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**Unformatted text preview: **OpenStax University Physics Volume I
Unit 1: Mechanics
Chapter 2: Vectors University Physics Volume I
Unit 1: Mechanics
Chapter 2: Vectors
Conceptual Questions
1. A weather forecast states the temperature is predicted to be the following day. Is this
temperature a vector or a scalar quantity? Explain.
Solution
scalar
3. Give a specific example of a vector, stating its magnitude, units, and direction.
Solution
answers may vary
r
5. Suppose you add two vectors A and B. What relative direction between them produces the
resultant with the greatest magnitude? What is the maximum magnitude? What relative direction
between them produces the resultant with the smallest magnitude? What is the minimum
magnitude?
Solution
parallel, sum of magnitudes, antiparallel, zero
7. Is it possible for two vectors of different magnitudes to add to zero? Is it possible for three
vectors of different magnitudes to add to zero? Explain.
Solution
no, yes
9. When a 10,000-m runner competing on a 400-m track crosses the finish line, what is the
runner’s net displacement? Can this displacement be zero? Explain.
Solution
zero, yes
11. Can a magnitude of a vector be negative?
Solution
no
13. If two vectors are equal, what can you say about their components? What can you say about
their magnitudes? What can you say about their directions?
Solution
equal, equal, the same
15. Give an example of a nonzero vector that has a component of zero.
Solution
a unit vector of the x-axis
17. If two vectors are equal, what can you say about their components?
Solution
They are equal.
19. If one of the two components of a vector is not zero, can the magnitude of the other vector
component of this vector be zero?
Solution
yes
rr
21. What is wrong with the following expressions? How can you correct them? (a) C = AB, (b)
r rr
r
r
r r
r
r r r r
r
C = AB, (c) C = A B, (d) C = AB, (e) C + 2A = B, (f) C = A B, (g) A B = A B, (h )
r
r r
r r
r
C = 2A B, (i) C = A B , and (j) C = A B .
Page 1 of 9 OpenStax University Physics Volume I
Unit 1: Mechanics
Chapter 2: Vectors Solution
a. C = A ×B , b. C = A ´ B or C = A - B , c. C = A ´ B , d. C = AB , e. C + 2A = B , f.
r
r r
r r r
C = A B, g. left side is a scalar and right side is a vector, h. C = 2A B , i. C = A B , j. C=A B
23. If the dot product of two vectors vanishes, what can you say about their directions?
Solution
They are orthogonal.
Problems
25. A scuba diver makes a slow descent into the depths of the ocean. His vertical position with
respect to a boat on the surface changes several times. He makes the first stop 9.0 m from the
boat but has a problem with equalizing the pressure, so he ascends 3.0 m and then continues
descending for another 12.0 m to the second stop. From there, he ascends 4 m and then descends
for 18.0 m, ascends again for 7 m and descends again for 24.0 m, where he makes a stop, waiting
for his buddy. Assuming the positive direction up to the surface, express his net vertical
displacement vector in terms of the unit vector. What is his distance to the boat?
Solution
r
h = −49 m uˆ , 49 m
27. Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from
your starting point and what is the compass direction of a line connecting your starting point to
your final position? Use a graphical method.
Solution
30.8 m, 35.7 west of north
29. A delivery man starts at the post office, drives 40 km north, then 20 km west, then 60 km
northeast, and finally 50 km north to stop for lunch. Use a graphical method to find his net
displacement vector.
Solution
134 km, 80
31. In an attempt to escape a desert island, a castaway builds a raft and sets out to sea. The wind
shifts a great deal during the day and he is blown along the following directions: 2.50 km and
45.0 north of west, then 4.70 km and 60.0 south of east, then 1.30 km and 25.0 south of west,
then 5.10 km straight east, then 1.70 km and 5.00 east of north, then 7.20 km and 55.0 south
of west, and finally 2.80 km and 10.0 north of east. Use a graphical method to find the
castaway’s final position relative to the island.
Solution
7.34 km, 63.5 south of east
33. A trapper walks a 5.0-km straight-line distance from his cabin to the lake, as shown in the
following figure. Use a graphical method (the parallelogram rule) to determine the trapper’s
displacement directly to the east and displacement directly to the north that sum up to his
resultant displacement vector. If the trapper walked only in directions east and north, zigzagging
his way to the lake, how many kilometers would he have to walk to get to the lake? Page 2 of 9 OpenStax University Physics Volume I
Unit 1: Mechanics
Chapter 2: Vectors Solution
3.8 km east, 3.2 km north, 7.0 km
35. A pedestrian walks 6.0 km east and then 13.0 km north. Use a graphical method to find the
pedestrian’s resultant displacement and geographic direction.
Solution
14.3 km, 65
37. Assuming the +x-axis is horizontal and points to the right, resolve the vectors given in the
following figure to their scalar components and express them in vector component form. Solution r
r
r
r
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
A
=
+
8.66
i
+
5.00
j
B
=
+
30.09
i
+
39.93
j
C
=
+
6.00
i
−
10.39
j
D
a.
, b.
, c.
, d. = −15.97 ˆi + 12.04ˆj , f. F = -17.32 ˆi -10.00ˆj
39. You drive 7.50 km in a straight line in a direction 15 east of north. (a) Find the distances
you would have to drive straight east and then straight north to arrive at the same point. (b) Show
that you still arrive at the same point if the east and north legs are reversed in order. Assume the
+x-axis is to the east.
Solution
Page 3 of 9 OpenStax University Physics Volume I
Unit 1: Mechanics
Chapter 2: Vectors a. 1.94 km, 7.24 km; b. proof
41. A trapper walks a 5.0-km straight-line distance from her cabin to the lake, as shown in the
following figure. Determine the east and north components of her displacement vector. How
many more kilometers would she have to walk if she walked along the component
displacements? What is her displacement vector? Solution D = (3.8 ˆi + 3.2 ˆj)km
43. Two points in a plane have polar coordinates P1 (2.500m, 6) and P2 (3.800m, 2 3).
3.8 km east, 3.2 km north, 2.0 km, Determine their Cartesian coordinates and the distance between them in the Cartesian coordinate
system. Round the distance to a nearest centimeter.
Solution P1 (2.165m, 1.250m) , P2 (−1.900m, 3.290m) , 5.27 m 45. Two points in the Cartesian plane are A(2.00 m, –4.00 m) and B(–3.00 m, 3.00 m). Find the
distance between them and their polar coordinates.
Solution
8.60 m, A(2 5 m,0.647 ) , B(3 2 m,0.75 )
r
r
47. For vectors B = − ˆi − 4 ˆj and A = −3 ˆi − 2 ˆj, calculate (a) A + B and its magnitude and
direction angle, and (b) A - B and its magnitude and direction angle.
Solution
r r
r r
r r
a. A + B = −4 ˆi − 6 ˆj, | A + B | = 7.211, = 213.7 ; b. A - B = 2 ˆi - 2 ˆj , | A − B | = 2 2, = −45 ˆ
49. Given two displacement vectors A = (3.00 ˆi - 4.00 ˆj+ 4.00 k)m
and
r
B = (2.00 ˆi + 3.00 ˆj − 7.00 kˆ ) m, find the displacements and their magnitudes for (a) C = A + B
r r
r
and (b) D = 2A − B.
Solution
r
a. C = (5.0 ˆi − 1.0 ˆj − 3.0 kˆ ) m, C = 5.92 m ; r b. D = (4.0 ˆi − 11.0 ˆj + 15.0 kˆ ) m, D = 19.03m
51. In an attempt to escape a desert island, a castaway builds a raft and sets out to sea. The wind
shifts a great deal during the day, and she is blown along the following straight lines: 2.50 km
and 45.0 north of west, then 4.70 km and 60.0 south of east, then 1.30 km and 25.0 south of
west, then 5.10 km due east, then 1.70 km and 5.00 east of north, then 7.20 km and 55.0 south
Page 4 of 9 OpenStax University Physics Volume I
Unit 1: Mechanics
Chapter 2: Vectors of west, and finally 2.80 km and 10.0 north of east. Use the analytical method to find the
resultant vector of all her displacement vectors. What is its magnitude and direction?
Solution
r
D = (3.3 ˆi − 6.6 ˆj) km, ˆi is to the east, 7.34 km, −63.5
53. Given
the vectors in the preceding figure, find vector R that solves equations (a) D + R = F
r
r
r
r
and (b) C − 2D + 5R = 3F. Assume the +x-axis is horizontal to the right. Solution a. R = -1.35 ˆi - 22.04 ˆj , b. R = -17.98 ˆi + 0.89 ˆj
55. An adventurous dog strays from home, runs three blocks east, two blocks north, and one
block east, one block north, and two blocks west. Assuming that each block is about a 100 yd,
use the analytical method to find the dog’s net displacement vector, its magnitude, and its
direction. Assume the +x-axis is to the east. How would your answer be affected if each block
was about 100 m?
Solution r
D = (200 ˆi + 300 ˆj) yd, D = 360.5 yd, 56.3 north of east; The numerical answers would stay the same but the physical unit would be meters. The physical meaning and distances would be about
the same because 1 yd is comparable with 1 m. r 57. Given the displacement vector D = (3 ˆi − 4 ˆj) m, find the displacement vector R so that r r
D + R = −4D ˆj. Solution R = -3ˆi -16 ˆj 59. At one point in space, the direction of the electric field vector is given in the Cartesian
ˆ = 1 5 ˆi − 2 5 ˆj. If the magnitude of the electric field vector is E =
system by the unit vector E Page 5 of 9 OpenStax University Physics Volume I
Unit 1: Mechanics
Chapter 2: Vectors 400.0 V/m, what are the scalar components
point? What is the direction angle
Solution E Ex , Ey , and Ez of the electric field vector r
E at this of the electric field vector at this point? r
E = E Eˆ , Ex = +178.9V m, Ey = −357.8V m, Ez = 0.0V m, E = − tan −1 (2) 61. In the control tower at a regional airport, an air traffic controller monitors two aircraft as their
positions change with respect to the control tower. One plane is a cargo carrier Boeing 747 and
the other plane is a Douglas DC-3. The Boeing is at an altitude of 2500 m, climbing at 10
above the horizontal, and moving 30 north of west. The DC-3 is at an altitude of 3000 m,
climbing at 5 above the horizontal, and cruising directly west. (a) Find the position vectors of
the planes relative to the control tower. (b) What is the distance between the planes at the
moment the air traffic controller makes a note about their positions?
Solution ˆ
ˆ
a. R B = (12.278 ˆi + 7.089 ˆj+ 2.500 k)km
, R D = (-0.262 ˆi + 3.000 k)km
; r r b. |R B − R D | = 23.131km
63. Assuming the +x-axis is horizontal to the right for the vectors in the preceding figure, find (a)
r
r
the component of vector A along vector C, (b) the component of vector C along vector A , (c)
r
the component of vector ˆi along vector F , and (d) the component of vector F along vector ˆi.
Solution
a. 0, b. 0, c. 0.866, d. 17.32
ˆ
65. Find the angles that vector D = (2.0 ˆi - 4.0 ˆj+ k)m
makes with the x-, y-, and z- axes.
Solution
i = 64.12, j = 150.79, k = 77.39
67. Assuming the +x-axis is horizontal to the right for the vectors in the previous figure, find the r r r r r r
r r
r r
r
following vector products: (a) A C, (b) A F, (c) D C, (d) A (F + 2C), (e) ˆi B, (f) ˆj B,
r
r
(g) (3 ˆi − ˆj) B, and (h) Bˆ B.
Solution a. −120 kˆ , b. 0 , c. −94 kˆ , d. −240 kˆ , e. 4.0kˆ , f. −3 kˆ , g. 15kˆ , h. 0 r r r r r r r 69. For the vectors in the earlier figure, find (a) ( A F) D, (b) ( A F) (D B), and (c) r r r r
(A F)(D B). Solution
a. 0, b. 0, c. −20,000 kˆ
Additional Problems
71. You fly 32.0 km in a straight line in still air in the direction 35.0 south of west. (a) Find the
distances you would have to fly due south and then due west to arrive at the same point. (b) Find
the distances you would have to fly first in a direction 45.0 south of west and then in a direction Page 6 of 9 OpenStax University Physics Volume I
Unit 1: Mechanics
Chapter 2: Vectors 45.0 west of north. Note these are the components of the displacement along a different set of
axes—namely, the one rotated by 45 with respect to the axes in (a).
Solution
a. 18.4 km and 26.2 km, b. 31.5 km and 5.56 km
73. If the polar coordinates of a point are (r, ) and its rectangular coordinates are ( x, y),
determine the polar coordinates of the following points: (a) (–x, y), (b) (–2x, –2y), and (c) (3x, –
3y).
Solution
a. (r, − ) , b. (2r, + 2 ) , (c) (3 r, − )
75. Starting at the island of Moi in an unknown archipelago, a fishing boat makes a round trip
with two stops at the islands of Noi and Poi. It sails from Moi for 4.76 nautical miles (nmi) in a
direction 37 north of east to Noi. From Noi, it sails 69 west of north to Poi. On its return leg
from Poi, it sails 28 east of south. What distance does the boat sail between Noi and Poi? What
distance does it sail between Moi and Poi? Express your answer both in nautical miles and in
kilometers. Note: 1 nmi = 1852 m.
Solution dPM = 6.2nmi = 11.4km, dNP = 7.2nmi = 13.3km
r r r 77. Show that when A + B = C, then C = A + B + 2 AB cos , where is the angle between
r
vectors A and B.
Solution
proof
79. A skater glides along a circular path of radius 5.00 m in clockwise direction. When he coasts
around one-half of the circle, starting from the west point, find (a) the magnitude of his
displacement vector and (b) how far he actually skated. (c) What is the magnitude of his
displacement vector when he skates all the way around the circle and comes back to the west
point?
Solution
a. 10.00 m, b. 5 m, c. 0
2 2 2 r 81. If the velocity vector of a polar bear is u = (−18.0 ˆi − 13.0 ˆj) km h , how fast and in what
geographic direction is it heading? Here, ˆi and ˆj are directions to geographic east and north,
respectively.
Solution
22.2 km/h, 35.8 south of west
83. A diver explores a shallow reef off the coast of Belize. She initially swims 90.0 m north,
makes a turn to the east and continues for 200.0 m, then follows a big grouper for 80.0 m in the
direction 30 north of east. In the meantime, a local current displaces her by 150.0 m south.
Assuming the current is no longer present, in what direction and how far should she now swim to
come back to the point where she started?
Solution
270 m, 4.2 north of west
85. Vectors A and B are two orthogonal vectors in the xy-plane and they have identical r
A
magnitudes. If = 3.0 ˆi + 4.0 ˆj, find Page 7 of 9 OpenStax University Physics Volume I
Unit 1: Mechanics
Chapter 2: Vectors Solution B = -4.0 ˆi + 3.0 ˆj or B = 4.0 ˆi - 3.0 ˆj
87. Show that (B ´ C)× A is the volume of the parallelepiped, with edges formed by the three
vectors in the following figure. Solution
proof
Challenge Problems
89. What is the component of the force vector r
H = (1.0 ˆi + 4.0 ˆj) N? ˆ along the force vector
G = (3.0 ˆi + 4.0 ˆj+10.0 k)N Solution
GH = 19 N / 17 4.6 N
91. Distances between points in a plane do not change when a coordinate system is rotated. In
other words, the magnitude of a vector is invariant under rotations of the coordinate system.
Suppose a coordinate system S is rotated about its origin by angle to become a new
coordinate system S' , as shown in the following figure. A point in a plane has coordinates (x, y)
in S and coordinates ( x' , y' ) in S'.
(a) Show that, during the transformation of rotation, the coordinates in S' are expressed in terms
of the coordinates in S by the following relations: x ' = x cos + y sin . y ' = − x sin + y cos (b) Show that the distance of point P to the origin is invariant under rotations of the coordinate
system. Here, you have to show that
x 2 + y 2 = x' 2 + y' 2 .
(c) Show that the distance between points P and Q is invariant under rotations of the coordinate
system. Here, you have to show that
( xP − xQ ) 2 + ( yP − yQ ) 2 = ( x 'P − x 'Q ) 2 + ( y 'P − y 'Q ) 2 . Page 8 of 9 OpenStax University Physics Volume I
Unit 1: Mechanics
Chapter 2: Vectors Solution
proof This file is copyright 2016, Rice University. All Rights Reserved. Page 9 of 9 OpenStax University Physics Volume I
Unit 1: Mechanics
Chapter 3: Motion Along a Straight Line University Physics Volume I
Unit 1: Mechanics
Chapter 3: Motion Along a Straight Line
Conceptual Questions
1. Give an example in which there are clear distinctions among distance traveled, displacement,
and magnitude of displacement. Identify each quantity in your example specifically.
Solution
You drive your car into town and return to drive past your house to a friend’s house.
3. Bacteria move back and forth using their flagella (structures that look like little tails). Speeds
of up to 50 μm/s (50 × 10−6 m/s) have been observed. The total distance traveled by a bacterium
is large for its size, whereas its displacement is small. Why is this?
Solution
If the bacteria are moving back and forth, then the displacements are canceling each other and
the final displacement is small.
5. Does a car’s odometer measure distance traveled or displacement?
Solution
Distance traveled
7. There is a distinction between average speed and the magnitude of average velocity. Give an
example that illustrates the difference between these two quantities.
Solution
Average speed is the total distance traveled divided by the elapsed time. If you go for a walk,
leaving and returning to your home, your average speed is a positive number. Since Average
velocity = Displacement/Elapsed time, your average velocity is zero.
9. If you divide the total distance traveled on a car trip (as determined by the odometer) by the
elapsed time of the trip, are you calculating average speed or magnitude of average velocity?
Under what circumstances are these two quantities the same?
Solution
Average speed. They are the same if the car doesn’t reverse direction.
11. Is it possible for speed to be constant while acceleration is not zero?
Solution
No, in one dimension constant speed requires zero acceleration.
13. Give an example in which velocity is zero yet acceleration is not.
Solution
A ball is thrown into the air and its velocity is zero at the apex of the throw, but acceleration is
not zero.
15. Plus and minus signs are used in one-dimensional motion to indicate direction. What is the
sign of an acceleration that reduces the magnitude of a negative velocity? Of a positive velocity?
Solution
Plus, minus
17. State two scenarios of the kinematics of single object where three known quantities require
two kinematic equations to solve for the unknowns.
Solution
If the acceleration, time, and displacement are the knowns, and the initial and final velocities are
the unknowns, then two kinematic equations must be solved simultaneously. Also if the final
velocity, time, and displacement are the knowns then two kinematic equations must be solved for
the initial velocity and acceleration.
Page 1 of 16 OpenStax University Physics Volume I
Unit 1: Mechanics
Chapter 3: Motion Along a Straight Line 19. An object that is thrown straight up falls back to Earth. This is one-dimensional motion. (a)
When is its velocity zero? (b) Does its velocity change direction? (c) Does the acceleration have
the same sign on the way up as on the way down?
Solution
a. at the top of its trajectory; b. yes, at the top of its trajectory; c. yes
21. The severity of a fall depends on your speed when you strike the ground. All factors but the
acceleration from gravity being the same, how many times higher could a safe fall on the Moon
than on Earth (gravitational acceleration on the Moon is about one-sixth that of the Earth)?
Solution
Earth v = v0 − gt = − gt ; Moon y = − 1 g
1
1 (6t ) 2 = − g 6t 2 = −6 gt 2 = −6 y
26
2
2 v = g
t v = v
6 − gt = − g
1
t t = 6t ; Earth y = − gt 2 Moon
6
2 23. When given the acceleration function, what additional information is needed to find the
velocity function and position function?
Solution
We must know the initial conditions on the velocity and position at t = 0 to solve for the
constants of integration.
Problems
25. A car is 2.0 km west of a traffic light at t = 0 and 5.0 km east of the light at t = 6.0 min.
Assume the origin of the coordinate system is the light and the positive x direction is eastward.
(a) What are the car’s position vectors at these two times? (b) What is the car’s displacement
between 0 min and 6.0 min?
Solution a. x1 = (-2.0 m) ˆi , x 2 = (5.0 m) ˆi ; b. x2 - x1 = 5.0m - (-2.0 m) = 7.0 m east
27. The position of a particle mo...

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- Spring '17
- Radia Redjimi
- Physics, mechanics, Acceleration, Velocity, m/s, Rice University, OpenStax University