Instructor Solutions Manual for
Physics 12th Edition By John
Cutnell, Kenneth Johnson,
David Young, Shane Stadler
(All Chapters 1-32, 100%
Original Verified, A+ Grade)
This is The Only Original and
Complete Instructor Solutions
Manual for 12th Edition, All
Other Files in the Market are
Fake/Old/Wrong Edition.
Part 1: Instructor Solutions Manual: Chapter 1-32
Part 2: Team Problems Solutions: Chapter 1-32
,Part 1: Instructor Solutions Manual Chapter 1-32
CHAPTER 1 INTRODUCTION AND
MATHEMATICAL CONCEPTS
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
1. (d) The resultant vector R is drawn from the tail of the first vector to the head of the
last vector.
2. (c) Note from the drawing that the magnitude R of the resultant vector R is equal to
the shortest distance between the tail of A and the head of B. Thus, R is less than the
magnitude (length) of A plus the magnitude of B.
3. (b) In this drawing the vector –C is reversed relative to C, while vectors A and B are
not reversed.
4. (c) In this drawing the vectors –B and –C are reversed relative to B and C, while vector
A is not reversed.
5. (c) When the two vector components Ax and Ay are added by the tail-to-head method,
the sum equals the vector A. Therefore, these vector components are the correct ones.
6. (b) The displacement vector A points in the –y direction. Therefore, it has no scalar
component along the x axis (Ax = 0 m) and its scalar component along the y axis is
negative.
7. (e) The scalar components are given by Ax′ = −(450 m) sin 35.0° = −258 m and
Ay′ = −(450 m) cos 35.0° = −369 m.
8. (d)
9. Rx = 0 m, Ry = 6.8 m
10. R = 7.9 m, θ = 21 degrees
11. (a) The triangle in the drawing is a right triangle. The lengths A and B of the two sides
are known, so the Pythagorean theorem can be used to determine the length R of the
hypotenuse.
⎛ 4.0 km ⎞
12. (b) The angle is found by using the inverse tangent function, θ = tan −1 ⎜ ⎟ = 53° .
⎝ 3.0 km ⎠
,2 INTRODUCTION AND MATHEMATICAL CONCEPTS
13. (e) These vectors form a closed four-sided polygon, with the head of the fourth vector
exactly meeting the tail of the first vector. Thus, the resultant vector is zero.
14. (b) The three vectors form a right triangle, so the magnitude of A is given by the
Pythagorean theorem as A = Ax2 + Ay2 . If Ax and Ay double in size, then the magnitude
of A doubles: ( 2A ) + ( 2A )
2 2
= 4 Ax + 4 Ay = 2 Ax + Ay = 2 A.
2 2 2 2
x y
⎛A ⎞
15. (a) The angle θ is determined by the inverse tangent function, θ = tan −1 ⎜ y ⎟ . If Ax and
⎝ Ax ⎠
Ay both become twice as large, the ratio does not change, and θ remains the same.
16. (d) The distance (magnitude) traveled by each runner is the same, but the directions
are different. Therefore, the two displacement vectors are not equal.
17. (c) Ax and Bx point in opposite directions, and Ay and By point in the same direction.
18. Ay = 3.4 m, By = 3.4 m
, Chapter 1 Problems 3
CHAPTER 1 INTRODUCTION AND
MATHEMATICAL CONCEPTS
PROBLEMS
1. REASONING We use the fact that 1 m = 3.28 ft to form the following conversion
factor: (1 m)/(3.28 ft) = 1.
SOLUTION To convert ft2 into m2, we apply the conversion factor twice:
⎛ 1 m ⎞⎛ 1 m ⎞
(
Area = 1330 ft 2 ⎜ ) ⎟⎜
⎝ 3.28 ft ⎠ ⎝ 3.28 ft
⎟ = 124 m
⎠
2
2. REASONING
a. To convert the speed from miles per hour (mi/h) to kilometers per hour (km/h), we
need to convert miles to kilometers. This conversion is achieved by using the relation
1.609 km = 1 mi (see the page facing the inside of the front cover of the text).
b. To convert the speed from miles per hour (mi/h) to meters per second (m/s), we must
convert miles to meters and hours to seconds. This is accomplished by using the
conversions 1 mi = 1609 m and 1 h = 3600 s.
SOLUTION
a. Multiplying the speed of 34.0 mi/h by a factor of unity, (1.609 km)/(1 mi) = 1, we
find the speed of the bicyclists is
⎛ mi ⎞ ⎛ mi ⎞ ⎛ 1.609 km ⎞ km
Speed = ⎜ 34.0 ⎟ (1) = ⎜ 34.0 ⎟⎜ ⎟ = 54.7
⎝ h ⎠ ⎝ h ⎠ ⎝ 1 mi ⎠ h
b. Multiplying the speed of 34.0 mi/h by two factors of unity, (1609 m)/(1 mi) = 1 and
(1 h)/(3600 s) = 1, the speed of the bicyclists is
⎛ mi ⎞ ⎛ mi ⎞ ⎛ 1609 m ⎞ ⎛ 1 h ⎞ m
Speed = ⎜ 34.0 ⎟ (1)(1) = ⎜ 34.0 ⎟⎜ ⎟⎜ ⎟ = 15.2
⎝ h ⎠ ⎝ h ⎠ ⎝ 1 mi ⎠ ⎝ 3600 s ⎠ s
3. SSM REASONING We use the facts that 1 mi = 5280 ft, 1 m = 3.281 ft, and
1 yd = 3 ft. With these facts we construct three conversion factors: (5280 ft)/(1 mi) = 1,
(1 m)/(3.281 ft) = 1, and (3 ft)/(1 yd) = 1.
Physics 12th Edition By John
Cutnell, Kenneth Johnson,
David Young, Shane Stadler
(All Chapters 1-32, 100%
Original Verified, A+ Grade)
This is The Only Original and
Complete Instructor Solutions
Manual for 12th Edition, All
Other Files in the Market are
Fake/Old/Wrong Edition.
Part 1: Instructor Solutions Manual: Chapter 1-32
Part 2: Team Problems Solutions: Chapter 1-32
,Part 1: Instructor Solutions Manual Chapter 1-32
CHAPTER 1 INTRODUCTION AND
MATHEMATICAL CONCEPTS
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
1. (d) The resultant vector R is drawn from the tail of the first vector to the head of the
last vector.
2. (c) Note from the drawing that the magnitude R of the resultant vector R is equal to
the shortest distance between the tail of A and the head of B. Thus, R is less than the
magnitude (length) of A plus the magnitude of B.
3. (b) In this drawing the vector –C is reversed relative to C, while vectors A and B are
not reversed.
4. (c) In this drawing the vectors –B and –C are reversed relative to B and C, while vector
A is not reversed.
5. (c) When the two vector components Ax and Ay are added by the tail-to-head method,
the sum equals the vector A. Therefore, these vector components are the correct ones.
6. (b) The displacement vector A points in the –y direction. Therefore, it has no scalar
component along the x axis (Ax = 0 m) and its scalar component along the y axis is
negative.
7. (e) The scalar components are given by Ax′ = −(450 m) sin 35.0° = −258 m and
Ay′ = −(450 m) cos 35.0° = −369 m.
8. (d)
9. Rx = 0 m, Ry = 6.8 m
10. R = 7.9 m, θ = 21 degrees
11. (a) The triangle in the drawing is a right triangle. The lengths A and B of the two sides
are known, so the Pythagorean theorem can be used to determine the length R of the
hypotenuse.
⎛ 4.0 km ⎞
12. (b) The angle is found by using the inverse tangent function, θ = tan −1 ⎜ ⎟ = 53° .
⎝ 3.0 km ⎠
,2 INTRODUCTION AND MATHEMATICAL CONCEPTS
13. (e) These vectors form a closed four-sided polygon, with the head of the fourth vector
exactly meeting the tail of the first vector. Thus, the resultant vector is zero.
14. (b) The three vectors form a right triangle, so the magnitude of A is given by the
Pythagorean theorem as A = Ax2 + Ay2 . If Ax and Ay double in size, then the magnitude
of A doubles: ( 2A ) + ( 2A )
2 2
= 4 Ax + 4 Ay = 2 Ax + Ay = 2 A.
2 2 2 2
x y
⎛A ⎞
15. (a) The angle θ is determined by the inverse tangent function, θ = tan −1 ⎜ y ⎟ . If Ax and
⎝ Ax ⎠
Ay both become twice as large, the ratio does not change, and θ remains the same.
16. (d) The distance (magnitude) traveled by each runner is the same, but the directions
are different. Therefore, the two displacement vectors are not equal.
17. (c) Ax and Bx point in opposite directions, and Ay and By point in the same direction.
18. Ay = 3.4 m, By = 3.4 m
, Chapter 1 Problems 3
CHAPTER 1 INTRODUCTION AND
MATHEMATICAL CONCEPTS
PROBLEMS
1. REASONING We use the fact that 1 m = 3.28 ft to form the following conversion
factor: (1 m)/(3.28 ft) = 1.
SOLUTION To convert ft2 into m2, we apply the conversion factor twice:
⎛ 1 m ⎞⎛ 1 m ⎞
(
Area = 1330 ft 2 ⎜ ) ⎟⎜
⎝ 3.28 ft ⎠ ⎝ 3.28 ft
⎟ = 124 m
⎠
2
2. REASONING
a. To convert the speed from miles per hour (mi/h) to kilometers per hour (km/h), we
need to convert miles to kilometers. This conversion is achieved by using the relation
1.609 km = 1 mi (see the page facing the inside of the front cover of the text).
b. To convert the speed from miles per hour (mi/h) to meters per second (m/s), we must
convert miles to meters and hours to seconds. This is accomplished by using the
conversions 1 mi = 1609 m and 1 h = 3600 s.
SOLUTION
a. Multiplying the speed of 34.0 mi/h by a factor of unity, (1.609 km)/(1 mi) = 1, we
find the speed of the bicyclists is
⎛ mi ⎞ ⎛ mi ⎞ ⎛ 1.609 km ⎞ km
Speed = ⎜ 34.0 ⎟ (1) = ⎜ 34.0 ⎟⎜ ⎟ = 54.7
⎝ h ⎠ ⎝ h ⎠ ⎝ 1 mi ⎠ h
b. Multiplying the speed of 34.0 mi/h by two factors of unity, (1609 m)/(1 mi) = 1 and
(1 h)/(3600 s) = 1, the speed of the bicyclists is
⎛ mi ⎞ ⎛ mi ⎞ ⎛ 1609 m ⎞ ⎛ 1 h ⎞ m
Speed = ⎜ 34.0 ⎟ (1)(1) = ⎜ 34.0 ⎟⎜ ⎟⎜ ⎟ = 15.2
⎝ h ⎠ ⎝ h ⎠ ⎝ 1 mi ⎠ ⎝ 3600 s ⎠ s
3. SSM REASONING We use the facts that 1 mi = 5280 ft, 1 m = 3.281 ft, and
1 yd = 3 ft. With these facts we construct three conversion factors: (5280 ft)/(1 mi) = 1,
(1 m)/(3.281 ft) = 1, and (3 ft)/(1 yd) = 1.
