Solution manual fundamentals of physics extended, 8th editionch03

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1. The x and the y components of a vector a lying on the xy plane are given by

ax = a cos θ ,

a y = a sin θ

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where a =| a | is the magnitude and θ is the angle between a and the positive x axis.
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(a) The x component of a is given by ax = 7.3 cos 250° = – 2.5 m.

(b) and the y component is given by ay = 7.3 sin 250° = – 6.9 m.
In considering the variety of ways to compute these, we note that the vector is 70° below
the – x axis, so the components could also have been found from ax = – 7.3 cos 70° and ay
= – 7.3 sin 70°. In a similar vein, we note that the vector is 20° to the left from the – y
axis, so one could use ax = – 7.3 sin 20° and ay = – 7.3 cos 20° to achieve the same results.


2. The angle described by a full circle is 360° = 2π rad, which is the basis of our
conversion factor.
(a)
20.0° = ( 20.0° )

2π rad
= 0.349 rad .
360°

50.0° = ( 50.0° )

2π rad
= 0.873 rad
360°

(b)

(c)
100° = (100° )

2π rad
= 1.75 rad
360°

(d)
0.330 rad = ( 0.330 rad )

360°
= 18.9°
2π rad

(e)
2.10 rad = ( 2.10 rad )

360°
= 120°
2π rad

7.70 rad = ( 7.70 rad )

360°

= 441°
2π rad

(f)


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3. A vector a can be represented in the magnitude-angle notation (a, θ), where

a = ax2 + a 2y
is the magnitude and

§ ay ·
¸
© ax ¹

θ = tan − 1 ¨
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is the angle a makes with the positive x axis.

(a) Given Ax = −25.0 m and Ay = 40.0 m, A = (− 25.0 m) 2 + (40.0 m) 2 = 47.2 m
(b) Recalling that tan θ = tan (θ + 180°), tan–1 [40/ (– 25)] = – 58° or 122°. Noting that
the vector is in the third quadrant (by the signs of its x and y components) we see that
122° is the correct answer. The graphical calculator “shortcuts” mentioned above are
designed to correctly choose the right possibility.


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4. (a) With r = 15 m and θ = 30°, the x component of r is given by rx = rcosθ = 15 cos
30° = 13 m.

(b) Similarly, the y component is given by ry = r sinθ = 15 sin 30° = 7.5 m.


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5. The vector sum of the displacements d storm and d new must give the same result as its
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originally intended displacement d o =120jˆ where east is i , north is j , and the assumed
length unit is km. Thus, we write

G
G
dstorm = 100 ˆi , d new = A ˆi + B ˆj.

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(a) The equation dstorm + d new = do readily yields A = –100 km and B = 120 km. The
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magnitude of d new is therefore A2 + B 2 = 156 km .
(b) And its direction is tan–1 (B/A) = –50.2° or 180° + ( –50.2°) = 129.8°. We choose the
latter value since it indicates a vector pointing in the second quadrant, which is what we
expect here. The answer can be phrased several equivalent ways: 129.8°
counterclockwise from east, or 39.8° west from north, or 50.2° north from west.


6. (a) The height is h = d sinθ, where d = 12.5 m and θ = 20.0°. Therefore, h = 4.28 m.
(b) The horizontal distance is d cosθ = 11.7 m.

7. The length unit meter is understood throughout the calculation.
(a) We compute the distance from one corner to the diametrically opposite corner:
d = 3.002 + 3.702 + 4.302 = 6.42 .

(b) The displacement vector is along the straight line from the beginning to the end point
of the trip. Since a straight line is the shortest distance between two points, the length of
the path cannot be less than the magnitude of the displacement.
(c) It can be greater, however. The fly might, for example, crawl along the edges of the
room. Its displacement would be the same but the path length would be
" + w + h = 11.0 m.
(d) The path length is the same as the magnitude of the displacement if the fly flies along
the displacement vector.
(e) We take the x axis to be out of the page, the y axis to be to the right, and the z axis to
be upward. Then the x component of the displacement is w =& 3.70, the y component of
the displacement is 4.30, and the z component is 3.00. Thus d = 3.70 i + 4.30 j + 3.00 k .
An equally correct answer is gotten by interchanging the length, width, and height.


(f) Suppose the path of the fly is as shown by the dotted lines on the upper diagram.
Pretend there is a hinge where the front wall of the room joins the floor and lay the wall
down as shown on the lower diagram. The shortest walking distance between the lower
left back of the room and the upper right front corner is the dotted straight line shown on
the diagram. Its length is
Lmin =

(w + h)

2

+ "2 =

( 3.70 + 3.00 )

2

+ 4.302 = 7.96 m .


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8. We label the displacement vectors A , B and C (and denote the result of their vector
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sum as r ). We choose east as the ˆi direction (+x direction) and north as the ˆj direction
(+y direction) All distances are understood to be in kilometers.
(a) The vector diagram representing the motion is shown below:

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A = 3.1 ˆj
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B = − 2.4 ˆi
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C = − 5.2 ˆj

(b) The final point is represented by
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r = A + B + C = − 2.4 ˆi − 2.1 ˆj

whose magnitude is
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r =

( −2.4 ) + ( −2.1)
2

2

≈ 3.2 km .

(c) There are two possibilities for the angle:
§ −2.1 ·
tan −1 ¨
¸ = 41°, or 221°.
© −2.4 ¹

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We choose the latter possibility since r is in the third quadrant. It should be noted that
many graphical calculators have polar ↔ rectangular “shortcuts” that automatically
produce the correct answer for angle (measured counterclockwise from the +x axis). We
may phrase the angle, then, as 221° counterclockwise from East (a phrasing that sounds
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peculiar, at best) or as 41° south from west or 49° west from south. The& resultant r is
not &shown in our sketch; it would be an arrow directed from the “tail” of A to the “head”
of C .


9. We find the components and then add them (as scalars, not vectors). With d = 3.40 km
and θ = 35.0° we find d cos θ + d sin θ = 4.74 km.

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10. We label the displacement vectors A , B and C (and denote the result of their vector
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sum as r ). We choose east as the ˆi direction (+x direction) and north as the ˆj direction
(+y direction).
All distances are understood to be in kilometers. We note that the angle
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between C and the x axis is 60°. Thus,

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A = 50 ˆi
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B = 30 ˆj
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C = 25 cos ( 60° ) ˆi + 25 sin ( 60° ) ˆj

(a) The total displacement of the car from its initial position is represented by
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r = A + B + C = 62.5 ˆi + 51.7 ˆj

which means that its magnitude is

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r = (62.5) 2 + (51.7)2 = 81 km.
(b) The angle (counterclockwise from +x axis) is tan–1 (51.7/62.5) = 40°, which is to say
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that it points 40° north of east. Although
the resultant& r is shown in our sketch, it would
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be a direct line from the “tail” of A to the “head” of C .


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11. We write r = a + b . When not explicitly displayed, the units here are assumed to be
meters.
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(a) The x and the y components of r are rx = ax + bx = 4.0 – 13 = –9.0 and ry = ay + by =
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3.0 + 7.0 = 10, respectively. Thus r = ( −9.0 m) ˆi + (10 m) ˆj .
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(b) The magnitude of r is

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r =| r |= rx2 + ry2 =

( − 9.0 )

2

+ (10 ) = 13 m .
2

(c) The angle between the resultant and the +x axis is given by

θ = tan–1(ry/rx) = tan–1 [10/( –9.0) ] = – 48° or 132°.
Since the x component of the resultant is negative and the y component is positive,
characteristic of the second quadrant, we find the angle is 132° (measured
counterclockwise from +x axis).

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12. The x, y and z components (with meters understood) of r = c + d are, respectively,
(a) rx = cx + d x = 7.4 + 4.4 = 12 ,
(b) ry = c y + d y = − 3.8 − 2.0 = − 5.8 , and
(c) rz = cz + d z = − 6.1 + 3.3 = − 2.8


13. Reading carefully, we see that the (x, y) specifications for each “dart” are to be
interpreted as ( ∆x , ∆y ) descriptions of the corresponding displacement vectors. We
combine the different parts of this problem into a single exposition.
(a) Along the x axis, we have (with the centimeter unit understood)
30.0 + bx − 20.0 − 80.0 = −140,
which gives bx = –70.0 cm.
(b) Along the y axis we have
40.0 − 70.0 + c y − 70.0 = − 20.0

which yields cy = 80.0 cm.
(c) The magnitude of the final location (–140, –20.0) is

(−140) 2 + (−20.0) 2 = 141 cm.

(d) Since the displacement is in the third quadrant, the angle of the overall displacement
is given by π + tan − 1[(− 20.0) /(− 140)] or 188° counterclockwise from the +x axis (172°
clockwise from the +x axis).


14. All distances in this solution are understood to be in meters.
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(a) a + b = (3.0 ˆi + 4.0 ˆj) + (5.0 ˆi − 2.0 ˆj) = 8.0 ˆi + 2.0 ˆj.

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(b) The magnitude of a + b is
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| a + b |= (8.0) 2 + (2.0) 2 = 8.2 m.

(c) The angle between this vector and the +x axis is tan–1(2.0/8.0) = 14°.
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(d) b − a = (5.0 i − 2.0 j) − (3.0 i + 4.0 j) = 2.0 i − 6.0 j.
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(e) The magnitude of the difference vector b − a is
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| b − a |= 2.02 + ( −6.0) 2 = 6.3m.

(f) The angle between this vector and the +x axis is tan-1( –6.0/2.0) = –72°. The vector is
72° clockwise from the axis defined by iˆ .


15. All distances in this solution are understood to be in meters.
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ˆ
(a) a + b = [4.0 + (−1.0)] ˆi + [(−3.0) + 1.0] ˆj + (1.0 + 4.0)kˆ = 3.0 ˆi − 2.0ˆj + 5.0 k.
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ˆ
(b) a − b = [4.0 − (−1.0)]iˆ + [(−3.0) −1.0]jˆ + (1.0 − 4.0)kˆ = 5.0 ˆi − 4.0 ˆj − 3.0 k.

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(c) The requirement a − b + c = 0 leads to c = b − a , which we note is the opposite of
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ˆ
what we found in part (b). Thus, c = −5.0 ˆi + 4.0 ˆj + 3.0 k.


16. (a) Summing the x components, we have 20 + bx – 20 – 60 = −140, which gives
bx = − 80 m.
(b) Summing the y components, we have 60 – 70 + cy – 70 = 30, which implies cy =110 m.
(c) Using the Pythagorean theorem, the magnitude of the overall displacement is given by
(− 140) 2 + (30) 2 ≈ 143 m.

(d) The angle is given by tan − 1 (30 /(− 140)) = − 12 ° , (which would be 12° measured
clockwise from the –x axis, or 168° measured counterclockwise from the +x axis)


17. Many of the operations are done efficiently on most modern graphical calculators
using their built-in vector manipulation and rectangular ↔ polar “shortcuts.” In this
solution, we employ the “traditional” methods (such as Eq. 3-6). Where the length unit is
not displayed, the unit meter should be understood.
(a) Using unit-vector notation,
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a = 50 cos(30°) ˆi + 50 sin(30°) ˆj
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b = 50 cos (195°) ˆi + 50sin (195°) ˆj
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c = 50 cos (315°) ˆi + 50sin (315°) ˆj
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a + b + c = 30.4 ˆi − 23.3 ˆj.

The magnitude of this result is

30.4 2 + ( −23.3) 2 = 38 m .

(b) The two possibilities presented by a simple calculation for the angle between the
vector described in part (a) and the +x direction are tan–1( –23.2/30.4) = –37.5°, and 180°
+ ( –37.5°) = 142.5°. The former possibility is the correct answer since the vector is in
the fourth quadrant (indicated by the signs of its components). Thus, the angle is –37.5°,
which is to say that it is 37.5° clockwise from the +x axis. This is equivalent to 322.5°
counterclockwise from +x.
(c) We find
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a − b + c = [43.3 − (−48.3) + 35.4] ˆi − [25 − ( − 12.9) + ( − 35.4)] ˆj = 127 ˆi + 2.60 ˆj

in unit-vector notation. The magnitude of this result is

(127)2 + (2.6)2 ≈ 1.30 ×102 m.

(d) The angle between the vector described in part (c) and the +x axis is
tan − 1 (2.6/127) ≈ 1.2° .
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(e) Using unit-vector notation, d is given by d = a + b − c = − 40.4 ˆi + 47.4 ˆj , which has

a magnitude of

( −40.4)2 + 47.4 2 = 62 m .

(f) The two possibilities presented by a simple calculation for the angle between the
vector described in part (e) and the +x axis are tan − 1 (47.4 /(− 40.4)) = − 50.0 ° , and
180° + (−50.0°) = 130° . We choose the latter possibility as the correct one since it
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indicates that d is in the second quadrant (indicated by the signs of its components).


18. If we &wish to use Eq. 3-5 in an unmodified fashion, we should note that the angle
between C and the +x axis is 180° + 20.0° = 200°.
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(a) The x component of B& is given by Cx – Ax = 15.0 cos 200° – 12.0 cos 40° = –23.3 m,
and the y component of B is given by Cy – Ay = 15.0 sin 200° – 12.0 sin 40° = –12.8 m.
Consequently, its magnitude is

(−23.3)2 + (−12.8)2 = 26.6 m .

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(b) The two possibilities presented by a simple calculation for the angle between B and
the +x axis are tan–1[( –12.8)/( –23.3)] = 28.9°, and 180° + 28.9°
& = 209°. We choose the
latter possibility as the correct one since it indicates that B is in the third quadrant
(indicated by the signs of its components). We note, too, that the answer can be
equivalently stated as − 151 ° .


19. It should be mentioned that an efficient way to work this vector addition problem is
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with the cosine law for general triangles (and since a , b and r form an isosceles triangle,
the angles are easy to figure). However, in the interest of reinforcing the usual systematic
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approach to vector addition, we note that the angle b makes with the +x axis is 30°
+105° = 135° and apply Eq. 3-5 and Eq. 3-6 where appropriate.
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(a) The x component of r is rx = 10 cos 30° + 10 cos 135° = 1.59 m.
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(b) The y component of r is ry = 10 sin 30° + 10 sin 135° = 12.1 m.
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(c) The magnitude of r is

(1.59) 2 + (12.1) 2 = 12.2 m.

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(d) The angle between r and the +x direction is tan–1(12.1/1.59) = 82.5°.


20. Angles are given in ‘standard’ fashion, so Eq. 3-5 applies directly. We use this to
write the vectors in unit-vector notation before adding them. However, a very differentlooking approach using the special capabilities of most graphical calculators can be
imagined. Wherever the length unit is not displayed in the solution below, the unit meter
should be understood.
(a) Allowing for the different angle units used in the problem statement, we arrive at
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E = 3.73 i + 4.70 j
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F = 1.29 i − 4.83 j
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G = 1.45 i + 3.73 j
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H = −5.20 i + 3.00 j
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E + F + G + H = 1.28 i + 6.60 j.

(b) The magnitude of the vector sum found in part (a) is

(1.28) 2 + (6.60) 2 = 6.72 m .

(c) Its angle measured counterclockwise from the +x axis is tan–1(6.60/1.28) = 79.0°.
(d) Using the conversion factor π rad = 180° , 79.0° = 1.38 rad.


21. (a) With ^i directed forward and ^j directed leftward, then the resultant is 5.00 ^i + 2.00
^

j . The magnitude is given by the Pythagorean theorem:
5.39 squares.
(b) The angle is tan−1(2.00/5.00) ≈ 21.8º (left of forward).

(5.00) 2 + (2.00) 2 = 5.385 ≈


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22. The strategy is to find where the camel is ( C ) by adding the two consecutive
displacements described in the problem, and then finding the difference between that
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location and the oasis ( B ). Using the magnitude-angle notation
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C = (24 ∠ − 15°) + (8.0 ∠ 90°) = (23.25 ∠ 4.41°)

so
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B − C = (25 ∠ 0°) − (23.25 ∠ 4.41°) = (2.5 ∠ − 45°)

which is efficiently implemented using a vector capable calculator in polar mode. The
distance is therefore 2.6 km.


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23. Let A represent the first part of Beetle 1’s trip (0.50 m east or 0.5 ˆi ) and C
represent the first part of Beetle 2’s trip intended voyage (1.6 m at 50º north of east). For
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their respective second parts: B is 0.80 m at 30º north of east and D is the unknown.
The final position of Beetle 1 is
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A + B = 0.5 ˆi + 0.8(cos30° ˆi + sin30° ˆj) = 1.19 ˆi + 0.40 ˆj.

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The equation relating these is A + B = C + D , where
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C = 1.60(cos 50.0 ° ˆi + sin50.0° ˆj)=1.03 ˆi+1.23 ˆj
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(a) We find D = A + B − C = 0.16 ˆi − 0.83 ˆj , and the magnitude is D = 0.84 m.

(b) The angle is tan − 1 (− 0.83 / 0.16) = − 79 ° which is interpreted to mean 79º south of
east (or 11º east of south).

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24. The desired result is the displacement vector, in units of km, A = 5.6, 90º (measured
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counterclockwise from the +x axis), or A = 5.6 ˆj , where ˆj is the unit vector along the
positive y axis (north). This consists of the sum of two displacements: during the
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whiteout, B = 7.8, 50° , or B = 7.8(cos 50 ° ˆi+sin50° ˆj) = 5.01 ˆi + 5.98 ˆj and the
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unknown C . Thus, A = B + C .
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(a) The desired displacement is given by C = A − B = − 5.01 ˆi − 0.38 ˆj . The magnitude is
(− 5.01) 2 + (− 0.38)2 = 5.0 km.

(b) The angle is tan − 1 (− 0.38 / − 5.01) = 4.3 °, south of due west.