Respuesta a problemas de serway (capitulo 3)

3
Vectors
CHAPTER OUTLINE
3.1 3.2 3.3 3.4 Coordinate Systems Vector and Scalar Quantities Some Properties of Vectors Components of a Vector and Unit Vectors

ANSWERS TO QUESTIONS
Q3.1 No. The sum of two vectors can only be zero if they are in opposite directions and have the same magnitude. If you walk 10 meters north and then 6 meters south, you won’t end up where you started. No, themagnitude of the displacement is always less than or equal to the distance traveled. If two displacements in the same direction are added, then the magnitude of their sum will be equal to the distance traveled. Two vectors in any other orientation will give a displacement less than the distance traveled. If you first walk 3 meters east, and then 4 meters south, you will have walked a total distance of7 meters, but you will only be 5 meters from your starting point.

Q3.2

Q3.3

The largest possible magnitude of R = A + B is 7 units, found when A and B point in the same direction. The smallest magnitude of R = A + B is 3 units, found when A and B have opposite directions. Only force and velocity are vectors. None of the other quantities requires a direction to be described. If thedirection-angle of A is between 180 degrees and 270 degrees, its components are both negative. If a vector is in the second quadrant or the fourth quadrant, its components have opposite signs. The book’s displacement is zero, as it ends up at the point from which it started. The distance traveled is 6.0 meters. 85 miles. The magnitude of the displacement is the distance from the starting point, the260-mile mark, to the ending point, the 175-mile mark. Vectors A and B are perpendicular to each other. No, the magnitude of a vector is always positive. A minus sign in a vector only indicates direction, not magnitude.

Q3.4 Q3.5

Q3.6 Q3.7 Q3.8 Q3.9

55

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Vectors

Q3.10 Q3.11 Q3.12 Q3.13

Any vector that points along a line at 45° to the x and y axes has components equal inmagnitude. A x = B x and A y = B y . Addition of a vector to a scalar is not defined. Think of apples and oranges. One difficulty arises in determining the individual components. The relationships between a vector and its components such as A x = A cos θ , are based on right-triangle trigonometry. Another problem would be in determining the magnitude or the direction of a vector from its components.Again,
2 2 A = A x + A y only holds true if the two component vectors, A x and A y , are perpendicular.

Q3.14

If the direction of a vector is specified by giving the angle of the vector measured clockwise from the positive y-axis, then the x-component of the vector is equal to the sine of the angle multiplied by the magnitude of the vector.

SOLUTIONS TO PROBLEMS
Section 3.1 P3.1 CoordinateSystems

a f a fa f y = r sin θ = a5.50 mf sin 240° = a5.50 mfa −0.866f =
(a)

x = r cos θ = 5.50 m cos 240° = 5.50 m −0.5 = −2.75 m −4.76 m

P3.2

x = r cos θ and y = r sin θ , therefore x1 = 2.50 m cos 30.0° , y1 = 2.50 m sin 30.0° , and

a

f

a

f

bx , y g = a2.17 , 1.25f m x = a3.80 mf cos 120° , y = a3.80 mf sin 120° , and bx , y g = a−1.90, 3.29f m .
1 1

2

2

22

(b) P3.3

d = ( ∆ x) 2 + ( ∆ y) 2 = 16.6 + 4.16 = 4.55 m

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 = x 2 + y 2 = (b)

a2.00 mf + a1.00 mf
2

2

= 5.00 m 2 = 2.24 m

θ = tan −1

FG 1 IJ = 26.6° ; r = H 2K

2.24 m, 26.6°

Chapter 357

P3.4

(a)

d=

bx

2

− x1

g + by
2

2

− y1

g

2

=

c2.00 − −3.00 h + a−4.00 − 3.00f
2

2

d = 25.0 + 49.0 = 8.60 m (b) r1 =

θ1

a2.00f + a−4.00f = 20.0 = F 4.00 IJ = −63.4° = tan G − H 2.00 K
2 2 −1

4.47 m

r2 =

a−3.00f + a3.00f
2

2

= 18.0 = 4.24 m

θ 2 = 135° measured from the +x axis.
P3.5 We have 2.00 = r cos 30.0° r= and y = r...