3 Chapter Solutions
*3.1
x = r cos = (5.50 m) cos 240 = (5.50 m)(0.5) = 2.75 m y = r sin = (5.50 m) sin 240 = (5.50 m)(0.866) = 4 .76 m
3.2
(a)
d= d=
(x 2 x 1 ) 2 + (y 2 y 1 ) 2 = 25.0 + 49.0 = 8.60 m (2.00)2 + (4.00)2 = 4.00 = 63.4 2.00
(2.00 [3.00]2) + (4.00 3.00)2
(b)
r1 =
20.0 = 4.47 m
1 = tan1
r2 =
(3.00)2 + (3.00)2 =
18.0 = 4.24 m
2 = 135 measured from + x axis.
3.3 We have 2.00 = r cos 30.0 r= 2.00 = 2.31 cos 30.0
and y = r sin 30.0 = 2.31 sin 30.0 = 1.15 3.4 (a) x = r cos and y = r sin , therefore x1 = (2.50 m) cos 30.0, y1 = (2.50 m) sin 30.0, and (x1, y1) = (2.17, 1.25) m x2 = (3.80 m) cos 120, y2 = (3.80 m) sin 120, and (x2, y2) = (1.90, 3.29) m (b) d= (x)2 + (y)2 = 16.6 + 4.16 = 4.55 m
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2
Chapter 3 Solutions
3.5
The x distance out to the fly is 2.00 m and the y distance up to the fly is 1.00 m. (a) We can use the Pythagorean theorem to find the distance from the origin to the fly, distance = x2 + y2 = (2.00 m)2 + (1.00 m)2 = 5.00 m2 = 2.24 m
(b)
1 = Arctan = 26.6; r = 2.24 m, 26.6 2 x2 + y2 and = Arctan y x (x)2 + y2 = x2 + y2 = r and its angle is
3.6
We have r =
(a)
The radius for this new point is Arctan y = 180 (x)
(b)
(2x)2 + (2y)2 = 2r 180 + .
This point is in the third quadrant if (x, y) is in the first
quadrant or in the fourth quadrant if (x, y) is in the second quadrant. It is at angle
(c)
(3x)2 + (3y)2 = 3r
This point is in the fourth quadrant if (x, y) is in the first
quadrant or in the third quadrant if (x, y) is in the second quadrant. It is at angle . 3.7 (a) The distance d from A to C is
300 km C
d
d=
x2 + y2
30 B
200 km A
where x = (200) + (300 cos 30.0) = 460 km and y = 0 + (300 sin 30.0) = 150 km
d = (460)2 + (150)2 = 484 km
(b) tan = y 150 = = 0.326 x 460
R 13 km
= tan-1(0.326) = 18.1 N of W
3.8 R 14 km
6 km
= 65 N of E
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Chapter 3 Solutions x 100 m
3
3.9
tan 35.0 =
x = (100 m)(tan 35.0) = 70.0 m
x
35.0 100 m
3.10
R = 310 km at 57 S of W
B
-R A
E base
0
100 km
200 km
3.11
(a)
Using graphical methods, place the tail of vector B at the head of vector A. The new vector A + B has a magnitude of 6.1 a t 112 from the x-axis. The vector difference A B is found by placing the negative of vector B at the head of vector A. The resultant vector A B has magnitude 14.8 units at an angle of 22 from the + x-axis.
y B --B
(b)
A A+B A--B x O
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4
Chapter 3 Solutions
3.12
Find the resultant F 1 + F 2 graphically by placing the tail of F 2 at the head of F 1. The resultant force vector F1 + F2 is of magnitude 9.5 N and at an angle of 57 above the x-axis .
y
F1 + F2
F2
F1 x
0
1
2
3 N
3.13
(a) B.
d = 10.0i = 10.0 m
since the displacement is a straight line from point A to point
C
(b)
The actual distance walked is not equal to the straight-line displacement. The distance follows the curved path of the semi-circle (ACB). 1 s = (2 r) = 5 = 15.7 m 2
B d 5.00 m
A
(c)
If the circle is complete, d begins and ends at point A. Hence, d = 0 .
3.14
Your sketch should be drawn to scale, and should look somewhat like that pictured below. The angle from the westward direction, , can be measured to be 4 N of W , and the distance R from the sketch can be converted according to the scale to be 7.9 m .
N
W
15.0 meters
R 3.50 meters
E 8.20 meters 30.0
S
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Chapter 3 Solutions 3.15 To find these vector expressions graphically, we draw each set of vectors. Measurements of the results are taken using a ruler and protractor. (a) A + B = 5.2 m at 60 (b) A B = 3.0 m at 330
5
B A+B
A -B
A
A-B
a
b
0
2m
4m
0
2m
4m
(c)
B A = 3.0 m at 150
(d)
A 2B = 5.2 m at 300
A
B-A B A - 2B -A
- 2B
c
d
0
2m
4m
0
2m
4m
*3.16
(a)
The large majority of people are standing or sitting at this hour. Their instantaneous foot-to-head vectors have upward vertical components on the order of 1 m and randomly oriented horizontal components. The citywide sum will be ~105 m upward .
(b)
Most people are lying in bed early Saturday morning. We suppose their beds are oriented north, south, east, west quite at random. Then the horizontal component of their total vector height is very nearly zero. If their compressed pillows give their height vectors vertical components averaging 3 cm, and if one-tenth of one percent of the population are on-duty nurses or police officers, we estimate the total vector height as ~ 105(0.03 m) + 102(1 m) ~103 m upward .
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6
Chapter 3 Solutions
y 135 ft 200 ft d 30.0 40.0 135 ft x
3.17
The scale drawing for the graphical solution should be similar to the figure at the right. The magnitude and direction of the final displacement from the starting point are obtained by measuring d and on the drawing and applying the scale factor used in making the drawing. The results should be d 420 ft and 3
N
3.18 x 0 km 1.41 4.00 2.12 4.71 R= y 3.00 km 1.41 0 2.12 2.29
4.00 km 45.0 3.00 km 2.00 km 45.0
R
3.00 km
|x|2 + |y|2 = 5.24 km y = 154 x or
t
E
= tan1
3.19
= 25.9 N of W
Call his first direction the x direction. R = 10.0 m i + 5.00 m(j) + 7.00 m(i) = 3.00 m i 5.00 m j = (3.00)2 + (5.00)2 m at Arctan 5 to the right 3
R = 5.83 m at 59.0 to the right from his original motion 3.20 Coordinates of super-hero are: x = (100 m) cos (30.0) = 86.6 m y = (100 m) sin (30.0) = 50.0 m
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Chapter 3 Solutions
7
y
30.0 100 m
x
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8
Chapter 3 Solutions
3.21
The person would have to walk 3.10 sin(25.0) = 1.31 km north , and 3.10 cos(25.0) = 2.81 km east .
3.22
+ x East, + y North x = 250 + 125 cos 30 = 358 m y = 75 + 125 sin 30 150 = 12.5 m d= (x)2 + (y)2 = (358)2 + (12.5)2 = 358 m
tan =
(y) 12.5 = = 0.0349 = 2.00 (x) 358
d = 358 m at 2.00 S of E *3.23 Let the positive x-direction be eastward, positive y-direction be vertically upward, and the positive z-direction be southward. The total displacement is then d = (4.80 cm i + 4.80 cm j) + (3.70 cm j 3.70 cm k) or (a) (b) 3.24 d = 4.80 cm i + 8.50 cm j 3.70 cm k The magnitude is d = (4.80)2 + (8.50)2 + (3.70)2 cm = 10.4 cm 8.50 , giving = 35.5 . 10.4
Its angle with the y-axis follows from cos =
B = Bxi + Byj + B2k B = 4.00i + 6.00j + 3.00k |B| = (4.00)2 + (6.00)2 + (3.00)2 = 7.81
= cos1 = cos1
4.00 = 59.2 7.81
6.00 = 39.8 7.81
3.00 = cos1 7.81 = 67.4
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Chapter 3 Solutions
y
9
3.25
Ax = 25.0 Ay = 40.0 A= Ax + Ay =
2 2
(25.0)2 + (40.0)2 = 47.2 units = 58.0, so that = 122
A 40.0 40.0
-25.0 -25.0
From the triangle, we find that
t
x
Goal Solution A vector has an x component of 25.0 units and a y component of 40.0 units. Find the magnitude and direction of this vector.
y 40 r 30 20 10 x --30--25--20--15--10 --5 0 5 10
G: First we should visualize the vector either in our mind or with a sketch. Since the hypotenuse of the right triangle must be greater than either the x or y components that form the legs, we can estimate the magnitude of the vector to be about 50 units. The direction of the vector appears to be about 120 from the +x axis. O: The graphical analysis and visual estimates above may be sufficient for some situations, but we can use trigonometry to obtain a more precise result. A: The magnitude can be found by the Pythagorean theorem: r = r= (25.0 units)2 + (40 units)2 = 47.2 units x2 + y2
We observe that tan =
y (if we consider x and y to both be positive) . x y 40.0 = tan1 = tan1 = tan1 (1.60) = 58.0 x 25.0
The angle from the +x axis can be found by subtracting from 180. = 180 58 = 122 L: Our calculated results agree with our graphical estimates. We should always remember to check that our answers are reasonable and make sense, especially for problems like this where it is easy to mistakenly calculate the wrong angle by confusing coordinates or overlooking a minus sign. Quite often the direction angle of a vector can be specified in more than one way, and we must choose a notation that makes the most sense for the given problem. If compass directions were stated in this question, we could have reported the vector angle to be 32.0 west of north or a compass heading of 328.
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10
Chapter 3 Solutions
*3.26
The east and north components of the displacement from Dallas (D) to Chicago (C) are the sums of the east and north components of the displacements from Dallas to Atlanta (A) and from Atlanta to Chicago. In equation form: dDCeast = dDAeast + dACeast = 730 cos 5.00 560 sin 21.0 = 527 miles. dDCnorth = dDAnorth + dACnorth = 730 sin 5.00 + 560 cos 21.0 = 586 miles. By the Pythagorean theorem, d = Then tan = (dDCeast)2 + (dDCnorth)2 = 788 mi
dDCnorth = 1.11 and = 48.0. dDCeast
Thus, Chicago is 788 miles at 48.0 north east of Dallas . 3.27 x = d cos = (50.0 m)cos(120) = 25.0 m y = d sin = (50.0 m)sin(120) = 43.3 m d = (25.0 m)i + (43.3 m)j 3.28 (a)
-B B A-B A A+B
B
(b)
C = A + B = 2.00i + 6.00j + 3.00i 2.00j = 5.00i + 4.00j C= 25.0 + 16.0 at Arctan 4 5
C = 6.40 at 38.7 D = A B = 2.00i + 6.00j 3.00i + 2.00j = 1.00i + 8.00j (1.00)2 + (8.00)2 at Arctan 8.00 (1.00)
D=
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Chapter 3 Solutions
11
D = 8.06 at (180 82.9) = 8.06 at 97.2
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12
Chapter 3 Solutions
3.29
d=
(x 1 + x 2 + x 3 ) 2 + (y 1 + y 2 + y 3 ) 2 52.0 = 7.21 m
= (3.00 5.00 + 6.00)2 + (2.00 + 3.00 + 1.00)2 =
= tan-1
3.30
6.00 = 56.3 4.00
A = 8.70i + 15.0j B = 13.2i 6.60j A B + 3C = 0 3C = B A = 21.9i 21.6j C = 7.30i 7.20j Cx = 7.30 cm Cy = 7.20 cm or
3.31
(a) (b) (c) (d)
(A + B) = (3i 2j) + (i 4j) = 2i 6j (A B) = (3i 2j) (i 4j) = 4i + 2j A + B = A B = 22 + 62 = 6.32 42 + 22 = 4.47
(e)
6 A + B = tan1 = 71.6 = 288 2 2 A B = tan1 = 26.6 4
3.32
Let i = east and j = north. R = 3.00b j + 4.00b cos 45 i + 4.00b sin 45 j 5.00b i R = 2.17b i + 5.83b j R= 2.17 2 + 5.832 b at Arctan
5.83 N of W 2.17
= 6.22 blocks at 110 counterclockwise from east
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Chapter 3 Solutions 3.33 x = r cos and y = r sin , therefore: (a) (b) (c) 3.34 (a) x = 12.8 cos 150, y = 12.8 sin 150, and (x, y) = (11.1i + 6.40j) m x = 3.30 cos 60.0, y = 3.30 sin 60.0, and (x, y) = (1.65i + 2.86j) cm x = 22.0 cos 215, y = 22.0 sin 215, and (x, y) = (18.0i 12.6j) in D = A + B + C = 2i + 4j D = (b) 22 + 42 = 4.47 m at = 63.4
13
E = A B + C = 6i + 6j E = 62 + 62 = 8.49 m at = 135
3.35
d1 = (3.50j) m d2 = 8.20 cos 45.0i + 8.20 sin 45.0j = (5.80i + 5.80j) m d3 = (15.0i) m R = d1 + d2 + d3 = (15.0 + 5.80)i + (5.80 3.50)j = (9.20i + 2.30j) m (or 9.20 m west and 2.30 m north) The magnitude of the resultant displacement is |R| = Rx + R y =
2 2
(9.20)2 + (2.30)2 = 9.48 m 2.30 = 166 9.20
The direction is = Arctan 3.36 Refer to the sketch
R = A + B + C = 10.0i 15.0j + 50.0i = 40.0i 15.0j R = [(40.0)2 + (15.0)2]1/2 = 42.7 yards
|A| = 10.0
|B| = 15.0
R
|C| = 50.0
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14
Chapter 3 Solutions
3.37
(a)
F = F1 + F2 F = 120 cos (60.0)i + 120 sin (60.0)j 80.0 cos (75.0)i + 80.0 sin (75.0)j F = 60.0i + 104j 20.7i + 77.3j = (39.3i + 181j) N F = (39.3)2 + (181)2 = 185 N ; = tan1 181 = 77.8 39.3
(b)
F3 = F = (39.3i 181j) N
Goal Solution The helicopter view in Figure P3.37 shows two people pulling on stubborn a mule. Find (a) the single force that is equivalent to the two forces shown and (b) the force that a third person would have to exert on the mule to make the resultant force equal to zero. The forces are measured in units of newtons. G: The resultant force will be larger than either of the two individual forces, and since the two people are not pulling in exactly the same direction, the magnitude of the resultant should be less than the sum of the magnitudes of the two forces. Therefore, we should expect 120 N < R < 200 N. The angle of the resultant force appears to be straight ahead and perhaps slightly to the right. If the stubborn mule remains at rest, the ground must be exerting on the animal a force equal to the resultant R but in the opposite direction.
120 N 80 N
75
60
O: We can find R by adding the components of the two force vectors. A: F1 = (120 cos 60)i N + (120 sin 60)j N = 60.0i N + 103.9j N F2 = (80 cos 75)i N + (80 sin 75)j N = 20.7i N + 77.3j N R = F1 + F2 = 39.3i N + 181.2j N R = |R| = (39.3)2 + (181.2)2 = 185 N
The angle can be found from the arctan of the resultant components. y 181.2 = tan1 = tan1 = tan1 (4.61) = 77.8 counterclockwise from the +x axis x 39.3 The opposing force that the either the ground or a third person must exert on the mule, in order for the overall resultant to be zero, is 185 N at 258 counterclockwise from +x. L: The resulting force is indeed between 120 N and 200 N as we expected, and the angle seems reasonable as well. The process applied to solve this problem can be used for other "statics" problems encountered in physics and engineering. If another force is added to act on a system that is already in equilibrium (sum of the forces is equal to zero), then the system may accelerate. Such a system is now a "dynamic" one and will be the topic of Chapter 5.
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Chapter 3 Solutions 3.38 East x 0m 1.41 0.500 +0.914 R = 3.39 x
2
15
North y 4.00 m 1.41 0.866 4.55 + y
2
= 4.64 m at 78.6 N of E
A = 3.00 m, A = 30.0, B = 3.00 m, B = 90.0 Ax = A cos A = 3.00 cos 30.0 = 2.60 m, Ay = A sin A = 3.00 sin 30.0 = 1.50 m so, A = Axi + Ayj = (2.60i + 1.50j) m
Bx = 0, By = 3.00 m so B = 3.00j m A + B = (2.60i + 1.50j) + 3.00j = (2.60i + 4.50j) m *3.40 The y coordinate of the airplane is constant and equal to 7.60 103 m whereas the x coordinate is given by x = vit where vi is the constant speed in the horizontal direction. At t = 30.0 s we have x = 8.04 103, so vi = 268 m/s. The position vector as a function of time is P = (268 m/s)t i + (7.60 103 m)j. At t = 45.0 s, P = [1.21 104 i + 7.60 103 j] m. The magnitude is P= (1.21 104)2 + (7.60 103)2 m = 1.43 104 m
and the direction is
= Arctan
7.60 103 = 1.21 104
y
32.2 above the horizontal 3.41 We have B = R A Ax = 150 cos 120 = 75.0 cm Ay = 150 sin 120 = 130 cm Rx = 140 cos 35.0 = 115 cm Ry = 140 sin 35.0 = 80.3 cm Therefore, B = [115 (75)]i + [80.3 130]j = (190i 49.7j) cm
A
B
120.0 35.0
R =A+B x
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16
Chapter 3 Solutions
B
= [1902 + (49.7)2]1/2 = 196 cm , = tan1
49.7 = 14.7 190
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Chapter 3 Solutions *3.42 Since A + B = 6.00j, we have (Ax + Bx)i + (Ay + By)j = 0i + 6.00 j giving Ax + Bx = 0, or Ax = Bx and Ay + By = 6.00 (1) (2)
A+B y
17
Since both vectors have a magnitude of 5.00, we also have: A x + Ay = Bx + By = (5.00)2
B
2 2 2 2
From Ax = Bx, it is seen that Ax = Bx . Therefore Ax + Ay = B x + By gives Ay = By . Then Ay = By, and Equation (2) gives Ay = By = 3.00. Defining
2 2 2 2
2
2
2
2
t
= 2t 2
t
A x
as the angle between either A or B and the y axis, it is seen that
Ay By 3.00 = = = 0.600 A B 5.00 and = 53.1 cos = The angle between A and B is then = 2 = 106 . 3.43 (a) (b) (c) 3.44 A = 8.00i + 12.0j 4.00 k B = A/4 = 2.00i + 3.00j 1.00k C = 3A = 24.0i 36.0j + 12.0k
R = 75.0 cos 240i + 75.0 sin 240j + 125 cos 135i + 125 sin 135j + 100 cos 160i + 100 sin 160j R = 37.5i 65.0j 88.4i + 88.4j 94.0i + 34.2j R = 220i + 57.6j (220)2 + 57.62 at Arctan 57.6 above the x-axis 220
R=
R = 227 paces at 165 3.45 (a) C = A + B = (5.00i 1.00j 3.00k) m |C| = (5.00)2 + (1.00)2 + (3.00)2 m = 5.92 m (b) D = 2A B = (4.00i 11.0j + 15.0k) m
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18
Chapter 3 Solutions
|D| = (4.00)2 + (11.0)2 + (15.0)2 m = 19.0 m
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Chapter 3 Solutions *3.46 The displacement from radar station to ship is S = (17.3 sin 136i + 17.3 cos 136j) km = (12.0i 12.4j) km From station to plane, the displacement is P = (19.6 sin 153i + 19.6 cos 153j + 2.20k) km, or P = (8.90i 17.5j + 2.20k) km. (a) From plane to ship the displacement is D = S P = (3.12i + 5.02j 2.20k) km (b) The distance the plane must travel is D = |D| = (3.12)2 + (5.02)2 + (2.20)2 km = 6.31 km
19
3.47
The hurricane's first displacement is displacement is
41.0 km (3.00 h) at 60.0 N of W, and its second h
25.0 km (1.50 h) due North. With i representing east and j representing h north, its total displacement is:
41.0 km cos 60.0 (3.00 h)(i) + 41.0 km sin 60.0 (3.00 h) j h h
+ 25.0
km (1.50 h) j = 61.5 km (i) + 144 km j h
with magnitude *3.48 (a)
(61.5 km)2 + (144 km)2 = 157 km
y
E = (17.0 cm) cos 27.0i + (17.0 cm) sin 27.0j E = (15.1i + 7.72j) cm
27.0
(b)
F = (17.0 cm) sin 27.0i + (17.0 cm) cos 27.0j
F
27.0 G E 27.0
F = (7.72i + 15.1j) cm (c) G = +(17.0 cm) sin 27.0i + (17.0 cm) cos 27.0j G = (+7.72i + 15.1j) cm
x
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20
Chapter 3 Solutions
3.49
Ax = 3.00, Ay = 2.00 (a) (b) A = Axi + Ayj = 3.00i + 2.00j |A| = tan = Ax + Ay =
2 2
(3.00)2 + (2.00)2 = 3.61
Ay 2.00 = = 0.667, tan1(0.667) = 33.7 Ax (3.00)
is in the 2nd quadrant, so = 180 + (33.7) = 146
(c) Rx = 0, Ry = 4.00, R = A + B thus B = R A and Bx = Rx Ax = 0 (3.00) = 3.00, By = Ry Ay = 4.00 2.00 = 6.00 Therefore, B = 3.00i 6.00j 3.50 Let +x = East, +y = North, x 300 175 0 125 (a) (b) 3.51 y 0 303 150 453 y = 74.6 N of E x
= tan1
R =
x2 + y2 = 470 km
Refer to Figure P3.51 in the textbook. (a) Rx = 40.0 cos 45.0 + 30.0 cos 45.0 = 49.5 Ry = 40.0 sin 45.0 30.0 sin 45.0 + 20.0 = 27.1 R = 49.5i + 27.1j (b) R = (49.4)2 + (27.1)2 = 56.4
= tan1
27.1 = 28.7 49.5
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Chapter 3 Solutions 3.52 Taking components along i and j , we get two equations: 6.00a 8.00b + 26.0 = 0 8.00a + 3.00b + 19.0 = 0 Solving simultaneously, a = 5.00, b = 7.00 Therefore, 5.00A + 7.00B + C = 0 *3.53
21
Let represent the angle between the directions of A and B. Since A and B have the same magnitudes, A, B, and R = A + B form an isosceles triangle in which the angles are 180 , /2, and /2. The magnitude of R is then R = 2A cos( /2). [Hint: apply the law of cosines to the isosceles triangle and use the fact that B = A.] Again, A, B, and D = A B form an isosceles triangle with apex angle . Applying the law of cosines and the identity (1 cos ) = 2 sin2( /2) gives the magnitude of D as D = 2A sin( /2). The problem requires that R = 100D. Thus, 2A cos( /2) = 200A sin( /2). This gives tan( /2) = 0.010 and = 1.15 .
R
/2 t/2 A
t
B D
A -B
t
*3.54
Let represent the angle between the directions of A and B. Since A and B have the same magnitudes, A, B, and R = A + B form an isosceles triangle in which the angles are 180 , /2, and /2. The magnitude of R is then R = 2A cos( /2). [Hint: apply the law of cosines to the isosceles triangle and use the fact that B = A.] Again, A, B, and D = A B form an isosceles triangle with apex angle . Applying the law of cosines and the identity (1 cos ) = 2 sin2( /2) gives the magnitude of D as D = 2A sin( /2). The problem requires that R = nD, or cos( /2) = nsin( /2), giving = 2 tan1 (1/n) .
R
/2 t/2 A
t
B D
A -B
t
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22
Chapter 3 Solutions
3.55
(a) (b)
Rx = 2.00 , Ry = 1.00 , Rz = 3.00 R = cos x = Rx + R y + R Z Rx R Ry R Rz R
2 2 2
=
4.00 + 1.00 + 9.00 Rx
=
14.0 = 3.74
(c)
x = cos1 y = cos1 z = cos1
= 57.7 from + x R = 74.5 from + y R
Rz R Ry
cos y =
cos z =
= 36.7 from + z
*3.56
Choose the +x-axis in the direction of the first force. The total force, in newtons, is then 12.0i + 31.0j 8.40i 24.0j = (3.60i) + (7.00j) N The magnitude of the total force is (3.60)2 + (7.00)2 N = 7.87 N and the angle it makes with our +x-axis is given by tan = (7.00) , = 62.8. Thus, its angle (3.60)
counterclockwise from the horizontal is 35.0 + 62.8 = 97.8
y x R 12 N 35.0 8.4 N horizontal
31 N
24 N
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Chapter 3 Solutions 3.57 d1 = 100i d2 = 300j
23
y
d3 = 150 cos (30.0)i 150 sin (30.0)j = 130i 75.0j d4 = 200 cos (60.0)i + 200 sin (60.0)j = 100i + 173j R = d1 + d2 + d3 + d4 = 130i 202j R = [(130)2 + (202)2]1/2 = 240 m 202 = 57.2 130
R
t
x
= tan1
= 180 + = 237
*3.58 dP/dt = d(4i + 3j 2t j)/dt = 0 + 0 2j = (2.00 m/s)j The position vector at t = 0 is 4i + 3j. At t = 1 s, the position is 4i + 1j, and so on. The object is moving straight downward at 2 m/s, so dP/dt represents its velocity vector . 3.59 v = vxi + vy j = (300 + 100 cos 30.0)i + (100 sin 30.0)j v = (387i + 50.0j) mi/h v 3.60 (a) = 390 mi/h at 7.37 N of E You start at r1 = rA = 30.0 m i 20.0 m j. The displacement to B is rB rA = 60.0i + 80.0j 30.0i + 20.0j = 30.0i + 100j You cover onehalf of this, 15.0i + 50.0j, to move to r2 = 30.0i 20.0j + 15.0i + 50.0j = 45.0i + 30.0j Now the displacement from your current position to C is rC r2 = 10.0i 10.0j 45.0i 30.0j = 55.0i 40.0j You cover onethird, moving to 1 r3 = r2 + r23 = 45.0i + 30.0j + (55.0i 40.0j) = 26.7i + 16.7j 3
2000 by Harcourt College Publishers. All rights reserved.
24
Chapter 3 Solutions
The displacement from where you are to D is rD r3 = 40.0i 30.0j 26.7i 16.7j = 13.3i 46.7j You traverse onequarter of it, moving to 1 1 r4 = r3 + (rD r3) = 26.7i + 16.7j + (13.3i 46.7j) = 30.0i + 5.00j 4 4 The displacement from your new location to E is rE r4 = 70.0i + 60.0j 30.0i 5.00j = 100i + 55.0j of which you cover onefifth, 20.0i + 11.0j, moving to r4 + r45 = 30.0i + 5.00j 20.0i + 11.0j = 10.0i + 16.0j. The treasure is at (10.0 m, 16.0 m) (b) Following the directions brings you to the average position of the trees. The steps we took numerically in part (a) bring you to rA + rB 1 rA + (rB rA) = 2 2 then to (rA + rB) rC (rA + rB)/2 (rA + rB + rC) + = 2 3 3 then to rA + rB + rC rD (rA + rB + rC) /3 (rA + rB + rC + rD) + = 3 4 4 and at last to rA + rB + rC + rD rE (rA + rB + rC + rD)/4 + 4 5 = rA + rB + rC + rD + rE 5
This center of mass of the tree distribution is in the same location whatever order we take the trees in.
2000 by Harcourt College Publishers. All rights reserved.
Chapter 3 Solutions
25
3.61
(a)
From the picture R1 = ai + bj and R1
= a2 + b2
(b)
R2 = ai + bj + ck. Its magnitude is
z a
R1
2
+ c2 =
a2 + b2 + c2
b
O x R1
R2
c y
3.62
(a) (b)
r1 + d = r2 defines the displacement d, so d = r2 r1.
r2 d
r1
3.63
The displacement of point P is invariant under rotation of the coordinates. Therefore, r = r' and r2 = (r')2 or, x2 + y2 = (x')2 + (y')2
y
Also, from the figure, = y' y tan1 x' = tan1 x
y
P
r
x
y' = x'
y tan x
y 1 + tan x
O
t
x
Which we simplify by multiplying top and bottom by x cos . Then, x' = x cos + y sin , y' = x sin + y cos
2000 by Harcourt College Publishers. All rights reserved.
26
Chapter 3 Solutions
2000 by Harcourt College Publishers. All rights reserved.