:50 s 1 y(t) = tan x(t) 1 3:00 m = 45:0 : 3:00 m (7 50 x(t) = (3:00 m) sin 220::0 s s) = 2:12 m and (7 50 y(t) = 3:00 m + (3:00 m) cos 220::0 s s) = 5:12 m : The magnitude of r(t) is then r(t) = x(t)2 + y(t)2 = (2:12 m)2 + (5:12 m)2 = 5:54 m ;
and the angle (t) that r(t) makes with the positive x axis is given by p p (t) = tan
At t = 10:0 s, we have 1 y(t) = tan x(t) 1 5:54 m = 67:5 : 2:54 m (10 x(t) = 3:00 m sin 220:0:0 s) = 0:00 m s 104 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS (10 y(t) = 3:00 m + 3:00 m cos 220:0:0 s) = 6:00 m : s The magnitude of r(t) is p p r(t) = x(t)2 + y(t)2 = (0:00 m)2 + (6:00 m)2 = 6:00 m ; and the angle (t) that r(t) makes with the positive x axis is given by 1 y (t) = tan 1 6:00 = 90:0 : (t) = tan x(t) 0:00 (b) The displacement vector, pointing from A to B in the gure, has a magnitude of p p r = 2r = 2(3:00 m) = 4:24 m and makes an angle of 135 with the positive x axis. (c) The magnitude of the average velocity is 24 m/s v = r = 4:5:00 s = 0:848 m/s ; t and the direction of the velocity is the same as that of the displacement. (d) At point A r v = 2T = 2(3::00sm) = 0:942 m/s : 20 0 The velocity is in the positive y direction. At point B , the magnitude of the velocity is still the same as that at point A, while the velocity is now pointing in the negative x direction. (e) The magnitude of the acceleration ais always given by 2 2 a = vr = (0:942 m/s) = 0:296 m/s2 ; 3:00 m while the direction of a is always towards the center of the circle. Thus a is in the negative x direction at point A and in the negative y direction at point B . To calculate the centripetal acceleration of the stone you need to know its speed while it is being whirled around. This the same as its initial speed when it ys o. Use the kinematic equations of projectile motion to nd that speed. Take the y axis to be upward and the x axis to be horizontal. Place the origin at the point where the stone leaves its circular orbit and take the time to be 0 when this occurs. Then the coordinates of the stone when it is a projectile are given by x = v0 t and y = 1 gt2 . It hits the ground when x = 10 m 2 and y = 2:0 m. Note that the initial velocity is horizontal. Solve the second equation for p the time: t = 2y=g. Substitute this expression into the rst equation and solve for v0 : q p v0 = x g=2y = (10 m) (9:8 m/s2 )=[2( 2:0 m)] = 15:7 m/s. The magnitude of the centripetal acceleration is a = v2 =R = (15:7 m/s)2 =(1:5 m) = 160 m/s2 .
71P and CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS 105 (a) Choose the upstream direction to be positive. Then 72E vbg = vbw + vwg = 14 km/h 9 km/h = 5 km/h :
Since vbg > 0, we know it is in the upstream direction. (b) vcg = vcb + vbg = 6 km/h + 5 km/h = 1 km/h ; where the minus sign indicates that the velocity is in the downstream direction. When the escalator is stalled the speed of the person is vp = `=t, where ` is the length of the escalator and t is the time the person takes to walk up it. This is vp = (15 m)=(90 s) = 0:167 m/s. The escalator moves at ve = (15 m)=(60 s) = 0:250 m/s. The speed of the person walking up the moving escalator is v = vp + ve = 0:167 m/s + 0:250 m/s = 0:417 m/s and the time taken to move the length of the escalator is t = `=v = (15 m)=(0:417 m/s) = 36 s. If the various times given are independent of the escalator length then the answer does not depend on that length either. In terms of ` (in meters) the speed (in meters per second) of the person walking on the stalled escalator is `=90, the speed of the moving escalator is `=60, and the speed of the person walking on the moving escalator is v = (`=90) + (`=60) = 0:0278`. The time taken is t = `=v = `=0:0278` = 36 s, independently of `. Choose the eastward direction to be positive, and let the velocity of the jetstream wind be vx . Then we have 2700 mi 2700 mi = 50 min ; 600 mi/h v 600 mi/h + v
x x
74E 73E which gives vx = 55 mi/h. Since vx > 0, the wind is due east (i.e., from the west).
75P From the perspective of the cameraman, the cheetah changes its velocity from 30 mi/h westward to (60 mi/h + 40 mi/h) = 100 mi/h eastward. Its acceleration is then mi/h)](1 h=3600 s) = 0:018 mi/s2 : a = [100 mi/h ( 30:0 s 2 From the perspective of the crew member, the cheetah changes its velocity from (30 mi/h+ 40 mi/h) = 70 mi/h westward to 40 mi/h eastward. The net change in its velocity is the same as viewed by both the cameraman and the crew member. Thus the acceleration of the cheetah is the same as viewed by both parties. 106 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS Let the speed of Peter be v1 and that of Paul be v2 . Also, denote the length of the corridor by L, then L=v1 = 150 s and L=v2 = 70 s. The time Mary takes is then 76P L 1 t = v + v = v =L + v =L = 1=150 s 1 1=70 s = 48 s : + 1 2 1 2
77E Let the velocity of the player be v1 and the relative velocity between the player and the ball be v2 . Then the velocity v of the ball relative to the eld is given by v = v1 + v2 . The smallest angle min corresponds to the case when v ? v1 . Thus v2 v goal v1 min = cos
= cos 1 v1 v2 1 4:0 m/s 130 :
78E 6:0 m/s (a) Denote the police and the motorist with subscripts p and m, respectively. The velocity of the motorist with respect to the police car is
vmp = vm vp = ( 60 j + 80 i) (km/h) : (b) vmp happens to be along the line of sight. This can be seen from Fig. 442. Notice that (60 km/h)=(600 m) = (80 km/h)=800 m. (c) No, they remain unchanged. Relative to the car the velocity of the snow
akes has a vertical component of 8:0 m/s and a horizontal component of 50 km/h = 13:9 m/s. The angle from the vertical is given by tan = vh =vv = (13:9 m/s)=(8:0 m/s) = 1:74: The angle is 60 .
79E CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS 107 Choose a coordinate system with the x axis pointing horizontally to the left and the y axis vertically upward. Then the velocity vs of the shopper is given by
vs = (v cos 40 ) i + (v sin 40 ) j ; 80E where v = 0:75 m/s. Similarly, for the daughter
vd = (v cos 40 ) i (v sin 40 ) j : The relative velocity is then
vsd = vs = [(v cos 40 ) i + (v sin 40 ) j] [(v cos 40 ) i (v sin 40 ) j] = 2(0:75 m/s)(sin 40 ) j = (0:96 j) m/s : vd (a) The initial speed of the package relative to the ground is vi = 12 m/s 6:2 m/s = 5:8 m/s: (b) The time of ight of the package is
s 81P t=
so the horizontal separation is 2H = g s (2)(9:5 m) = 1:4 s ; 9:8 m/s2 x = (12 m/s)(1:4 s) = 16:7 m : (c) At the instant before the impact, the horizontal component of the velocity vector of the package is still vx = vi = 5:8 m/s, while the (downward) vertical component is given by vy = 2gH = 2(9:80 m/s2 )(9:5 m) = 14 m/s :
Thus the angle in question is given by p p = tan
82P 1 vy = tan vx 1 14 m/s = 67 : 5:8 m/s Since the raindrops fall vertically relative to the train, the horizontal component of the velocity of a raindrop is vh = 30 m/s, the same as the speed of the train. If vv is the vertical component of the velocity and is the angle between the direction of motion and 108 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS the vertical then tan = vh =vv . Thus vp= vh = tan = (30 m/s)= tan 22 = 74:3 m/s. The v p 2 + v 2 = (30 m/s)2 + (74:3 m/s)2 = 80 m/s. speed of a raindrop is v = vh v From the gure we see that vw sin = vp sin 20 ; vw cos + vp cos = vpa = D : t
83P W N
20o Substituting vp = 500 km/h, D = 800 km and t = 2:00 h into the above equations, we nd vw = 185 km/h and = 112 , which means that the wind velocity is 22 south of west. vp E vw (a) The velocity vector for ships A and B are given by vA = (vA sin 45 ) i + (vA cos 45 ) j and vB = (vB sin 40 ) i (vB cos 40 ) j ; 84P vA
45o N W
40o E vB respectively. Thus the relative velocity is vAB = vA vB = (vB sin 40 vA sin 45 ) i + (vB cos 40 + vA cos 45 ) j ; the magnitude of which is p vAB = (vB sin 40 vA sin 45 )2 + (vB cos 40 + vA cos 45 )2 p = (28 sin 40 24 sin 45 )2 + (28 cos 40 + 24 cos 45 )2 (knots) = 38 knots : The angle which vAB makes with the south direction is given by 24 sin 45 1 vAB;x = tan 1 28 sin 40 = tan v 28 cos 40 + 24 cos 45 = 1:5 ; AB;y i.e., vAB is 1:5 east of north. CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS 109 (b) The time t it takes is given by t = 160 nautical miles = 160 nautical miles = 4:2 h : vAB 38 knots (c) Since vAB is unchanged, ship B will still be located 1:5 west of south as viewed from ship A. be the velocity of the plane relative to the ground, vGA be the velocity of the air relative to the ground, and vAP be the velocity of the plane relative to the air. The vector diagram is shown to the right, from which we get vGP = vAP + vGA . Since the magnitudes vGP and vAP are equal the triangle is isosceles, with two sides of equal length. Consider either one of the right triangles formed when the bisector of is drawn (the dotted line). It bisects vGA , so (70:0 sin(=2) = 2vvGA = 2(135 mi/h) = 0:259 : mi/h) GP
85P Let vGP N vAG vPG /2 /2 vPA W S E This means = 30:1 . Now vGA makes the same angle with the EW line as the dotted line does with the NS line. The wind is blowing in the direction 15 north of west. (b) The plane is headed in the direction 30 east of north.There is another solution, with the plane headed 30 west of north and the wind blowing 15 north of east. Let the time the bullet spends inside the boxcar be t and its speed inside the boxcar be v3 . Then from the gure shown
86P v = sin 1 v1 t 3t = sin
1 v1t v1 = 2:6 : 85 km/h(1000 m/km)(1 h=3600 s) (1 20%)(650 m/s) v3t Thus the sniper should re at an angle of (90 + 2:6 ) = 93 from the direction of motion of the boxcar. v3 110 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS (a) Let the velocities of the woman and the current relative to the ground be vw and vc , respectively. The resultant velocity v of the woman relative to the ground is then v = vw + vc . Also in this case v ? vc : Thus the woman must row in the direction shown, with 87P vw v (b) The time it takes is i.e., she has to row at 30 upstream. 4:0 mi t = 4:0vmi = v4:0 mi = (4:0 mi/h)(cos 30 ) = 1:14 h = 69 min : cos
w = sin 1 vc = sin vw 1 2:0 mi/h = 30 ; 4:0 mi/h v0 (c) As she rows downstream, her speed is vw + vc . As she rows back, her speed reduces to vw vc . Thus the total time it takes to go downstream a distance L = 2 mi and then back is mi t = v L v + v L v = 4:0 mi/h2+ 2:0 mi/h + 4:0 mi/h2 mi2:0 mi/h = 80 min : w+ c w c (d) This time she moves upstream with speed vw vc and moves back with speed vw + vc , so t = v L v + v L v = 80 min : +
w c w c (e) To cross the river in the shortest possible time, she has to achieve the largest possible velocity perpendicular to the current. This can be done by heading directly across the river (i.e., vw ? vc ), in which case the time it takes will be 0 tmin = 4:v mi = 4:40:0 mi = 1:0 h = 60 min : mi/h
w
88E (a) and (b) In Eq. 432, you can see easily that if one of the two speeds on the righthandside (vPB or vBA ) is equal to c, then vPA is always equal to c as well. For example, suppose vPB = c, then as well, regardless of the speed of the ship itself. (In fact, this remarkable and somewhat v c vPA = 1 +PB +vvBA 2 = 1 + + vBA 2 = c 1 + vBA =c = c : vPB BA =c cvBA =c 1 + vBA =c Since in our case the signal speed we detect is c, it will remain c relative to the spaceship CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS 111 counterintuitive observation about the constancy of the speed of light is one of the key postulates in Einstein's special theory of relativity, as you will see later.) The speed of the electron as measured by observer A is + 0:63c v = 1 0:42c:42)(0:63) = 0:83c : + (0 Denote our own spaceship, the proton, and the other spaceship with subscripts A, B and C , respectively. Then according to the problem statement vBA = 0:9800c and vBC = 0:9800c. Thus the speed of the other ship relative to ours is v + 0 9800 vCA = 1 +CB +vvBA 2 = 01:9800c9800:c)2 =cc = 0:9998c ; vCB BA =c + (0: where we used vCB = vBC = 0:9800c. (a)
91P 90P 89E Denote our galaxy as and the speed of relative to as v , etc. Obviously, v = v = 0:35c . (b) v v = 1 + +vv 2 = 1 + (0::35cc)(0::35cc))=c2 = 0:62c : v =c (0 35 )(0 35 92 our galaxy(
) ! for v0 = 9:0 m/s: = 39:2 and x = 10:15 m; for v0 = 5:0 m/s: = 31:6 and x = 4:15 m; for v0 = 15 m/s: = 42:6 and x = 25:0 m for launch angles from 5 to 70 , it always move away from the launch site. for a 75 launch angle, it moves toward the site from 11:5 s to 18:5 s after launch. for an 80 launch angle, it moves toward the site from 10:5 s to 20:5 s after launch. for an 85 launch angle, it moves toward the site from 10:5 s to 20:5 s after launch. for a 90 launch angle, it moves toward the site from 10 s to 20:5 s after launch.
93 112 CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS (a) min = 33 and max = 58 ; (b) 33:4 m/s; (c) 33:1 m/s at 45:4 ; (d) 32:5 m/s at 48 (a) 1:6 s; (b) yes; (c) 14 m/s; (d) yes (a) 2(40:0 m) NW; (b) (1:9 m) NW; (c) (0:47 m/s2 ) NE; (d) (7:1 m/s) SE (a) D = (1:0 m) i (2:0 m) j + (1:0 m) k; (b) 2:4 m; (c) v = (0:025 m/s) i (0:050 m/s) j + (0:025 m) k; (d) cannot be determined without additional information
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