chap3

CHAPTER 3 THE VECTOR DESCRIPTION OFMOTION

ActivPhysics can help with these problems:

Activity 4.1

Section 3-2: Vector Arithmetic

Problem

1. You walk west 220 m, then north 150 m. What arethe magnitude and direction of your displacementvector?

Solution

The triangle formed by the two displacement vectorsand their sum is a right triangle, so the PythagoreanTheorem gives the magnitude C =

√A2 + B2 =

√

(220 m)2 + (150 m)2 = 266 m, and the basic definition of the tangent gives β = tan−1(150 m/220 m)= 34.3◦. The direction of C can be specified as34.3◦ N of W, or 55.7◦ W of N, or by the azimuth304.3◦ (CW from N), etc.

Problem 1 Solution.

Problem

2. An ion in a mass spectrometer (a device that sortsatomic-size particles) follows a semicircular path ofradius 15.2 cm. What are (a) the distance it travelsand (b) the magnitude of its displacement?

Solution

(a) The length of the semicircle is 12(2πr) = πr =

π(15.2 cm) = 47.8 cm. (b) The magnitude of thedisplacement vector, from the start of the semicircle toits end, is just a diameter, or 2(15.2 cm) = 30.4 cm.

Problem

3. A migrating whale follows the west coast of Mexicoand North America toward its summer home inAlaska. It first travels 360 km due northwest to justoff the coast of Northern California and then turnsdue north and travels 400 km toward itsdestination. Determine graphically the magnitudeand direction of its displacement vector.

Solution

We can find the magnitude and direction of the vectorsum of the two displacements either using geometryand a diagram, or by adding vector components.From the law of cosines:

C =√

A2 + B2− 2AB cos γ

=√

(360 km)2 + (400 km)2 − 2(360 km)(400 km) cos 135◦

= 702 km.

From the law of sines: C/ sinγ = B/ sinβ, or

β=sin−1

(

B sin γ

C

)

= sin−1

[(

400 m

702 m

)

sin 135◦]

= 23.7◦.

The direction of C can be specified as 45◦ + 23.7◦ =68.7◦ N of W, or 180◦ − 68.7◦ = 111◦ CCW from thex-axis (east) in the illustration.

Problem 3 Solution.

In a coordinate system with x-axis east and y-axisnorth, the first displacement is 360 km (ı cos 135◦+ sin 135◦) and the second simply 400 km . Their sumis (−255ı+ 255+ 400) km = (−255ı+ 655) km,

2 CHAPTER 3

which is the total displacement. Its magnitude is√

(−255)2 + (655)2 km/702 km and its direction(measured CCW from the x-axis) is θx = cos−1

(−255 km/702 km) = 111◦, as above. (Note that sinceCy > 0 and Cx < 0, θx is in the second quadrant.)

Problem

4. A city’s streets are laid out with its north-southblocks twice as long as its east-west blocks. Youwalk 8 blocks east and 3 blocks north. Determine(a) the total distance you’ve walked and (b) themagnitude of your displacement vector. Express inunits of east-west blocks.

Solution

(a) Eight blocks east is 8 units, but three blocks northis 3 × 2 = 6 units, so the total distance walked is14 units. (b) The magnitude of your displacementvector is the hypotenuse of a right triangle, with sidesof 8 units and 6 units; its length is

√

(8 u)2 + (6 u)2 =10 units.

Problem

5. Vector A has magnitude 3.0 m and points to theright; vector B has magnitude 4.0 m and pointsvertically upward. Find the magnitude anddirection of a vector C such that A+B+C=0.

Problem 5 Solution.

Solution

The vectors A,B, and C form a 3-4-5 right triangle,as shown in the sketch. Therefore, C = 5 m, and thedirection of C, measured CCW from the direction ofA, is 180◦ + tan−1 (B/A) = 180◦ + 53.1◦ = 233◦

(other angles could have been chosen). C could alsobe determined algebraically from components, withx-axis parallel to A and y-axis parallel to B. Then

A = (3.0 m)ı, B = (4.0 m), and C=−(A+B) =(−3.0 m)ı + (−4.0 m) = Cxı + Cy . The magnitude of

C is√

C2x + C2

y =√

(−3.0 m)2 + (−4.0 m)2, and the

angle that C makes with the x-axis is cos−1(Cx/C) =cos−1(−3.0 m/5.0 m), which is in the third quadrant,as calculated above. (Note: the angle of C could alsobe specified as −127◦, or CW from the x-axis.)

Problem

6. Two vectors A and B have the same magnitude, A,and are at right angles. Find the magnitude of thevectors (a) A+ 2B, (b) 3A−B.

Solution

Any two vectors (V1 and V2) and their sum (V3 =

V1 +V2) form a triangle. If V1 is perpendicular toV2, the triangle is a right triangle, and themagnitudes are related by the Pythagorean Theorem,V3 =

√

V21 + V2

2. (a) For V1 = A and V2 = |2B| =2A, V3 =

√5A. (b) For V1 = 3A and V2 = | − B| = A,

V3 =√

10A.

Problem

7. Vectors A and B in Fig. 3-22 have the samemagnitude, A. Find the magnitude and direction of(a) A − B and (b) A+B.

figure 3-22 Problem 7 Solution.

Solution

The vectors A and B in Fig. 3-22 form two sides of aparallelogram, in which A−B and A+B are thediagonals, as shown. Since the magnitudes of A and B

are equal, the parallelogram is a rhombus, and thediagonals are perpendicular (the converse of this isalso true; see Problem 60). Then A+B is along theperpendicular bisector of the base A−B of anisosceles triangle, and vice versa. Using the givenangles, we find the magnitudes (a) |A−B| =2A sin 20◦ = 0.684A, and (b) |A+B| = 2A cos 20◦ =1.88A. In Fig. 3-22, (a) A−B is up, and (b) A+B isto the right, but the directions could be specifiedrelative to A,B, or some other coordinate system.(This problem can also be readily solved with

CHAPTER 3 3

components and unit vectors. Figure 3-22 suggests acoordinate system with x-axis to the right and y-axisup, as shown. Then A=A(ı cos 20◦ + sin 20◦) andB=A(ı cos 20◦− sin 20◦), from which A±B areeasily obtained.)

Problem

8. Vector A has magnitude 1.0 m and points at 35◦

clock-wise from the x-axis. Vector B has magnitude1.8 m. What angle should B make with the x-axisin order that A + B be purely vertical?

Solution

The vectors A,B, and A+B form a triangle, butgiven only that B = 1.8A and A+B areperpendicular to the x-axis, two are possible. In one,the angle opposite side B is β = 125◦, in the otherβ = 55◦, as sketched. For either, the law of sines givesα = sin−1(A sin β

B) = sin−1( sin 125◦

1.8) = sin−1( sin 55◦

1.8) =

27.1◦. Therefore, B makes an angle of ± 62.9◦ withthe negative x-axis or ± 117◦ with the x-axis (look atsuitable right triangles in the sketch). (This result canalso be obtained from components: since A+B isvertical, Ax + Bx = 0, or Bx = B cos θx = −Ax =−A cos 35◦, so θx = cos−1(−A cos 35◦/B) =cos−1(− cos 35◦/1.8) = ±117◦, where By could bepositive or negative.)

Problem 8 Solution.

Problem

9. Three vectors A, B, and C have the samemagnitude L and form an equilateral triangle, asshown in Fig. 3-23. Find the magnitude anddirection of the vectors (a) A+B, (b) A−B,

(c) A+B+C, (d) A+B−C.

figure 3-23 Problems 9, 16, and 22.

Solution

(c) Since the vectors form a closed figure (a triangle),their sum is zero, i.e., A+B+C= 0. (a) The vectorequation in part (c) has solution A+B= −C. Thus|A+B|= |−C|= |C|=L and the direction of A+B

is opposite to the direction of C, or 180◦ from C.(d) Similarly (A+B−C= (−C)−C=−2C, andso has magnitude 2L and direction opposite to C.(b) Finally, (A−B)=A− (−A−C)= 2A+C, sothese vectors form a 30◦ − 60◦ − 90◦ right triangle, asshown. Then |A−B|=

√3|C|=

√3L, and its

direction is 90◦ CCW from C.Of course, this problem can be solved readily using

components and a coordinate system with x-axisparallel to C and y-axis perpendicular, as shownsuperposed on Fig. 3-23. Then A= |A|(ı cos 120◦+ sin 120◦) = L(−ı+

√3)/2, B =L(−ı−

√3)/2, and

C = Lı. It is a simple matter to find(a) A+B = −Lı, (b) A−B =

√3L,

(c) A+B+C= 0, and (d) A+B−C = − 2Lı.The magnitudes and directions are as above.

Problem

10. A direct flight from Orlando, Florida, to Atlanta,Georgia, covers 660 km and heads at 29◦ west ofnorth. Your flight, however, stops at Charleston,South Carolina, on the way to Atlanta.Charleston is 510 km from Orlando, in a direction9.3◦ east of north. Use graphical techniques tofind the magnitude and direction of theCharleston-to-Atlanta leg of your flight.

Solution

The displacements between the three cities are asshown in the diagram. Graphical techniques can be

4 CHAPTER 3

confirmed by calculations using components and unitvectors, or trigonometry. Evidently, the displacementfrom Charleston to Atlanta is C=A−B =

(660 km)(ı cos(90◦ + 29◦)+ sin 119◦) −

(510 km)(ı cos(90◦ − 9.3◦)+ sin 80.7◦) =(−320ı+577− 82.4ı− 503) km = (−420ı+74.0) km.Its magnitude is

√

(−402)2 + (74.0)2 km = 409 kmand its direction is θ = tan−1(74.0/(−402)) =170◦ CCW from the x-axis, or θ − 90◦ = 79.6◦ CCWfrom the y-axis. For those solving this problemfrom the laws of cosines and sines√

(660)2 + (510)2 − 2(660)(510) cos38.3◦ = 409,sin−1(660 sin38.3◦/409) = 91.1◦, and90◦ − (91.1◦ − 80.7◦) = 79.6◦.

Problem 10 Solution.

Section 3-3: Coordinate Systems, Vector

Components, and Unit Vectors

Problem

11. Vector V represents a displacement of 120 km at29◦ counterclockwise from the x-axis. Write V inunit vector notation.

Solution

Take the y-axis 90? CCW from the x-axis, as inFigs. 3-10 and 11. Then V=Vxı+Vy =V (ı cos θx +

cos θy) = V (ı cos θx + sin θx)= (120 km)×(ı cos 29◦ + sin 29◦)= (105ı+ 58.2) km. (Note: Thecomponent of a vector along an axis is defined in termsof the cosine of the angle it makes with that axis. Intwo dimensions, θy = |θx − 90◦|, and cos θy = sin θx.)

Problem

12. Find the magnitude of the vector 34ı+ 13 m, anddetermine the angle it makes with the x axis.

Solution

Equation 3-1 gives the magnitude of a vector in terms

of its Cartesian components: A =√

A2x + A2

y =√

(34 m)2 + (13 m)2 = 36.4 m. Equation 3-3 gives theangle A makes with the x-axis: θx = cos−1(Ax/A) =cos−1(34 m/36.4 m) = 20.9◦. (Since both Ax and Ay

are positive, we know that θx is in the first quadrant.)Of course, Equation 3-2 could also have been usedhere, but Equation 3-3 holds in three dimensions,whereas Equation 3-2 does not.

Problem

13. Express each of the vectors of Fig. 3-24 in unitvector notation, with the x-axis horizontally to theright and the y-axis vertically upward.

figure 3-24 Problems 13, 19, and 20.

Solution

Take the x-axis to the right and the y-axis 90◦

counterclockwise from it. Then

A = 10(ı cos 35◦ + sin 35◦)= 8.19ı+ 5.74

B = 6(ı cos 235◦ + sin 235◦)= − 3.44ı− 4.91

C = 8(ı cos 115◦ + sin 115◦)= − 338ı+ 7.25

Problem

14. (a) What is the magnitude of ı + ? (b) Whatangle does it make with the x-axis?

Solution

The same reasoning as in Problem 12 shows that(a) the magnitude of ı + is

√12 + 12 =

√2, and that

(b) the angle it makes CCW from the x-axis isθx = cos−1(1/

√2) = 45◦. (Thus, n= (ı + )/

√2 is a

unit vector midway between the x- and y-directions;see Problem 63.)

Problem

15. Repeat Problem 3, using unit vector notation.

CHAPTER 3 5

Solution

See solution to Problem 3.

Problem

16. Express the vectors of Fig. 3-23 in unit vectornotation, taking the x-axis horizontal and they-axis vertical. Each vector has length A.

Solution

See the solution to Problem 9.

Problem

17. Let A= 15ı− 40 and B= 31+ 18k. Find avector C such that A+B+C= 0.

Solution

C=−A−B=−(15ı− 40)− (31+ 18k)=−15ı+ 9− 18k. (Since A and B are specified interms of unit vectors, this form is also appropriate forC.)

Problem

18. A proton travels in a circular path around the2.0-km-diameter accelerator at Fermilab, nearChicago. Write expressions for the proton’sdisplacement vector from the center of the circlewhen it is at (a) 0◦; (b) 30◦; (c) 45◦; (d) 60◦;(e) 90◦; (f) 180◦, as measured counterclockwisefrom the x-axis.


Solution

The position of any point on the circumference of acircle, relative to the center, is r= xı+ y = r(ı cos θ +

sin θ), where θ is the angle measured CCW from thex-axis. At Fermilab, r = 1 km (half the diameter), sofor the values of θ given, (a) r= (1 km)(ı cos 0◦ +

sin 0◦)= ı km; (b) r = (1 km)(ı cos 30◦ +

sin 30◦) = 12(√

3ı + ) km; (c) r= (ı+ ) km/√

2;

(d) r= 12(ı+

√3) km; (e) r= km; and

(f) r= −ı km .

Problem

19. Use the result of Problem 13 to find the vectors(a) A+B+C, (b) A−B+C, and(c) A+ 1.5B− 2.2C.

Solution

The components of A,B, and C are given in thesolution to Problem 13. A component of the sum (ordifference) is the sum (or difference) of thecomponents; the component of a scalar multiple is thescalar multiple of the component. For example, thex-component of A+ 1.5B− 2.2C is Ax + 1.5Bx−2.2Cx. Therefore (a) A+B+C = 1.37ı+ 8.07 =8.19 (ı cos 80.4◦ + sin 80.4◦) (b) A−B+C =8.25ı+ 17.9 = 19.7(ı cos 65.3◦ + sin 65.3◦) (c) A+

1.5B− 2.2C= 10.5ı− 17.6 = 20.5(ı cos 301◦ +

sin 301◦) The first form is the vector in components,

the second gives the magnitude (√

x2 + y2) anddirection θx = tan−1(y/x).

Problem

20. Find a vector D such that A+B+C+D= 0 forthe vectors of Fig. 3-24.

Solution

The vector D=−(A+B+C) is the negative of thevector given in part (a) of the preceeding solution, soD=− 1.37ı− 8.07= 8.19(ı cos 260◦ + sin 260◦).(Note that 180◦ + θ is the opposite direction to θ, inthe x-y plane.)

Problem

21. You’re trying to reach a pond that lies 3.5 km tothe northeast of your starting point. You firstfollow a logging road that runs east for 0.80 km.Then you follow a deer trail heading northeast for2.1 km. From there you bushwack straight to thepond. Describe your final displacement vector,(a) in unit vector notation; and (b) as amagnitude and compass direction.

Solution

The desired total displacement, R= 3.5 km NE, isthe sum of three displacements, R1 = 0.80 kmE,R2 = 2.1 km NE, and R3 =R−R1 −R2 to be found.(a) With x-axis E and y-axis N, R1 = 0.80km,R2 = (2.1 km)(ı cos 45◦ + sin 45◦)= 1.48ı+ 1.48 km,and R= 2.47ı+ 2.47 km. Therefore R3 =

6 CHAPTER 3

(2.47 − 0.80 − 1.48)ı+ (2.47 − 1.48) km =0.190ı+ 0.990 km= (1.01 km)(ı cos 79.1◦ +

sin 79.1◦). (See Equations 3-1 and 3-2 or the solutionto Problem 19 for the last step.)

Problem

22. For the vectors of Fig. 3-23, find two values for thescalar c such that A+ cB has magnitude 2.18L.

Solution

The vectors A and c B form two sides of atriangle with included angle of 60◦ if c > 0 and120◦ if c < 0, as shown. The magnitudes of thesides, which are given as |A|=L, |cB|= |c|L, and|A+ cB| = 2.18L, are related by the law ofcosines. Since |c|L cos 60◦ = 1

2cL for c > 0 is the

same as |c|L cos 120◦ = (−c)L(− 12) for c < 0, the

law of cosines for both possibilities is (2.18)2L2 =L2 + c2L2 − cL2, or c2 − c − 3.75 = 0. Thequadratic formula gives the two solutions asc = [1 ±

√

1 + 4(3.75)]/2 = 2.50 or −1.50,respectively.


Problem

23. In Fig. 3-14 the angle between x- and x′-axesis 21◦, the angle between the vector A and thex-axis is 54◦, and A’s magnitude is 10 units.(a) Find the components of A in both coordinatesystems shown. (b) Verify that the magnitude ofA, computed using Equation 3-1, is the same inboth coordinate systems.

Solution

(a) In the x-y system, Ax =A cos θ = 10 cos 54◦ = 5.88and Ay = 10 sin54◦ = 8.09. In the x′-y′ system,

θ′ = 54◦ − 21◦ = 33◦ so A′

x = 10 cos 33◦ = 8.39 andA′

y = 10 sin 33◦ = 5.45 (b) Direct calculation shows

that√

(5.88)2 + (8.09)2 =√

(8.39)2 + (5.45)2 = 10,which reflects the fact that sin2 + cos2 = 1. (Themathematical definition of a two-dimensional vector isa pair of numbers (Vx, Vy) which transform like theposition vector (x, y) when the coordinate axes arerotated.)

Problem

24. Vector A is 10 units long and points 30◦ counter-clockwise (CCW) from horizontal. What are the xand y components on a coordinate system (a) withthe x-axis horizontal and the y-axis vertical;(b) with the x-axis at 45◦ CCW from horizontaland the y-axis 45◦ CCW from vertical; and(c) with the x-axis at 30◦ CCW from horizontaland the y-axis 90◦ CCW from the x-axis?

Solution

The component of a vector along any direction equalsthe magnitude of the vector times the cosine of theangle (≤ 180◦). Thus,(a) Ax = 10 cos 30◦ = 8.66, Ay = 10 cos 60◦ = 5.00(b) A′

x = 10 cos 15◦ = 9.66, A′

y = 10 cos105◦ = −2.59(c) A′′

x = 10 cos 0◦ = 10, A′′

y = 10 cos 90◦ = 0.


Problem

25. Express the sum of the unit vectors ı, , and k inunit vector notation, and determine its magnitude.

Solution

r= ı+ + k; |r|=√

12 + 12 + 12 =√

3.

Problem

26. A mountain expedition starts a base camp at analtitude of 5500 m. Four climbers then establishan advance camp at an altitude of 7400 m; theadvance camp is southeast of the base camp, at ahorizontal distance of 8.2 km. From the advancecamp, two climbers head directly north to an8900-m summit, a horizontal distance of 2.1 km.

CHAPTER 3 7

Using a coordinate system with the x-axiseastward, the y-axis northward, and the z-axisupward, and with origin at the base camp, expressthe positions of the advance camp and summit inunit vector notation, and determine thestraight-line distance from base camp to summit.

Solution

The vector from base to advance camps isA= (8.2 km)(ı cos 45◦ − sin 45◦) + (7.4 km−

5.5 km)k= (5.80ı− 5.80+ 1.90k) km, and thatfrom advance camp to summit B= (2.1+ 1.5k) km.Therefore, the vector from base camp to summit isC=A+B= (5.80ı− 5.80+ 1.90k) km+

(2.1+ 1.5k) km+ (5.80ı− 3.70+ 3.40k) km. Thestraight-line distance isC =

√

(5.8)2 + (−3.7)2 + (3.4)2km = 7.67 km.


Problem

27. In Fig. 3-15, suppose that vectors A and C bothmake 30◦ angles with the horizontal while B

makes a 60◦ angle, and that A = 2.3 km,B = 1.0 km, and C = 2.9 km. (a) Express thedisplacement vector ∆r from start to summit ineach of the coordinate systems shown, and(b) determine its length.

Solution

Using the x-y system in Fig. 3-15, we have θA

θC = 30◦, and θB = 60◦. Then A=A(ı cos θA +

sin θA)= (2.3 km)(ı cos 30◦ + sin 30◦)=(1.99ı+ 1.15)km, B= (1 km)(ı cos 60◦ + sin 60◦)=(0.50ı+ 0.87) km, and C= (2.9 km)(ı cos 30◦ +

sin 30◦)= (2.51ı+ 1.45)km. Thus,∆r=A+B+C= (5.00ı+ 3.47) km, and∆r =

√

(5.00)2 + (3.47)2 km = 6.09 km. A similarcalculation in the x′ -y′ system, with θ′A = θ′C = 0 andθ′B = 30◦, yields A = 2.3ı′ km, B= (1 km)×(ı′ cos 30◦ + ′ sin 30◦)= (0.87ı′ + 0.50′) km, C=

2.9ı′ km, ∆r= (6.07ı′ + 0.50′) km, and√

(6.07)2 + (0.50)2 = 6.09, of course.

Section 3-4: Velocity and Acceleration Vectors

Problem

28. An object is moving at 18 m/s at an angle ofcounterclockwise from the x-axis. What are thex and y components of its velocity?

Solution

vx = (18 m/s) cos 220◦ = −13.8 m/s.vy = (18 m/s) cos(220◦ − 90◦) = (18 m/s) sin 220◦ =−11.6 m/s.

Problem

29. A car drives north at 40 mi/h for 10 min, thenturns east and goes 5.0 mi at 60 mi/h. Finally, itgoes southwest at 30 mi/h for 6.0 min. Draw avector diagram and determine (a) the car’sdisplacement and (b) its average velocity for thistrip.

Solution

Take a coordinate system with x-axis east, y-axisnorth, and origin at the starting point. The firstsegment of the trip can be represented by adisplacement vector in the y direction of length(40 mi/h)(10 min), or r1 = (20/3) mi. For the secondsegment, r2 = 5ı mi. The time spent on this segment ist2 = 5 mi/(60 mi/h) = 5 min. The final segment haslength (30 mi/h)(6 min). A unit vector in thesouthwest direction is ı cos 225◦ + sin 225◦ =

− (ı + )/√

2, so r3 = − (3/√

2)(ı+ ) mi. Thesedisplacements and their sum are shown in the sketch.(a) The total displacement is rtot =r1 +r2 +r3 =

[(20/3)+ 5ı− (3/√

2)(ı+ )] mi= (2.88ı+ 4.55) mi.(b) The total time is 10 min + 5 min +6 min = 21 min,so the average velocity for the trip is v =rtot/ttot =(2.88ı + 4.55) mi/(21/60) h= (8.22ı+ 13.0) mi/h.(Note: Instead of unit vector notation, rtot and vcould be specified by their magnitudes√

(2.88)2 + (4.55)2 mi = 5.38 mi and 15.4 mi/h,respectively, and common direction,θ = tan−1(4.55/2.88) = 57.7◦ N of E.)

Problem

30. A biologist studying the motion of bacteria notes abacterium at position r1 = 2.2ı+ 3.7− 1.2k µ m(1 µ m = 10−6 m). After 6.2 s the bacterium is atr2 = 4.6ı+ 1.9k µ m. What is its average velocity?Express in unit vector notation, and calculate themagnitude.

Solution

v = (r2 − r1)/(t2 − t1)

8 CHAPTER 3


= [(4.6ı+ 1.9k)− (2.2ı+ 3.7− 1.2k)]µ m/6.2 s

= (0.387ı− 0.597+ 0.500k)µ m/s

|v| =√

(0.387)2 + (−0.597)2 + (0.5)2µ m/s

= 0.869 µ m/s.

Problem

31. The Orlando-to-Atlanta flight described inProblem 10 takes 2.5 h. What is the averagevelocity? Express (a) as a magnitude anddirection, and (b) in unit vector notation with thex-axis east and the y-axis north.

Solution

(b) The displacement from Orlando to Atlantacalculated in Problem 10 was A= (−320ı+ 577) kmin a coordinate system with x-axis east and y-axisnorth. If this trip took 2.5 h, the average velocity wasv=A/2.5 h = (−128ı+ 231) km/h. (a) This hasmagnitude

√

(−128)2 + (231)2 km/h = 264 km/h anddirection θ = tan−1(231/−128) = 119◦ (which wasgiven).

Problem

32. The minute hand of a clock is 5.5 cm long. Whatis the average velocity vector for the tip of thehand during the interval from the hour to20 minutes past the hour, expressed in acoordinate system with the y-axis toward noonand x-axis toward 3 o’clock?

Solution

At the hour, the tip of the minute hand has positionr1 = (5.5 cm), while at 20 min past the hour, it hasposition r2 = (5.5 cm)(ı cos (− 30◦)+ sin (− 30◦))=

(4.76ı− 2.75) cm. The average velocity is v =(r2 − r1)/20 min = (4.76ı− 8.25) cm/20 min =(0.238ı− 0.413) cm/min.


Problem

33. A hot-air balloon rises vertically 800 m over aperiod of 10 min, then drifts eastward 14 km in27 min. Then the wind shifts, and the balloonmoves northeastward for 15 min, at a speed of24 km/h. Finally, it drops vertically in 5 min untilit is 250 m above the ground. Express theballoon’s average velocity in unit vector notation,using a coordinate system with the x-axiseastward, the y-axis northward, and the z-axisupward.

Solution

The displacement for the first segment of the balloon’sexcursion is r1 = 0.8k km, in the coordinate systemspecified, and for the second segment r2 = 14ı km.The third segment has length (24 km/h)(15 min) =6 km in the northeast direction ı cos 45◦ + sin 45◦,so r3 = (6 km)(ı+ )/

√2 = (4.24ı+ 4.24) km.

Finally, the last segment’s displacement is r4 =

(250 m − 800 m)k= − 0.55k km (this is a drop of550 m from the preceding altitude). The totaldisplacement ∆r=r1 +r2 +r3 +r4 =[(14+ 4.24)ı+ 4.24+ (0.8 − 0.55)k] km= (18.2ı+4.24+ 0.25k) km is accomplished in total time∆t = (10 + 27 + 15 + 5)min = 0.950 h, so the averagevelocity is ∆r/∆t = (19.2ı+ 4.47+ 0.263k) km/h.

Problem

34. Figure 3-25 shows the path of a bug as it crawlsaround a tabletop. Dots mark the position of thebug at each second. Determine the averagevelocity of the bug over the interval (a) from 1.0 sto 2.0 s; (b) from 2.0 s to 4.0 s; (c) 0 to 6.0 s.

CHAPTER 3 9

x (mm)

y (mm)

1 32–3 –2 –1–4 4

–3

–1

–4

1

2

3

4

–2

5.0 s

0 s

1.0 s

2.0 s

3.0 s

4.0 s

6.0 s

figure 3-25 Problem 34.

Solution

The position vectors for the bug, at each second oftime, can be read off Fig. 3-25, so the average velocityfor any interval can be calculated from Equation 3-5,v = ∆r/∆t. (a) va = [r(2 s)− r(1 s)]/(2 s − 1 s)= [(−3 mm)− (3ı− ) mm]/1 s = −(3ı+ 2) mm/s.(b) vb = [r(4 s)− r(2 s)]/(4 s − 2 s)=[(− ı +

0.5) mm −(−3 mm)]/2 s = (−0.5ı+ 1.75) mm/s.(c) vc = [r(6 s)− r(0)]/6 s = [(− ı + 2 ) mm− 0]/6 s=

(−0.167ı+ 0.333) mm/s.

Problem

35. An object’s position as a function of time is givenby r= 12tı+ (15t− 5.0t2) m, where t is time in s.(a) What is the object’s position at t = 2.0 s? (b)What is its average velocity in the interval fromt = 0 to t = 2.0 s? (c) What is its instantaneousvelocity at t = 2.0 s?

Solution

(a) The object’s position is given as a function of time,so when t = 2 s, this is r(2 s) = (12 m/s)(2 s)ı+

[(15 m/s)(2 s)− (5.0 m/s2)(2 s)2 ]= 24ı+ 10 m,

where we explicitly displayed the units of thecoefficients in the intermediate step. (b) Sincer(0) = 0, the average velocity for this interval isv= [r(2 s) − r(0)]/(2 s − 0) = 12ı+ 5 m/s. (c) Theinstantaneous velocity at any time is dr/dt =

(12 m/s)l + [(15 m/s) − (5.0 m/s2)2t]=v(t) (seeAppendix A-2 for the derivative of tn), so whent = 2 s, v(2s) = 12ı− 5 m/s.

Problem

36. A supersonic aircraft is traveling east at2100 km/h. It then begins to turn southward,emerging from the turn 2.5 min later heading duesouth at 1800 km/h. What are the magnitude anddirection of its average acceleration during the

turn?

Solution

In a coordinate system with x-axis east and y-axisnorth, the initial velocity of the airplane at thebeginning of its turn is v1 = 2100ı km/h, and the finalvelocity at the end is v2 = −1800 km/h. The averageacceleration (Equation 3-8) is a= (v2 − v1)÷(t2 − t1) = (−1800− 2100ı)(km/h)/2.5 min =

−(3.89ı + 3.33) m/s2, with magnitude

√

(−3.89)2 + (−3.33)2 m = 5.12 m/s2

and directionθ = tan−1(−3.33/3.89) = 221◦ (in the third quadrant,nearly southwest).

Problem

37. A car, initially going eastward, rounds a 90◦ bendand ends up heading southward. If thespeedometer reading remains constant, what is thedirection of the car’s average acceleration vector?

Solution

Since the speed is constant, the change in velocity forthe 90◦ turn is ∆v=−v−(vı) = −v(ı+ ), where ı iseast and is north. The direction of the averageacceleration is the same as that of ∆v, which isparallel to −(ı+ ) or southwest.(θ = tan−1(−1/−1) = 225◦.)

Problem

38. Earth moves in a nearly circular orbit about theSun. What is the magnitude of its averageacceleration over the intervals (a) from January 1to July 1 and (b) from January 1 to April 1?(c) What is the angle between the two averageacceleration vectors for the intervals given?Consult Appendix E for Earth’s orbital speed.

Solution

(a) In six months (half a year or 12×3.156×107 s) the

Earth has traveled half way around its orbit, so itsvelocity has merely reversed direction, vJul=−vJan.The magnitude of the average acceleration isaa = |vJul −vJan| /∆t = 2 |vJul| / 1

2y. The Earth’s

orbital speed is nearly constant at 30km/s; therefore

aa = 4(30 km/s)/3.156×107 s = 3.80 mm/s2. The

direction of aa is parallel to vJul. (b) In just threemonths, the Earth covers one fourth of its orbit, so itsvelocity changes by almost 90◦, i.e., vApr⊥vJan. Then∆v = vApr − vJan forms the hypotenuse of an isoscelesright triangle, as shown in the sketch, with magnitude√

2|v|. Therefore, the magnitude of the averageacceleration is ab =

√2|v|/ 1

4y = 4

√2(30 km/s)÷

3.156×107 s = 5.38 mm/s2. (c) From the sketch, one

10 CHAPTER 3

can see that the angle between the ∆v’s in parts(a) and (b) is 45◦ (which would be exact only for acircular orbit and equal-length months).


Problem

39. What are (a) the average velocity and (b) theaverage acceleration of the tip of the 2.4-cm-longhour hand of a clock in the interval from 12 p.m.

to 6 p.m.? Express in unit vector notation, withthe x-axis pointing toward 3 p.m. and the y-axistoward 12 p.m.

Solution

The position and velocity of the tip of the hour handare shown in the diagram for 12 p.m. and 6 p.m., inthe coordinate system specified. The magnitude of theposition is a constant, namely, the 2.4-cm radius. Themagnitude of the velocity is also constant, namely, thecircumference divided by 12 h, or 2π(2.4 cm)/12 h =1.26 cm/h. (a) v= (r2 − r1)/6 h =(−2.4− 2.4) cm/6 h=− 0.8 cm/h. (b) a =(v2 − v1)/6 h = (−1.26ı− 1.26ı)(cm/h)/6 h. =

−0.419ı cm/h2.


Problem

40. Attempting to stop on a slippery road, a carmoving at 80 km/h skids across the road at a30◦ angle to its initial motion, coming to a stop in3.9 s. Determine the average acceleration in m/s2,using a coordinate system with the x-axis in thedirection of the car’s original motion and they-axis toward the side of the road to which the carskids.


Solution

The car’s acceleration is opposite to the direction ofthe skid, since it comes to a stop (if v= 0,a=−v0/∆t). Its magnitude is given by Equation 3-8,|a| = |v − v0| /∆t = |−v0| /∆t = (80 m/3.6 s)/3.9 s =

5.70 m/s2. In relation to the coordinate system

specified, a = (5.70 m/s2)(ı cos 210◦ + sin 210◦) =

−(5.70 m/s2)(√

3ı + )/2 = −(4.93ı+ 2.85) m/s2.(Note that v0, the velocity at the start of the skid, isnot in the direction of the initial motion before theskid.)

Problem

41. An object undergoes acceleration of 2.3ı+3.6 m/s

2over a 10-s interval. At the end of this

time, its velocity is 33ı + 15 m/s. (a) What wasits velocity at the beginning of the 10-s interval?(b) By how much did its speed change? (c) Byhow much did its direction change? (d) Show thatthe speed change is not given by the magnitude ofthe acceleration times the time. Why not?

Solution

(a) v=v0 + at, so v0 =v −at, or v0 =

(33ı+ 15) m/s − (2.3ı+ 3.6) m/s2(10 s) =(10ı− 21) m/s. (b) v0 =

√

(10)2 + (−21)2 = 23.3 m/s,

and v =√

(33)2 + (15)2 = 36.2 m/s, so the change inspeed is ∆v = v − v0 = 13.0 m/s (we did not round offbefore subtracting). (c) θ = tan−1(15/33) = 24.4◦ andθ0 = tan−1(−21/10) = 295◦ = −64.5◦ (positive anglesCCW, negative angles CW, from x-axis) so thedirection changed by ∆θ = θ − θ0 = 89.0◦. (d) at =

CHAPTER 3 11

√

(2.3)2 + (3.6)2 m/s2(10 s) = 42.7 m/s 6= ∆v. The

difference between at = |v − v0| and ∆v = v − v0 canbe seen from the triangle inequality:| v − v0 | ≤ | v − v0| ≤ v + v0.


Problem

42. An object’s position as a function of time is givenby r= (bt3 + ct)ı+ dt2 + (et + f)k, where b, c, d,e, and f are constants. Determine the velocity andacceleration as functions of time.

Solution

Differentiating each term using Equation 2-3, we findv= dr= dt= (3bt2 + c)ı+ 2dt+ ek, anda= dv/dt= 6btı= 2d.

Problem

43. The position of an object is given byr= (ct − bt3)ı+ dt2, with constants c = 6.7 m/s,

b = 0.81 m/s3, and d = 4.5 m/s2. (a) Determinethe object’s velocity at time t = 0. (b) How longdoes it take for the direction of motion to changeby 90◦? (c) By how much does the speed changeduring this time?

Solution

(a) v(t)= dr/dt = (c − 3bt2)ı+ 2dt, sov(0)= (6.7 m/s)ı. (b) The direction of v(t) isθ(t) = tan−1(2dt/(c − 3bt2)), measured CCW fromthe x-axis, so θ(0) = 0◦. θ(t) = 90◦ when theargument of the arctan is ∞, or t =

√

c/3b. Thus,

t =

√

(6.7 m/s)/3(0.81 m/s3) = 1.66 s. (The direction

of motion was −90◦ at t = −1.66 s.) (c) Since vx = 0when t =

√

c/3b the speed at t = 1.66 s is

2d√

c/3b = 14.9 m/s. The speed at t = 0 isc = 6.7 m/s, so the change was 8.24 m/s.

Problem

44. For the object of Problem 35, determine (a) theaverage acceleration in the interval from t = 0 tot = 2.0 s and (b) the instantaneous acceleration att = 2.0 s.

Solution

(a) The average acceleration during the interval usedin Problem 35 is a= [v(2 s) − v(0)]/(2 s − 0)=

[(12ı− 5) m/s− (12ı+ 15) m/s]/(2 s)= − 10 m/s2

(see part (c) of the solution to Problem 35 for v(t)).(b) The instantaneous acceleration at any time is

a(t)= dv/dt= − (10 m/s2), which is constant,

including when t = 2 s.

Section 3-5: Relative Motion

Problem

45. A dog paces around the perimeter of a rectangularbarge that is headed up a river at 14 km/hrelative to the riverbank. The current in the wateris at 3.0 km/h. If the dog walks at 4.0 km/h, whatare its speeds relative to (a) the shore and (b) thewater as it walks around the barge?

Solution

(a) Let S be a frame of reference fixed on the shore,with x-axis upstream, and let S′ be a frame attachedto the barge. The velocity of S′ relative to S isV= (14 km/h)ı. The velocity of the dog relative tothe shore is (from Equation 3-25) v=v′ +V, and itsspeed is v = |v′ +V|, where v′ = 4 km/h. When thedog is walking upstream, v′ ‖ V, v= (4 + 14) km/h =18 km/h, and when walking downstream, −v′ ‖ V,and v = (14 − 4) km/h = 10 km/h. Whenv′ ⊥ V, v=

√142 + 42 = 14.7 km/h. (In general,

v2 = v′2 + V 2 + 2v′V cos θ′, where θ′is the anglebetween v′ and V in S′.) (b) Since the current flowsdownstream, according to Equation 3-25:

(

vel. of bargerel. to water

)

=

(

vel. of bargerel. to shore

)

−(

vel. of waterrel. to shore

)

= 14ı− (−3ı) km/h.

Going through the same steps as in part (a), for a newframe S moving with the water, with a new relativevelocity V= (17 km/h)ı, we find the speed of the dogrelative to the water to be (4 + 17) = 21 km/h,(17 − 4) = 13 km/h, and

√42 + 172 = 17.4 km/h, for

the corresponding segments of the barge’s perimeter.

12 CHAPTER 3


Problem

46. A jetliner with an airspeed of 1000 km/h sets outon a 1500-km flight due south. To maintain asouthward direction, however, the plane must bepointed 15◦ west of south. If the flight takes100 min, what is the wind velocity?

Solution

Using the same reference frames as specified inExample 3-7, we are given that v=

(1500 km/100 min)(−)= −900 km/h, andv′ = −(1000 km/h)(ı sin 15◦ + cos 15◦). The windvelocity is V=v −v′ = (−900+ 259ı+ 966)=(259ı + 65.9) km/h. Therefore, the wind speed isV =

√2592 + 65.92 = 267 km/h, and the angle V

makes with the x-axis is (tan− 1 65.9/259=14.3◦)(N of E). (The wind direction, by convention, is thedirection facing the wind, in this case 14.3◦ S of W.)


Problem

47. A spacecraft is launched toward Mars at theinstant Earth is moving in the +x direction at itsorbital speed of 30 km/s, in the Sun’s frame ofreference. Initially the spacecraft is moving at40 km/s relative to Earth, in the +y direction.At the launch time, Mars is moving in the

−y direction at its orbital speed of 24 km/s. Findthe spacecraft’s velocity relative to Mars.

Solution

Equation 3-10 says that the velocity of the spacecraftrelative to Mars, vCM, equals the difference of thevelocities of each relative to the Sun, vCS,−vMS. (Ournotation vCM means the velocity of C relative to M.)vMS is given as −(24 km/s) in the Sun’s referenceframe, the x-y coordinates in the problem. The Earthhas velocity vES = (30 km/s)ı in the Sun’s frame, andvCE = (40 km/s), where the y-axes in the Earth’s andSun’s frames are assumed to be parallel. A secondapplication of Equation 3-10 gives vCS =vCE +vES =

(40+ 30ı) km/s; therefore vCM =vCS −vMS =

(30ı= 40) km/s − (−24) km/s= (30ı+ 64) km/s.

Problem

48. You wish to row straight across a 63-m-wide river.If you can row at a steady 1.3 m/s relative to thewater, and the river flows at 0.57 m/s, (a) in whatdirection should you head? (b) How long will ittake you to cross the river?

Solution

The current is perpendicular to the direction in whichyou wish to cross, as shown in the sketch. V is thecurrent velocity (velocity of the water relative to theground), v is the velocity of the boat relative to theground, and v′ is the velocity of the boat relative tothe water. These three vectors satisfy Equation 3-10.(a) Evidently, sin θ = |V| / |v′|, or θ =sin−1(0.57/1.3) = 26.0◦, which is your headingupstream. (b) |v| = |v′| cos θ = (1.3 m/s) cos 26.0◦ =1.17 m/s is your speed across the river, so the crossingtime is t = r/v = 63 m/(1.17 m/s) = 53.9 s.


Problem

49. You’re on an airport “people mover,” a conveyorbelt going at 2.2 m/s through a level section of theterminal. A button falls off your coat and dropsfreely 1.6 m, hitting the belt 0.57 s later. What

CHAPTER 3 13

are the magnitude and direction of the button’sdisplacement and average velocity during its fall in(a) the frame of reference of the “people mover”and (b) the frame of reference of the airportterminal? (c) As it falls, what is its acceleration ineach frame of reference?

Solution

(c) Let S be the frame of reference of the airport andS′ the frame of reference of the conveyor belt. If thevelocity of S′ relative to S is a constant, 2.2 m/s in thex-x′ direction, then the acceleration of gravity is thesame in both systems. (a) In S′, the initial velocity ofthe button is zero, and it falls vertically downward. Itsdisplacement is simply ∆y′ = −1.6 m, and its averagevelocity is ∆y′/∆t = −1.6 m/0.57 s = 2.81 m/s.(b) In S, the initial velocity of the button is not zero(it is 2.2 m/s in the x direction), and so the buttonfollows a projectile trajectory to be described inChapter 4. Here, we observe that while the buttonfalls vertically through a displacement ∆y =∆y′ = −1.6 m, it also moves horizontally (in thedirection of the conveyor belt) through adisplacement ∆x = (2.2 m/s)(0.57 s) = 1.25 m. Itsnet displacement in S is ∆r=∆xı+∆y =(1.25ı− 1.60) m, so its average velocity isv=∆r/∆t= (1.25ı− 1.60) m/0.57 s =(2.20ı− 2.8) m/s. These have magnitudes |∆r| =√

(1.25)2 + (−1.60)2 m = 2.03 m and

|v| =√

(2.20)2 + (−2.81)2 m/s = 3.57 m/s, and thesame direction θ = tan−1(−1.60/1.25) = −51.9◦ fromthe horizontal, or 38.1◦ from the downward vertical.

Problem

50. An airplane with airspeed of 370 km/h fliesperpendicularly across the jet stream. To achievethis flight, the plane must be pointed into the jetstream at an angle of 32◦ from the perpendiculardirection of its flight. What is the speed of the jetstream?


Solution

The velocity of the jet stream, V (air relative toground, also called the “windspeed”), the velocity ofthe airplane relative to the ground, v (the “groundspeed”), and the velocity of the airplane relative to theair, v′ (the “air speed”) form the sides of a trianglegiven by Equation 3-10, v′ +V=v, as shown. We aregiven that the triangle is a right triangle (v ⊥ V),that the angle between the airspeed and groundspeedis 32◦, and the hypotenuse (magnitude of airspeed) is370 km/h. From trigonometry, the magnitude of thewindspeed is |V| = (370 km/h) sin 32◦ = 196 km/h.

Paired Problems

Problem

51. A rabbit scurries across a field, going eastward21.0 m. It then turns and darts southwestward for8.50 m. Then it pops down a rabbit hole, 1.10 mvertically downward. What is the magnitude ofthe displacement from its starting point?

Solution

In a coordinate system with x-axis east, y-axis north,and z-axis up, the three displacements can be writtenin terms of their components and the unit vectors:r1 = 21.0ı m, r2 = (8.50 m)(ı cos 225◦ + sin 225◦)=−(6.01)(ı+ ) m, and r3 = −1.10k m. (Note thatsouthwest is 180◦ + 45◦ CCW from east.) The totaldisplacement is the vector sum r1 +r2 +r3 =[(21.0− 6.01)ı− 6.01− 1.10k] m, and its magnitude isthe square root of the sum of its components,√

(15.0)2 + (−6.01)2 + (−1.10)2 m = 16.2 m.

Problem

52. A cosmic ray plows into Earth’s upperatmosphere, liberating a shower of subatomicparticles. One of these particles moves downward3.2 km, then undergoes a collision, after which itmoves 1.6 km at 27◦ northward of the vertical. Itthen undergoes another collision that sends itmoving horizontally eastward for 2.1 km before itannihilates with another particle. What is themagnitude of its overall displacement?

Solution

Take a standard coordinate system with x-axis east,y-axis north, and z-axis up. The particle beginsmoving downward (say along the z-axis) through adisplacement r1 =−3.2k km. It is then deflected inthe y-z plane (northward of the vertical) by 27◦,so r2 + (1.6 km)( sin 27◦ − k cos 27◦), andfinally eastward through r3 = 2.1ı km.

14 CHAPTER 3

Thus, |r1 +r2 +r3| =√

(2.1)2 + (1.6 sin 27◦)2 + (−3.2 − 1.6 cos 27◦)2 km= 5.13 km.


Problem

53. A car is heading into a turn at 85 km/h. It entersthe turn, slows to 55 km/h, and emerges 28 s laterat 35◦ to its original direction, still moving at55 km/h. What are (a) the magnitude and(b) the direction of its average acceleration, thelatter measured with respect to the car’s originaldirection?

Solution

The initial and final velocities have magnitudes of85 km/h and 55 km/h, respectively, and make anangle of 35? as shown, where we chose the x-axisparallel to vi and the y-axis in the directionof the turn. The change in velocity is∆v=vf −vi = (55 km/h)(ı cos 35◦ + sin 35◦) −(85 km/h)ı = (−39.9ı + 31.5) km/h: (a) Themagnitude of the average acceleration isa = |∆v| /∆t =

√

(−39.9)2 + (31.5)2 (km/h)/28 s =

1.82 km/h/s = 0.505 m/s2. (b) The direction of

a is the same as the direction of ∆v, or θ =tan−1(31.5/− 39.9) = 142◦. (This problem could alsohave been solved using the laws of cosines and sines;see the solution to the next problem.)

Problem

54. The Galileo space probe was originally to belaunched directly toward its destination, Jupiter.But the 1986 explosion of the space shuttleChallenger led to a decision that Galileo’sliquid-fueled booster rocket was unsafe to fly onthe shuttle. As a result, Galileo’s trajectorybecame a complicated path through the inner


solar system, picking up speed through a so-called“gravity assist” maneuver involving closeencounters with the planets Venus and Earth. Thefirst Earth encounter occurred on December 8,1990, when Galileo, outbound from Venus, passed200 miles from Earth. Thirty days before theencounter, Galileo was approaching Earth at2.99×104 m/s. Thirty days after the encounter,Galileo was moving at 54◦ to its pre-encounterdirection, at a speed of 3.51×104 m/s. What were(a) the magnitude and (b) the direction ofGalileo’s average acceleration during this interval,the latter measured with respect to itspre-encounter direction?


Solution

The initial and final velocities for the 60 d interval aregiven, as indicated in the diagram. Instead of usingunit vectors, as in the previous problem, let’suse the laws of cosines and sines. (a) ∆v=√

(29.9)2 + (35.1)2 − 2(29.9)(35.1) cos 54◦ km/s =29.9 km/s. Therefore a = |∆v| /∆t (29.9 km/s)÷(60×86,400 s) = 5.76 mm/s

2. (b) θ = 180◦−

sin−1(35.1 sin 54◦/29.9) = 180◦ − 71.9◦ = 108◦.

Problem

55. The sweep-second hand of a clock is 3.1 cm long.What are the magnitude of (a) the averagevelocity and (b) the average acceleration of thehand’s tip over a 5.0-s interval? (c) What is the

CHAPTER 3 15

angle between the average velocity andacceleration vectors?

Solution

There will be numerous occasions to use vectorcomponents to analyze circular motion in laterchapters (or see the solutions to Problems 18, 32, 38,and 39), so let’s use geometry to solve this problem.(a) The angular displacement of the hand during a 5 sinterval is θ = (5/60)(360◦) = 30◦. The positionvectors (from the center hub) of the tip at thebeginning and end of the interval, r1 and r2, form thesides of an isosceles triangle whose base is themagnitude of the displacement, | ∆r | = 2 |r| sin 1

2θ =

2(3.1 cm) sin(30◦/2) = 1.60 cm, and whose base angleis 1

2(180◦ − 30◦) = 75◦. Thus, the average velocity has

magnitude | ∆r | /∆t = 1.60 cm/5 s = 0.321 cm/s anddirection 180◦ − 75◦ = 105◦CW from r1. (b) Theinstantaneous speed of the tip of the second-hand is aconstant and equal to the circumference divided by 60s, or v = 2π(3.1 cm)/60 s = 0.325 cm/s. The directionof the velocity of the tip is tangent to thecircumference, or perpendicular to the radius, in the


direction of motion (CW). The angle between twotangents is the same as the angle between the twocorresponding radii, so v1, v2 and ∆v form anisosceles triangle similar to the one in part (a). Thus|∆v| = 2 |v| sin 1

2θ = 2(0.325 cm/s) sin(1

2× 30◦) =

0.168 cm/s. The magnitude of the averageacceleration is |∆v| /∆t = (0.168 cm/s)/5 s =

3.36×10−2 cm/s2, and its direction is 105◦ CW from

the direction of v1, or 195◦ CW from the direction ofr1. (c) The angle between a and v, from parts (a) and(b), is 195◦ − 105◦ = 90◦. (Note: This is the geometryused in Section 4.4 to discuss centripetal acceleration.)

Problem

56. A proton in a cyclotron follows a circular path23 cm in diameter, completing one revolution in0.17 µs. What are the magnitude of (a) theaverage velocity and (b) the average accelerationas the proton sweeps through one-twelfth of thefull circle? (c) What is the angle between theaverage velocity and acceleration vectors?

Solution

One-twelfth of a revolution is 30◦, so the geometry ofthis problem is the same as that for the previous one,and the directions of v, a and the angle between themare the same as in Problem 55. The magnitudes,however, are different. (a) |∆r| = 2 |r| sin 1

2θ =

(23 cm) sin 15◦ = 5.95 cm and ∆t = 0.17 µs/12 =14.2 ns. Thus, |v| = |∆r| /∆t = 4.20 Mm/s. (b) Theinstantaneous constant speed is π(23 cm)/(0.17 µs) =4.25 Mm/s, so |∆v| = 2(4.25 Mm/s) sin 15◦ =

2.20 Mm/s. Then |a| = |∆v| /∆t = 1.55×1014 m/s2.

Problem

57. A ferryboat sails between two towns directlyopposite one another on a river. If the boat sailsat 15 km/h relative to the water, and if thecurrent flows at 6.3 km/h, at what angle shouldthe boat head?


16 CHAPTER 3

Solution

The velocity of the boat relative to the ground, v, isperpendicular to the velocity of the water relative tothe ground, the current velocity V, which form a righttriangle with hypotenuse v′equal to the velocity of theboat relative to the water, as shown in the diagramand as required by Equation 3-10. The headingupstream is θ = sin−1(|V| / |v′|) = sin−1(6.3/15) =24.8◦.

Problem

58. A flock of geese is attempting to migrate duesouth, but the wind is blowing from the west at5.1 m/s. If the birds can fly at 7.5 m/s relative tothe air, in what direction should they head?

Solution

If the windspeed is perpendicular to the geese’s desiredgroundspeed, their airspeed must be inclined upwindby θ = sin−1(5.1/7.5) = 42.8◦. (See the diagram andthe solution to Problem 50 for a definition of terms.)


Supplementary Problems

Problem

59. A satellite is in a circular orbit 240 km aboveEarth’s surface, moving at a constant speed of7.80 km/s. A tracking station picks up thesatellite when it is 5.0◦ above the horizon, asshown in Fig. 3-26. The satellite is tracked until itis directly overhead. What are the magnitudes of(a) its displacement, (b) its average velocity, and(c) its average acceleration during the trackinginterval? Is the value of the average accelerationapproximately familiar?

Solution

Once we know the angular displacement, ∆θ, of thesatellite in its orbit, we can calculate the magnitudes

figure 3-26 Problem 59 (figure is not to scale).

of the displacement from its initial position P1 to itsfinal position P2, and its average velocity andacceleration, during the same tracking interval, byusing the geometrical analysis in the solution toProblem 55, to which the reader is referred. In thediagram based on Fig. 3-26, O is the center of theEarth, RE = 6370 km is the average radius of theEarth, A is the tracking station, andr = RE + h = 6370 km + 240 km = 6610 km is theradius of the orbit. We apply the law of cosines totriangle OAP1 to find AP1, and then the law of sinesto find ∆θ. Thus,

(6610 km)2 = (6370 km)2 + (AP1)2

− 2(6370 km)(AP1) cos 95◦.

This is a quadratic equation with (positive) solution

AP1 = (6370 km) cos 95◦

+√

(6370 km)2 cos2 95◦ + (6610 km)2 − (6370 km)2

= 1295 km.

(We are keeping four significant figures in theintermediate results, but will round off to three at theend.) Then sin∆θ/AP1 = sin 95◦/r gives∆θ = sin−1(1295 sin95◦/6610) = 11.26◦.

(a) Now that we know ∆θ, the magnitude of thedisplacement can be found from the isosceles triangleOP1P2 as in Problem 55. P1P2 = 2r sin 1

2∆θ =

2(6610 km) sin 12(11.26◦) = 1296 km ≈ 1300 km.

(b) To find the magnitude of the average velocity,we first need to find the tracking interval ∆t . Thetime for a complete orbit (called the period) is theorbital circumference divided by the speed, orT = 2πr/v = 2π(6610 km)/(7.8 km/s) = 5.325×103 s.During the tracking interval, the satellite completesonly a fraction ∆θ/360◦ of a complete orbit, so∆t = (11.26◦/360◦)(5.325×103 s) = 166.5 s. Thus,|v| =P1P2/∆t = 1296 km/166.5 s = 7.78 km/s.

CHAPTER 3 17

(c) As shown in the solution to Problem 55, thevelocity vectors at P1 and P2 form an isosceles trianglesimilar to OP1P2. Thus | ∆v| = 2 |v| sin 1

2∆θ =

2(7.8 km/s) sin 12(11.26◦) = 1.530 km/s, and the

magnitude of the average acceleration is|a| = |∆v| /∆t = (1.530 km/s)/166.5 s = 9.19 m/s

2.

When we recall that the Earth’s gravity holds thesatellite in its orbit, it is not surprising that thismagnitude is close to g. Of course, at an altitude of240 km, the Earth’s gravitational field is only9.11 m/s2 (compared to 9.81 m/s2 at the surface). Wemust also remember that there is a discrepancybetween the average and instantaneous accelerationsdue to the finite size of ∆t .

Problem 59 Solution

Problem

60. The sum, A+B, of two vectors is perpendicularto the difference, A−B. How do the magnitudesof the two vectors compare?

Solution

The vectors A+B and A−B are the two diagonalsof the parallelogram formed by sides A and B. If thediagonals are perpendicular, the parallelogram is arhombus; hence |A| = |B|. (Students are advised toredo this problem after the “dot” product is defined inSection 7.2.)

Problem

61. Find two vectors in the x-y plane that areperpendicular to the vector aı+ b.

Solution

One vector perpendicular to the given vector,A= aı+ b, is a vector B of the same magnitude,


|B| = |A|, but pointing at an angle 90◦ more than theangle of A with the x-axis, as shown in the sketch, i.e.,θB = 90◦ + θA. The components of A areAx = |A| cos θA = a and Ay = |A| sin θA = b, whilethe components of B are Bx = |B| cos θB =|A| cos(90◦ + θA) = − |A| sin θA = −b andBy = |B| sin(90◦ + θA) = |A| cos θA = a. Thus,B = −bı + a is perpendicular to A = aı+ b. Clearly,−B= bı− a is also perpendicular to A, as is anyscalar multiple of B. (Another way to do this problemis to use the scalar product defined in Section 7.2.)


Problem

62. Find the angle between the vectors 3.0ı+ 1.7 and6.1ı− 4.2.

Solution

The angle a vector makes with the x-axis can be foundfrom its components, θ = tan−1(Ay/Ax) (seeEquation 3-2 and Fig. 3-11(a)). For the first vector,θ1 = tan−1(1.7/3.0) = 29.5◦ (in the first quadrant) andfor the second, θ2 = tan−1(−4.2/6.1) = 325◦ or −34.6◦

(in the fourth quadrant). The angle between them isthe difference θ1 − θ2 = 29.5◦ − (−34.6◦) = 64.1◦.(Note: 29.5◦ − 325◦ = −296◦ + 360◦ = 64.1◦. Thereare two possible angles between any two vectors, oneless than, the other greater than 180◦. By convention,the smaller one is always used.)

Problem

63. Write an expression for a unit vector that lies at45◦ between the positive x- and y-axes.

18 CHAPTER 3

Solution

A vector of unit magnitude, making a 45◦ angle CCWwith the x-axis, can be expressed as 1· cos 45◦ı +1· sin 45◦ = (ı+ )/

√2. (A unit vector in any

direction in the x-y plane is therefore n=

ı cos θ + sin θ.)

Problem

64. A vector A has components Ax and Ay in acoordinate system with axes x and y. Find itscomponents A′

x and A′

y in a coordinate systemwhose axes x′ and y′ are rotated counterclockwisethrough an angle θ from the x- and y-axes. Testyour result for the cases θ = 0 and θ = 90◦.

Aysin θ

Axcosθ Axsin θ

Aycosθ

x′

y′

Ay′

AZ ′

Ay

AZ

x

y

θ

θî

î′ J

J′


Solution

By drawing an extra line on Fig. 3-14 (through Ax onthe x-axis perpendicular to the dotted line through A′

y

on the y′-axis), one sees that A′

x = Ax cos θ + Ay sin θand A′

y = −Ax sin θ + Ay cos θ. For θ = 0,A′

x = Ax and A′

y = Ay (no change), while for θ = 90◦,A′

x = Ay and A′

y = −Ax (a 90◦ CW rotation of A).

Problem

65. Figure 3-27 shows two arbitrary vectors A and B

that sum to a third vector C. By working withcomponents, prove the law of cosines:C2 = A2 + B2 − 2AB cos γ.

Solution

Adding some lines parallel to the axes and labeling theangles of the vectors in Fig. 3-27, one sees thatCx = Ax + Bx = A cos θA + B cos θB andCy = A sin θA + B sin θB. Squaring, adding, and usingtwo trigonometric identities from Appendix A, onegets C2 = C2

x + C2y = A2(cos2 θA + sin2 θA)+

B2(cos2 θB + sin2 θB) + 2AB(cos θA cos θB +sin θA sin θB) = A2 + B2 + 2AB cos(θB − θA). This isthe law of cosines when we replace θB − θA by180◦ − γ, where γ is the angle between sides A and B

in the vector triangle.

θB

θA x

y

B

A

C g


Problem

66. You wish to paddle a canoe perpendicularly acrossa river of width w and back. If the river’s flowspeed is c, and you can paddle at speed v relativeto the water, show that your round-trip traveltime is given by ∆t = 2w/

√v2 − c2.

Solution

Crossing perpendicular to the current, your velocityrelative to the ground, u, and the current velocity, c,form a right triangle with hypotenuse equal to yourvelocity relative to the water, v. (See Equation 3-10and the sketch.) Thus, u =

√v2 − c2, and the time for

a round-trip of length 2w is ∆t = 2w/u.


Problem

67. Town B is located across the river from town Aand at a 40.0◦ angle upstream from A, as shown inFig. 3-28. A ferryboat travels from A to B; it sailsat 18.0 km/h relative to the water. If the currentin the river flows at 5.60 km/h, at what angleshould the boat head? What will be its speedrelative to the ground? Hint: Set upEquation 3-10 for this situation. Each componentof Equation 3-10 yields two equations in the

CHAPTER 3 19

unknowns v, the magnitude of the boat’s velocityrelative to the ground, and φ, the unknown angle.Solve the x equation for cosφ and substitute intothe second equation, using the relationsin φ =

√

1 − cos2 φ. You can then solve for v,then go back and get φ from your first equation.

A

40°

BV

V

u 9


Solution

The diagram shows the velocities and coordinate axesadded to Fig. 3-28. v′ (the velocity of the boatrelative to the water), v (the velocity of the boatrelative to the ground), and V (the velocity of thewater relative to the ground) are related byEquation 3-10, v′ =v −V. This vector equation isequivalent to two scalar equations, one for eachcomponent, v′x = vx − Vx and v′y = vy − Vy. Thecomponents of each vector, in terms of itsmagnitude and angle, are v′x = v′ cos θ′×v′y = v′ sin θ′, vx = v cos 40◦, vy = v sin 40◦, andVx = 0, Vy = −V. Therefore, the x and y componentequations are v′ cos θ′ = v cos 40◦ andv′ sin θ′ = v sin 40◦ + V. We can eliminate θ′ bysquaring and adding (since sin2 + cos2 = 1):v′2(cos2 θ′ + sin2 θ′) = v2(cos2 40◦ + sin2 40◦)+2vV sin 40◦ + V 2, or v2 + 2vV sin 40◦ − v′2 + V 2 = 0.The positive root of this quadratic for v (appropriatefor a magnitude) is v = −V sin 40◦ +√

V 2 sin2 40◦ + v′2 − V 2. If the given values v′ =18.0 km/h and V = 5.60 km/h are substituted, we findthe speed relative to the ground is v = 13.9 km/h.Going back to the x component equation, wefind the heading θ′ = cos−1(v cos 40◦/v′) =

cos−1(13.9 cos40◦/18.0) = 53.8◦. (Equations similar tothese will be solved when collisions in two-dimensionsand the conservation of momentum are discussed inChapter 11.)

20 CHAPTER 3

Problem

68. A space shuttle orbits the Earth at 27,000 km/h,while at the equator Earth rotates at 1300 km/h.The two motions are in roughly the same direction(west to east), but the shuttle orbit is inclined at25◦ to the equator. What is the shuttle’s velocityrelative to scientists tracking it from the equator?

Solution

Let v and V be the velocities of the space shuttle andthe tracking station, respectively, relative to areference frame fixed at the center of the Earth. Theshuttle’s velocity relative to the station is v′ =v −V.In general, this is a problem in spherical trigonometry,in which v′ depends on the relative positions of theshuttle and the station. Suppose that the shuttle ispassing directly over the tracking station (as at pointA). Then the angle between v and V is 25◦, whilemagnitudes are given, so v′ can be found from the lawof cosines and the law of sines:

v′ =√

(1300)2 + (27, 000)2 − 2(1300)(27, 000) cos 25◦

= 26.5×103 km/h,

θ = 180◦ − sin−1(27 sin 25◦/26.5) = 25.5◦.


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