rrChapter 3: Vectors in PhysicsAnswers to Even-Numbered Conceptual Questions2. Vectors rA, rG, and rJ are all equal to one another. In addition, vector rI is the same as vector rL.4. No. The component and the magnitude can be equal if the vector has only a single component. If the vectorhas more than one nonzero component, however, its magnitude will be greater than either of itscomponents.6. No. If a vector has a nonzero component, the smallest magnitude it can have is the magnitude of thecomponent.8. The vectors rA and rB must point in the same direction.10. The vector rA can point in the following directions: 45˚, 135˚, 225˚, and 315˚. In each of these directions Ax= Ay.12. Two vectors of unequal magnitude cannot add to zero, even if they point in opposite directions. Threevectors of unequal magnitude can add to zero if they can form a triangle.14. When sailing upwind, your speed relative to the wind is greater than the speed of the wind itself. If you saildownwind, however, you move with the wind, and its speed relative to you is decreased.Solutions to Problems and Conceptual Exercises1. Picture the Problem: Each component of a vector is doubled in magnitude.Strategy: Note the relationship between the components of a vector and its magnitude and direction to answer the conceptual question.Solution: 1. (a) Doubling each of the components of a vector will double its magnitude, or increase its magnitude by a multiplicative factor of 2. You can picture this in your head or confirm it mathematically with a calculation like: A = Ax2+Ay2 ⇒ 2Ax( )2+ 2Ay( )2= 4 Ax2+Ay2( )=2 Ax2+Ay2= 2A2. (b) Doubling each of the components of a vector will not change its direction at all; the direction changes by a multiplicative factor of 1. You can picture this in your head or confirm it mathematically with a calculation like: θ =tan−1AyAx( ) ⇒ tan−12Ay2Ax( )=tan−1AyAx( )=θInsight: You can change a vector’s direction only by changing the relative magnitudes of its components. In this exercise each component was changed by the same multiplicative factor, so the relative magnitudes were unchanged.2. Picture the Problem: Compare the magnitudes of the vectors depicted in the figure.Strategy: Concentrate on the lengths of the vectors as drawn and ignore their direction. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.3 – 1Chapter 3: Vectors in Physics James S. Walker, Physics, 4th EditionSolution: By comparing the lengths of the vectors as drawn we can arrive at the ranking: B < C < A < DInsight: Note that the symbol B refers to the magnitude of the vector and Burrefers to both magnitude and direction.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.3 – 2Chapter 3: Vectors in Physics James S. Walker, Physics, 4th Edition3. Picture the Problem: Compare the magnitudes of the x components of the vectors depicted in the figure.Strategy: Concentrate on the value of the x component of each vector. A vector that is oriented vertically in the diagram has an x component of zero, whereas horizontal vectors have x components with large magnitudes, either positive (to the right) or negative (to the left).Solution: Note that the x component of Dur is large and negative. The value of its x component is therefore the smallest even though the magnitude of its x component is the largest. By comparing the values of the x components of the vectors as drawn we can arrive at the ranking: Dx<Cx<Bx<AxInsight: Note that the symbol Bx refers to the value of the x component of vector Bur.4. Picture the Problem: Compare the magnitudes of the y components of the vectors depicted in the figure.Strategy: Concentrate on the value of the y component of each vector. A vector that is oriented horizontally in the diagram has a y component of zero, whereas vertical vectors have y components with large magnitudes, either positive (upward) or negative (downward).Solution: Note that the y component of Dur is small and negative. The value of its y component is therefore even smallerthan the zero y component of Bur. By comparing the values of the y components of the vectors as drawn we can arrive atthe ranking: Dy<By<Ay<CyInsight: Note that the symbol By refers to the value of the y component of vector Bur.5. Picture the Problem: The press box is 32.0 ft above second base and an unknown horizontal distance away.Strategy: Use the tangent function to determine the horizontal distance.Solution: Use the tangent function to find x: x =ytanθ=38.0 fttan15.0°=119 ftInsight: Dividing distances into right triangles in this manner is an important strategy for solving physics problems.6. Picture the Problem: You drive 1.2 miles along an inclined roadway, gaining 530 ft of altitude.Strategy: Use the sine function to determine the angle and then the additional distance x along the hypotenuse. Solution: 1. (a) Apply the sine function: sinθ =ys=530 ft1.2 mi × 5280 ft/mi 2. Now solve for θ =sin−1530 ft6300 ft/mi⎛⎝⎜⎞⎠⎟= 4.8°3. (b) Use the known angle together with the sine function to find x: x =Δysinθ=150 ftsin4.8°=1800 ft =0.34 miInsight: It may be helpful for you to review the trigonometric functions sine, cosine, and tangent before tackling other problems in this chapter.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.3 – 315.0°32.0 ftx530 ft1.2 mi150 ftxChapter 3: Vectors in Physics James S. Walker, Physics, 4th Edition7. Picture the Problem: The road gains 6 ft in altitude for every 100 ft it spans in the horizontal direction.Strategy: Use the tangent function to determine the angle. Solution: Apply the tangent function: sinθ =yx ⇒ θ =tan−16 ft100 ft⎛⎝⎜⎞⎠⎟= 3° Insight: It may be helpful for you to review the trigonometric functions sine,
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