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Unformatted text preview: CHAPTER 3 THE VECTOR DESCRIPTION OF
MOTION ActivPhysics can help with these problems: arablem Activity 4.1 3. A migrating whale follows the west coast of Mexico
and North America toward its summer home in
Section 32: Vector Arlth metlc Alaska. It ﬁrst travels 360 km due northwest to just
off the coast of Northern California and then turns
Problem due north and travels 400 km toward its
You walk west 220 m, then north 150 In. What are destination. Determine graphically the magnitude
the magnitude and direction of your displacement and direction 0f its displacement vector ~
vector? '
. Solution
SO'Ut'O" We can ﬁnd the magnitude and direction of the vector
The triangle formed by the two displacement vectors sum of the two displacements either using geometry
and their sum is a right triangle, so the Pythagorean and a di ram, or by adding vector components.
Theorem gives the magnitude C = \/A5 + B2 = From the 3W 0f cosines: «(220 m)2 + (150 m) = 266 m, and the basic deﬁ C = m nition of the tangent gives ,6 = tan—1050 m/ 220 m)
= 343°. The direction of C can be speciﬁed as = (360 km? + (400 km)2 — 2(360 km)(400 km) ”5135‘, 34.3° N of W, or 55.7" W of N, or by the azimuth = 702 km'
304.3° (CW from N), etc. From the law of Sines: C’/ sin7 = B/ sin ,6, or
_ B sin’y 400 m
.. ' 1 _____  —l  o = . o.
ﬁ—sm ( C ) srn [(702 m) sm135 ] 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 the :raxis (east) in the illustration. Problem 1 Solution. Problem 2. An ion in a mass spectrometer (a device that sorts
atomicsize particles) follows a semicircular path of
radius 15.2 cm. What are (a) the distance it travels
and (b) the magnitude of its displacement? Solution
(a) The length of the semicircle is %(27rr) = 7rr = 1r(15.2 cm) = 47.8 cm. (b) The magnitude of the . . . _ .
displacement vector, from the start of the semicircle to In a coordinate system w} t1? f'akxx': 8381; and 3’ axrs
its end, is just a diameter, or 205.2 cm) = 30. 4 cm‘ north, the ﬁrst displacement Is 360 (1 cos 135 + Problem 3 Solution. CHAPTER 3 31 135°) and the second simply 400 km 3. Their sum A, is 180° + tan"1 (B/A) = 180° + 53.1° 2 233°
{7255'1' +2553 +4003) km = (~255i+655j) km, (other angles could have been chosen). C could also is the total displacement. Its magnitude is be determined algebraically from components, with
255)2 + (655)2 km/702 km and its direction zaxis parallel to A and y—axis parallel to B. Then ed CCW from the wards) is 0, = cos‘1 A = (3.0 m)i, B = (4.0 m)j, and C: ——(A+B) = .
km/702 km) == 111°, as above. (Note that since (—3.0 m)i + (—4.0 m)j = Czi + Cyj. The magnitude of 0 and 0:, < 0, 9,, is in the second quadrant.) C is (0?, + C: = /_______(3.0 m)2 + (——4.0 m—"i, and the angle that C makes with the mantis is cos—1(C'z/C) =
cos"l(—3.0 m/5.0 in), which is in the third quadrant, city’s streets are laid out with its northsouth as culated above (Note the a
. . cal . . ngle of C could also
;;Hocks tw1ce as long as its eastwest blocks. You b iﬁ _ 2 h _ .
mil: 8 blocks east and 3 blocks north. Determine e spec ed as 1 70’ or CW from t e a; 5') ) the total distance you’ve walked and (b) the Problem
“magnitude of your displacement vector. Express in cuts of eastwest blocks. 6. Two vectors A and B have the same magnitude, A, and are at right angles. Find the magnitude of the
vectors (a) A+ZB, (b) 3A — B. t blocks east is 8 units, but three blocks north Solution x 2 = 6 umts, so the total distance walked 13 Any two vectors (V1 and V2) and their sum (V3 = units (b) The magnitude of your displacement . . .
, . . . . . V1 +V2) form a triangle. If V1 is perpendicular to
.. is the hypotenuse of a right triangle, With Sides V2, the triangle is a right tri 1e, and the :‘  « .  F"T"‘—"2_
'3' units and 6 units, ‘ts length Is (8 u) + (6 u) — magnitudes are related by the Pythagorean Theorem,
' V3—  «(I 01:— +62 (a) For V1: A and V2‘ — ’23]: 2A,V3= 5A (b)ForV1—3AandV2—[——Bl
, ' v3~ — x/l—OA. War A has magnitude 3.0 m and points to the right; vector B has magnitude 4.0 in and points Problem yertically upward. Find the magnitude and ”E , 'on of a vector C such that A +3 + 0:0. Vectors A and B in Fig. 322 have the same magnitude, A. Find the magnitude and direction of
(a) A —B and (b) A+B. FIGURE 322 Problem 7 Solution. Solution The vectors A and B in Fig. 322 form two sides of a
parallelogram, in which A — B and A +B are the
diagonals, as shown. Since the magnitudes of A and B
are equal, the parallelogram is a rhombus, and the
diagonals are perpendicular (the converse of this is ion
, . also true; see Problem 60). Then A + B is along the
vectors A, B, and C form a 3—4—5 right triangle, perpendicular bisector of the base A _ B of an "shown in the sketch. Therefore C’ = 5 m and the . . . . .
' . ’ . ’ . isosceles triangle, and Vice versa. Usmg the given
non of C, measured CCW from the direction of les, we ﬁn d the magnitudes (a) l A __ B] = Problem 5 Solution. 32 CHAPTER 3 2A sin 20° = 0.684A, and (b) M + Bl = 2A cos 20° = 1.88A. In Fig. 322, (a) A—B is up, and (b) A+B is to the right, but the directions could be speciﬁed
relative to A, B, or some other coordinate system.
(This problem can also be readily solved with
components and unit vectors. Figure 322 suggests a
coordinate system with zaxis to the right and yaxis
up, as shown. Then A = A(i‘ cos 20° + 3 sin 20°) and
B=A(icos20°—— jsin 20°), from which A:l:B are
easily obtained.) Problem
8. Vector A has magnitude 1.0 m and points at 35° clockwise from the xaxis. Vector B has magnitude 1.8 In. What angle should B make with the z—axis
in order that A + B be purely vertical? Solution The vectors A, B, and A+B form a triangle, but
given only that B = 1.811 and A +B are
perpendicular to the z—axis, two are possible. In one,
the angle opposite side B is ﬁ— — 125°, in the other
[3— — 55°, as sketched. For either, the law of Sines gives
= sin —1(ﬂ§'ﬂ) = Sin—1(s_i_____n1 125°)_ = sm_1(°i°+°°)— __
27.1°. Therefore, B makes an angle of 3:62. 9° with
the negative :raxis or :i: 117° with the z—axis (look at
suitable right triangles in the sketch). (This result can
also be obtained from components: since A + B is
vertical, A;: + B;I = 0, or B, = Bcosﬂa, = —A1 =
—Acos35°, so 9; = cos—1(—A cos35°/B) =
cos‘l(— cos 35°/1.8) = :l:117°, where 3,, could be
positive or negative.) * Problem 8 Solution. Problem Three vectors A, B, and C have the same
magnitude L and form an equilateral triangle, as
shown in Fig. 323. Find the magnitude and
direction of the vectors (a) A +B, (b) A — B,
(c) A+B+C, (d) A+B—C. 4‘3 mum; a23 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 vector
equation in part (c) has solution A+B: —C. Thus
[A +Bl = l —Cl = [CI = 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 = —ZC, and
so has magnitude 2L and direction opposite to C. (b) Finally, (A—B)=A(—A—C)=2A+C, so
these vectors form a 30°  60° — 90° right triangle, as
shown. Then lA—Bl=f [Cl f L and its
direction is 90° CCW from C. Of course, this problem can be solved readily using
components and a coordinate system with x—axis
parallel to C and yaxis perpendicular, as shown
superposed on Fig 323. Then A: lAl(‘i cos 120°+
jsin 120°)— — L(—i‘ + Jim/2, B :L(—i — fl)/2 and
C: Li. It is a simple matter to ﬁnd
(a) A+B = —Li, 0)) A B: ﬁLj, (c) A+B+C=0, and (d) A+B—C = —2Li.
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 of
north. Your ﬂight, however, stops at Charleston, South Carolina, on the way to Atlanta.
Charleston is 510 km from Orlando, in a direction 9.3° east of north. Use graphical techniques to ﬁnd the magnitude and direction of the
CharlestontOAtlanta leg of your ﬂight. Solution The displacements between the three cities are as
shown in the diagram. Graphical techniques can be
conﬁrmed by calculations using components and unit
vectors, or trigonometry. Evidently, the displacement
from Charleston to Atlanta is C = A  B = (660 km)(‘i cos(90° + 29°) +j sin 119°)  (510 km)(‘icos(90° — 9.3°) +3 sin 807°) = (—320‘i+ 5773 — 82.4‘i — 503: km = —420i + 74.05) km.
Its magnitude is (—402) + (74.0)2 km = 409 km and its direction is 9 = tan‘1(74.0/(——402)) = 170° CCW from the z—axis, or 9 — 90° = 79.6° CCW
from the yaxis. Fbr those solving this problem
from the laws of casinos and Sines «(660)§ + (510)2 — 2(660)(510) cos 38.3° = 409,
sin‘1(660 sin 38.3°/409) = 91.1°, and 90° — (91.1° — 807°) = 79.6°. Problem 10 Solution. Section 3—3: Coordinate Systems, Vector
Components, and Unit Vectors P oblem Vector V represents a displacement of 120 km at
29° counterclockwise from the maxis. Write V in
unit vector notation. Solution Take the yaxis 90? CCW from the zaxis, as in
Figs. 310 and 11. Then V: V,’i+ 14,3: V(icos9¢ +
jcosﬂy) = V(‘icos93 +jsin9,) = (120 km)x CHAPTER 3 33 (‘1‘ cos 29° + 3 sin 29°) 2: (105i + 58.23) km. (Note: The
component of a vector along an axis is deﬁned in terms
of the cosine of the angle it makes with that axis. In
two dimensions, 0,, = (9..., — 90°], and cos 0,, = sin 03.) Problem Find the magnitude of the vector 34i+ 13j m, and
determine the angle it makes with the :1: axis. Solution
Equation 31 gives the magnitude of a vector in terms of its Cartesian components: A = 1IA"; + A3, = (/(34 m)2 + (13 m)2 = 36.4 in. Equation 3—3 gives the
angle A makes with the m—axis: 9, = cos—1(Az/A) =
cos—1(34 m/36.4 m) = 20.9°. (Since both AI and A”
are positive, we know that 9... is in the ﬁrst quadrant.)
Of course, Equation 32 could also have been used here, but Equation 33 holds in three dimensions,
whereas Equation 32 does not. Problem 13. Express each of the vectors of Fig. 3—24 in unit
vector notation, with the asaxis horizontally to the
right and the yaxis vertically upward. FIGURE 3—24 Problems 13, 19, and 20. Solution
Take the z~axis to the right and the y—axis 90°
counterclockwise from it. Then
A 10(icos 35° +jsin 35°) = 8.19i'+ 5.743
B 6(i cos 235° +3 sin 235°) 2 — 3.443 —— 4.913
C = 8(icos 115° +3 sin 115°) = — 3383+ 7.253 Problem 14. (a) What is the magnitude of i+j? (b) What
angle does it make with the :raxis? Solution The same reasoning as in Problem 12 shows that
(a) the magnitude ofi+j is «12 + 15 = «i, and that . ..
a;
3?,
2'»
it; 34 CHAPTER 3 (b) the angle it makes CCW from the z—axis is a. = cos1(1Ni) = 45°. (Thus, 71:6 +3)/\/2' is a
unit vector midway between the :z: and ydirections;
see Problem 63.) Problem
15. Repeat Problem 3, using unit vector notation. Solution
See solution to Problem 3. Problem 16. Express the vectors of Fig. 323 in unit vector
notation, taking the z—axis horizontal and the
yaxis vertical. Each vector has length A. Solution
See the solution to Problem 9. Problem Let A: 153— 403 and B=3lj+18lc Find a
vector 0 such that A+B+C=O. Solution (1: art—3:. (15‘i—40j) — (313+18E)= —15‘i+ 93 — 18k. (Since A and B are speciﬁed in
terms of unit vectors, this form is also appropriate for
C.) oblem
18. A proton travels in a. circular path around the 2.0—km—diarneter accelerator at Fermilab, near
Chicago. Write expressions for the proton’s displacement vector from the center of the circle
when it is at (a) 0°; (b) 30"; (c) 45°; ((21) 60°;
(e) 90°; (f) 180°, as measured counterclockwise from the z—axis. _
4‘? Problem 18 Solution. Solution , The position of any point on the circumference of a circle, relative to the center, is r = xi + yj = r(‘i cos 9 +
j sin 0), where 9 is the angle measured CCW from the
z—axis. At Fermilab, r = 1 km (half the diameter), so
for the values of 9 given, (a) r: (1 km)(icos 0° +.
jsinO") =i km; (b) r = (1 km)('icos30° + imam) = §(\/§i+j) km; (c) r: (i+j‘) km/ﬁ; (d) r= %(i+\/53) km; (6) r=.i km; and (f) r: —i 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 — 2.20. Solution The components of A,B, and C are given in the
solution to Problem 13. A component of the sum (or
difference) is the sum (or difference) of the
components; the component of a scalar multiple is the
scalar multiple of the component. For example, the
z—component of A + 1.53 — 2.20 is A: + 1.533—
2205. Therefore (a) A + B +C = 1.37'i +8.07j =
8.19 (icos 80.4° +jsin 804") (b) A — B + C =
8.25‘i+ 17.93 = 19.7(‘icos 65.3° +jsin 653°) (c) A +
1.5B — 2.2C = 10.5‘i — 17.63 = 20.5(‘i cos 301° + 5 sin 301°) The ﬁrst form is the vector in components,
the second gives the magnitude (V532 + 3/2) and
direction 9:: = tan“1(y/:z:). Problem 20. Find a vector D such that A+B+C+D=0 for
the vectors of Fig. 3—24. Solution The vector D = —(A + B + C) is the negative of the
vector given in part (a) of the preceeding solution, So
D= — 1.37i— 8.07j=8.19(icos 260° +jsin260°).
(Note that 180° + 0 is the opposite direction to 9, in
the x—y plane.) Problem 21. You’re trying to reach a pond that lies 3.5 km to
the northeast of your starting point. You ﬁrst
follow a logging road that runs east for 0.80 km.
Then you follow a deer trail heading northeast for
2.1 km. From there you bushwack straight to the
pond. Describe your ﬁnal displacement vector,
(3) in unit vector notation; and (b) as a
magnitude and compass direction. Solution The desired total displacement, R: 3.5 km NE, is
the sum of three displacements, R1 = 0.80 kmE,
R2=2.1 km NE, and R3=R—R1 —R2 to be found.
(a) With z—axis E and y—axis N, R1 =0.80km, R2 = (2.1 km)(icos45° + jsin 45°) 2 1.48i+ 1.483 km,
and R: 2.47i+2.47j km. Therefore R3 = (2.47 — 0.80 — 1.48)i+ (2.47 — 1.48)j km =
0.190i+0.9903 km=(1.01 km)(‘icos79.1° + jsin 791°). (See Equations 31 and 32 or the solution
to Problem 19 for the last step.) ’ oblem
a or the vectors of Fig. 323, ﬁnd two values for the
scalar c such that A + cB has magnitude 2.18L. Solution The vectors A and c B form two sides of a
triangle with included angle of 60° if c > 0 and
120° if c < 0, as shown. The magnitudes of the
sides, which are given as lAl=L, IcBI = chL, and
IA +cBl = 2.18L, are related by the law of
cosines. Since IclL cos 60° = écL for c > 0 is the
same as lclL cos 120° = (—c)L(—%) for c < 0, the
law of cosines for both possibilities is (2.18)2L2 =
L2 + 02L2 —~ 0L2, or c2 — c— 3.75 = 0. The
quadratic formula ives the two solutions as c = [1 i 1 + 4(3.75)]/2 = 2.50 or —1.50,
respectively. Problem 22 Solution. CHAPTER 3 35 Problem 23. In Fig. 3—14 the angle between :c and f—axes ' is 21°, the angle between the vector A and the
zaxis is 54°, and A’s magnitude is 10 units.
(a) Find the components of A in both coordinate
systems shown. (b) Verify that the magnitude of
A, computed using Equation 3—1, is the same in
both coordinate systems. Solution (a) In the x—y system, A: =Acos 0 = 10 cos 54° = 5.88
and Au = 10 sin 54° = 8.09. In the :L"y’ system, 0’ = 54° — 21° = 33° so A; = 10cos 33° = 8.39 and
A; = 10 sin 33° = 5.45 (b) Direct calculation shows
that ‘/(5.88)2 + (8.09 7 = ,/(8.39)2 + (5.45)2 = 10,
which reﬂects the fact that sin2 + cos2 = 1. (The
mathematical deﬁnition of a twodimensional vector is
a pair of numbers (V3, V”) which transform like the
position vector (I, y) when the coordinate axes are
rotated.) Problem 24. Vector A is 10 units long and points 30° counter
clockwise (CCW) from horizontal. What are the a:
and 31 components on a coordinate system (a) with
the z—axis horizontal and the yaxis vertical; (b) with the x—axis at 45° CCW from horizontal
and the y—axis 45° CCW from vertical; and (c) with the xaxis at 30° CCW from horizontal
and the y—axis 90° CCW from the x—axis? Solution The component of a vector along any direction equals
the magnitude of the vector times the cosine of the
angle (3 180°). Thus, (a) A, = 10 cos 30° = 8.66, A1, = 10 c0560° = 5.00 (b) A; = 10cos 15° = 9.66,A; = 10 cos 105° = —2.59
(c) Ag = 10 cos 0° = 10, A: = 10cos 90° = 0. Problem 24 Solution. 36 CHAPTER 3 Problem @ Express the sum of the unit vectors ‘i, j, and k in
unit vector notation, and determine its magnitude. Solution
r=i+j+k; r=\/12+12+12=~/§.
Problem 26. A mountain expedition starts a base camp at an
altitude of 5500 in. Four climbers then establish
an advance camp at an altitude of 7400 In; the
advance camp is southeast of the base camp, at a.
horizontal distance of 8.2 km. From the advance
camp, two climbers head directly north to an
8900m summit, a horizontal distance of 2.1 km.
Using a coordinate system with the :caxis
eastward, the y—axis northward, and the z—axis upward, and with origin at the base camp, express the positions of the advance camp and summit in
unit vector notation, and determine the
straightline distance from base camp to summit. Solution The vector from base to advance camps is A: (8.2 km)(icos45° —jsin45°) + (7.4 km— 5.5 km)k= (5.80i — 5.803 + more km, and that
from advance camp to summit B = (2.13 + 1.5k) km.
Therefore, the vector from base camp to summit is
C=A+B= (5.80i— 5303+ 1.90k) km+ (215+ 1.51?) km + (5.80i — 3.70j+3.40k) km. The
straightline distance is C = «(5.8)? + (—3.7')2 + (3.4)2km = 7.67 km. Problem 26 Solution. Problem 27. In Fig. 3—15, suppose that vectors A and C both
make 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. (8.) Express the
displacement vector Ar from start to summit in
each of the coordinate systems shown, and (b) determine its length. Solution Using the 123} system'in Fig. 3—15, we have 0,, 9c = 30°, and 93 = 60°. Then A=A(icos 0A +
jsin 9,4) = (2.3 km)(‘icos 30° +jsin 30°) = (1993+ 1.1sj)km, B: (1 km)(icos 60° +jsin 60°) =
(0.50il— 0.873) km, and C = (2.9 km)(i cos 30° + 3 sin 30°) = (2.51‘i + 1.453)km. Thus, Ar=A + B +0 = 5.oor+ 3.47:) km, and
Ar = (5.00)2 + (3.47)2 km = 6.09 km. A similar calculation in the 1’ y' system, with 0;, = 0’0 = 0 and
9;, = 30°, yields A = 2.3? km, B=(1km)x
(1" cos 30° +3, sin 30°) = (0.87!" + 0.503,) km, C = 2.91" km, Ar: 6.07? + 0.50)") km, and
(6.07)2 + (0.50)2 = 6.09, of course.
Section 3—4: Velocity and Acceleration Vectors Problem 6? An object is moving at 18 m/s at an angle of counterclockwise from the zaxis. What are the
:1: and y components of its velocity? Solution 1),; = (18 m/s) cos 220° = —l3.8 m/s. 11y = (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, then
turns east and goes 5.0 mi at 60 mi/h. Finally, it
goes southwest at 30 mi/h for 6.0 min. Draw a
vector diagram and determine (a) the car’s
displacement and (b) its average velocity for this
trip. Solution Take a coordinate system with z—axis east, y—axis
north, and origin at the starting point. The ﬁrst
segment of the trip can be represented by a
displacement vector in the y direction of length (40 mi/h)(10 min), or r1 = (20/3)? mi. For the second
segment, r2 = 5i mi. The time spent on this segment is
#2 = 5 mi/ (60 mi/h) = 5 min. The ﬁnal segment has
length (30 mi/h)(6 min). A unit vector in the
southwest direction is i cos 225° +jsin 225° = —— (i+j)/\/2, so r3 = — (3/\/2)(‘i+j) mi. These displacements and their sum are shown in the sketch. (a) The total displacement is rm —. r1 + r2 + 13 .—
[(20/3)j + 51 —(3/,/‘)(1+3)] mi: (2. 881‘ +4. 553) mi
(b) The total time is 10 min + 5 min+6min= 21 min,
so the average velocity for the trip is {1 = not/trot =
(2.88i + 4.553) mi/(21/60) h =(8.221+13.0j) mi/h.
(Note: Instead of unit vector notation, rm and ii
could be speciﬁed by their magnitudes (2.88)2 + (4.55)2 mi = 5.38 mi and 15.4 mi/h,
respectively, and common direction,
9 = tan1(4.55/2.88) = 57.7° N of E.) Problem 29 Solution. Problem I 30. A biologist studying the motion of bacteria notes a
bacterium at position r1— = 2. 21‘ +3 73— 1. 2k [.1 m (1 n m— .— 10’6 Am). After 6. 2 s the bacterium is at
r;— — 4 61‘ + 1. 9k 11 m. What is its average velocity? Express' in unit vector notation, and calculate the
magnitude. Solution V = (1‘2 — I‘ll/(t'z — t1)
= =.[(46i+1. 912) —(2 21+3 73—1. 212)];1 m/6.2 s
(.0 387i 0. 597j+0 500k)p m/s M: 1/ (0. 387)2 + (0. 597)2 + (0. 5)211 m/s = 0.869 )1 m/s. roblem
31. The Orlandoto—Atlanta ﬂight described in
Problem 10 takes 2.5 h. What is the average velocity? Express (a) as a magnitude and
direction, and (b) in unit vector notation with the :caxis east and the yaxis north. CHAPTER 3 37 Solution (b) The displacement from Orlando to Atlanta
calculated in Problem 10 was A: (~320i+577j) km
in a coordinate system with :raxis east and y—axis
north. If this trip took 2.5 h, the average velocity was
e=A/2.5 h = 128i+23lj) km/h. (a) This has
magnitude \/(~—128)2 + (231)2 km/ h = 264 km/h and
direction 0 = tan1(231/—128) = 119° (which was
given). Problem 32. The minute hand of a clock is 5.5 cm long. What
is the average velocity vector for the tip of the
hand during the interval from the hour to
20 minutes past the hour, expressed in a
coordinate system with the yaxis toward noon
and maxis toward 3 o’clock? Solution At the hour, the tip of the minute hand has position
r1 = (5.5 cm)j, while at 20 min past the hour, it has
position r2 = (5.5 cm)(i cos (— 30°) +3 sin ( — 30°)) =
(4.76i — 2.75j) cm. The average velocity is v = (r2 — r1)/20 min = (4.761 — 8.253) cm/20 min =
(0.238‘1‘ — 0.4133) cm/min. (“1‘
direction
of V >X ~30° \ Problem 32 Solution. Problem 33. A hotair balloon rises vertically 800 m over a.
period of 10 min, then drifts eastward 14 km in
27 min. Then the wind shifts, and the balloon
moves northeastward for 15 min, at a speed of
24 km/h. Finally, it drops vertically in 5 min until
it is 250 m above the ground. Express the
balloon’s average velocity in unit vector notation,
using a coordinate system with the m—axis
eastward, the y—axis northward, and the z—axis
upward. 38 CHAPTER 3 Solution The displacement for the ﬁrst segment of the balloon's
excursion is r1 = 0.8l} km, in the coordinate system
speciﬁed, and for the second segment 1'; = 141‘ km.
The third segment has length (24 km/h)(15 min) = 6 km in the northeast direction icos 45° +3 sin 45°, so r3 = (6 km)(i+5)/\/§ = (aural4.2.5) km.
Finally, the last segment’s displacement is r4 = (250 m — 800 m)1‘c= — 05512 km (this is a drop of
550 m from the preceding altitude). The total
displacement Ar = r1 + r2 + r3 + r}. = [(14+424liﬁ‘ 4,243+ (0.8 — 0.55)k] km: (18.2i+
4.24j+0.25k) km is accomplished in total time At = (10 +27 + 15 + 5)min = 0.950 h, sothe average
velocity is Ar/At = (19.2i+4.47j+0.263k) km/h. Problem 34. Figure 325 shows the path of a bug as it crawls
around a tabletop. Dots mark the position of the
bug at each second. Determine the average
velocity of the bug over the interval (a) from 1.0 s
to 2.0 s; (b) from 2.0 s to 4.0 s; (c) 0 to 6.0 s. FIGURE 325 Problem 34. Solution The position vectors for the bug, at each second of
time, can be read off Fig. 3—25, so the average velocity
for any interval can be calculated from Equation 3—5,
v = Ar/At. (a) Ta, = [r(2 s) — r(1 s)]/(2 s — 1 s)
=[(—3j mm)  (Sij) mm]/1 s = —(3i‘+23) mm/s.
(b) a: ir(4 s) — r(2 s)1/(4 s  2 s) =l( —i+ 0.53) mm (—3j mm)]/2 s = (0.5‘i+ 1.753) mm/s. (c) irc=lr(6 s) — r(0)]/6 s=i( —i+2s> mm—Ol/6 s=
(—0.167i + 0.3333) mm/s. Problem $3 An object’s position as a function of time is given
by r: 12ti+ (15t — 5.0t2)j 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 from t = 0 to t = 2.0 s? (c) What is its instantaneous
velocity at t = 2.0 s? Solution (8.) The object’s position is given as a function of time,
so when t = 2 5, this is r(2 s) = (12 m/s)(2 s)i+
[(15 m/s)(2 s) (5.0 m/s2)(2 s)2]j= 24r+ 103 m,
where we explicitly displayed the units of the
coefﬁcients in the intermediate step. (b) Since r(0) = 0, the average velocity for this interval is
V7: [r(2 s) — r(0)]/(2 s — 0) = 12i+5j m/s. (c) The
instantaneous velocity at any time‘ is dr/dt = (12 m/s)l + [(15 m/s)  (5.0 m/sz)2t]j=v(t) (see
Appendix A2 for the derivative of t"), so when t = 2 s, v(23) = 12i— 53 m/s. Problem 36. A supersonic aircraft is traveling east at
2100 km/h. It then begins to turn southward,
emerging from the turn 2.5 min later heading due
south at 1800 km/h. What are the magnitude and
direction of its average acceleration during the
turn? Solution In a coordinate system with zaxis east and y—axis
north, the initial velocity of the airplane at the
beginning of its turn is V; = 2100i km/h, and the ﬁnal
velocity at the end is V2 = —1800j km/h. The average
acceleration (Equation 08) is 5: (v2 — v1)—:— (t2 — t1) = (—18003  2100i)(km/h)/2.5 min = —(3.89‘i + 3.333) m/sz, with magnitude ‘/(—3.89)5 + (—3.33)2 m = 5.12 m/s2 and direction 0 = tan"(——3.33/3.89) = 221° (in the third quadrant,
nearly southwest). Problem 37. A car, initially going eastward, rounds a 90° bend
and ends up heading southward. If the
speedometer reading remains constant, what is the
direction of the car’s average acceleration vector? Solution Since the speed is constant, the change in velocity for
the 90° turn is Av = —1)j—(vi) = —v(i+j), where i is
east and 3 is north. The direction of the average
acceleration is the same as that of Av, which is parallel to —(i+j) or southwest.
(9 = tan—1(—1/—1)= 225°.) 40 CHAPTER 3 (a) = (v — Vol /At = l—vul /At = (80 m/3.6 s)/3.9 s =
5.70 m/sz. In relation to the coordinate system
speciﬁed, a = (5.70 m/s2)(iees210° +3sin 210°) =
—(5.70 m/s”)(¢§i +3)/2 = —(4.93i+ 2.853) m/sz.
(Note that v0, the velocity at the start of the skid, is
not in the direction of the initial motion before the
skid.) Problem 41. An object undergoes acceleration of 2.3i+
3.63 rn/s2 over a 10—s interval. At the end of this
time, its velocity is 33i+ 155 m/s. (8) What was
its velocity at the beginning of the 10—3 interval?
(b) By how much did its speed change? ((2) By
how much did its direction change? (d) Show that
the speed change is not given by the magnitude of
the acceleration times the time. Why not? Solution
(a) v=vo +81, so vo=vat, or vo= (33i+15j) m/s— (2.3i+3.63 m s2 10s = _
(lO‘i— 213) 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 in
speed is Av = v — v0 = 13.0 m/s (we did not round off before subtracting). (c) 9 = tan—1(15/33) = 24.4° and
90 = ten1(—21/10) = 295° = 64.5° (positive angles
CCW, negative angles CW, from xaxis) so the
direction changed by A0 = 0 — 90 = 89.0°. (d) at =
«(2.3)2 + (3.6)2 m/sz(10 s) = 42.7 m/s at Av. The
diﬁ'erence between at = lv — Vol and Au = v — no can
be seen from the triangle inequality:
lvvolSlvVolsv+vo. Problem 41 Solution. Problem 42. An object’s position as a function of time is given
by 1'=(bt3 + ct)i+dt23+ (at + f)k, where b, c, d,
e, and f are constants. Determine the velocity and
acceleration as functions of time. Solution Differentiating each term using Equation 23, we ﬁnd
v=dr=dt= (3bt2 +c)i+2dtj+ ck, and a: dv/dt == 6bti = 2115. Problem The position of an object is given by 1': (ct — bt3)i + dtz“, with constants c = 6.7 m/s, b = 0.81 m/s3, and d = 4.5 m/sz. (a) Determine the object’s velocity at time t = 0. (b) How long
, does it take for the direction of motion to change by 90°? (c) By how much does the speed change during this time? Solution (a) v(t) = dr/dt = (c — 3bt2)i+ 2dt3, so v(0) = (6.7 m/s)‘i. (b) The direction of v(t) is an) = tan‘1(2dt/(c  36?». measured cow from
the xards, so 0(0) = 0°. 0(t) = 90° when the
argument of the arctan is 00, or t = ‘/ c/ 3b. Thus, t = (/(61 m/s)/3(0.81 m/s3) = 1.66 s. (The direction of motion was ——90° at t = —1.66 s.) (c) Since '03 = 0
when t = ‘/c/3b the speed at t = 1.66 s is 2d‘/c/3b = 14.9 m/s. The speed at t = 0 is
c = 6.7 m/s, so the change was 8.24 m/s. Problem 44. For the object of Problem 35, determine (a) the
average acceleration in the interval from t = 0 to
t = 2.0 s and (b) the instantaneous acceleration at
t = 2.0 5. Solution (:1) The average acceleration during the interval used
in Problem 35 is 5: [W2 s)  V7(0)]/ (2 s — 0) =
((12i— 53) m/s:— (12i+ 153) m/s]/(2 s) = — 103 m/s2
(see part (c) of the solution to Problem 35 for em).
(b) The instantaneous acceleration at any time is
5(t) =dv/dt: — (10 m/s2)j, which is constant,
including when t = 2 8. Section 35: Relative Motion Problem 45. A dog paces around the perimeter of a rectangular
barge that is headed up a river at 14 km/ h
relative to the riverbank. The current in the water
is at 3.0 km/h. If the dog walks at 4.0 km/h, what
are its speeds relative to (a) the shore and (b) the
water as it walks around the barge? Solution (a) Let S be a frame of reference ﬁxed on the shore,
with zaxis upstream, and let S" be a frame attached
to the barge. The velocity of 5" relative to S is V = (14 km/h)i. The velocity of the dog relative to
the shore is (from Equation 325) v=v’ +V, and its speed is v = (v’+V, where v’ = 4 km/h. When the
dog is walking upstream, v’ H V, v = (4 + 14) km/h =
18 km/h, and when walking downstream, —v’ H V,
and v = (14  4) km/h = 10 km/h. When v’ .L V, v: \/145 + 45 = 14.7 km/h. (In general, 122 = 0’2 + V2 + 2v’V cos9’, where 9’is the angle
between v’ and V in 5".) (b) Since the current ﬂows
downstream, according to Equation 325: (vei. of barge) __ (val. of barge) __ (vel. of water) rel. to water — rel. to shore rel. to shore
= 14i— (3‘i) km/h. Going through the same steps as in part (a), for a new
frame 5’ moving with the water, with a new relative
velocity V = (17 km/h)i, we find the speed of the dog
relative to the water to be (4 + 17) = 21 km/h, (17 — 4) = 13 km/h, and V42 + ii = 17.4 km/h, for
the corresponding segments of the barge’s perimeter. Problem 45 Solution. Problem 46. A jetliner with an airspeed of 1000 km/h sets out
on a 1500—km ﬂight due south. To maintain a
southward direction, however, the plane must be
pointed 15° west of south. If the ﬂight takes
100 min, what is the wind velocity? Solution Using the same reference frames as speciﬁed in
Example 37, we are given that v = (1500 km/100 min)(j) = 9003 km/h, and v’ = —(1000 km/h)(isin 15° +jcos15°). The wind
velocity is V = v — v’ = (—9003 + 2591‘ + 9663) =
(259i + 65.9j) km/h. Therefore, the wind speed is
V: V2592 + 6.5.9i =267 km/h, and the angle V
makes with the zaxis is (tan  1 65.9/259: 14.3°)
(N of E). (The wind direction, by convention, is the
direction facing the wind, in this case 14.3° S of W.) CHAPTER 3 41 o  V' ”143°
. 4!???
Problem 46 Solution. Problem 47. A spacecraft is launched toward Mars at the
instant Earth is moving in the +7: direction at its
orbital speed of 30 km/s, in the Sun’s frame of
reference. Initially the spacecraft is moving at
40 km/s relative to Earth, in the +3; direction. At the launch time, Mars is moving in the
—y direction at its orbital speed of 24 km/s. Find
the spacecraft’s velocity relative to Mars. Solution Equation 310 says that the velocity of the spacecraft
relative to Mars, VCM, equals the diﬁerence of the
velocities of each relative to the Sun, Vcs, —VM5. (Our
notation VCM means the velocity of C relative to M.)
vMs is given as —(24 km/s)j in the Sun’s reference
frame, the :ry coordinates in the problem. The Earth
has velocity was = (30 km/s)i in the Sun’s frame, and
V0}; = (40 km/s)j, where the yaxw in the Earth’s and
Sun’s frames are assumed to be parallel. A second
application of Equation 310 gives Vcs = vcE +VES =
(403 + 30$) km/s; therefore VCM = Vcs — vMs =
(30i=40j) km/s ~— (—243) km/s: (30i+643) km/s. Problem You wish to row straight across a 63mwide river. If you can row at a steady 1.3 m/s relative to the
water, and the river ﬂows at 0.57 m/s, (a) in what
direction should you head? (b) How long will it
take you to cross the river? Solution The current is perpendicular to the direction in which
you wish to cross, as shown in the sketch. V is the
current velocity (velocity of the water relative to the
ground), v is the velocity of the boat relative to the
ground, and v’ is the velocity of the boat relative to
the water. These three vectors satisfy Equation 310.
(a) Evidently, sine = V( / (v’l, or 9 = sm1(o.57/1.3) = 26.0°, which is your heading
upstream. (b) (v) = v’cos€ = (1.3 m/s) cos 260" = 42 CHAPTER 3 1.17 m/s is your speed across the river, so the crossing
time is t = r/v = 63 m/(1.17 m/s) = 53.9 s. a
V a
7’ V
Problem 48 Solution. Problem You’re on an airport “people mover,” a conveyor
belt going at 2.2 m/s through a level section of the
terminal. A button falls off your coat and drops
freely 1.6 m, hitting the belt 0.57 s later. What
are the magnitude and direction of the button’s
displacement and average velocity during its fall in
(a) the frame of reference of the “people mover”
and (b) the frame of reference of the airport
terminal? (c) As it falls, what is its acceleration in each frame of reference? Solution (c) Let .S' be the frame of reference of the airport and
S" the frame of reference of the conveyor belt. If the
velocity of 5" relative to .S’ is a constant, 2.2 m/s in the
zz’ direction, then the acceleration of gravity is the
same in both systems. (a) In S’, the initial velocity of
the button is zero, and it falls vertically downward. Its
displacement is simply Ay’ = «1.6 In, and its average
velocity is Ay’/At = —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 a: direction), and so the button
follows a projectile trajectory to be described in
Chapter 4. Here, we observe that while the button
falls vertically through a displacement Ay : Ay’ = —1.6 m, it also moves horizontally (in the
direction of the conveyor belt) through a, displacement An: = (2.2 m/s)(0.57 s) = 1.25 In. Its
net displacement in S is Ar = Ami + ij = (1251‘ — 1.603) In, so its average velocity is ‘7: Ar/At: (1.25i— 1.603) m/0.57 s = (2.20i— 2.83 m 5. These have magnitudes Ar =
(1.25)2 + (—1.60)2 m = 2.03 m and
Iv] = «(2.20)2 + (—2.81)2 m/s = 3.57 m/s, and the same direction 0 = tan1(1.60/1.25) = —51.9° from
the horizontal, or 38.1° from the downward vertical. Problem 50. An airplane with airspeed of 370 km/h ﬂies
perpendicularly across the jet stream. To achieve this ﬂight, the plane must be pointed into the jet stream at an angle of 32° from the perpendicular
direction of its ﬂight. What is the speed of the je stream?
'7" v
32"
a»
U
Problem 50 Solution. _
Solution The velocity of the jet stream, V (air relative to
ground, also called the “windspeed”), the velocity of
the airplane relative to the ground, v (the “ground
spwd”), and the velocity of the airplane relative to the
air, v’ (the “air speed”) form the sides of a triangle
given by Equation 310, v’ +V=v, as showu. We are
given that the triangle is a right triangle (v .L V),
that the angle between the airspeed and groundspeed
is 32°, and the hypotenuse (magnitude of airspeed) is
370 km /h. Rom trigonometry, the magnitude of the
windspeed is [VI = (370 km/h) sin 32° = 196 km/h. Paired Problems Problem 51. A rabbit scurries across a ﬁeld, going eastward
21.0 m. It then turns and darts southwestward for 8.50 m. Then it pops down a rabbit hole, 1.10 m
vertically downward. What is the magnitude of the displacement from its starting point? Solution In a coordinate system with xaxis east, yaxis north,
and zaxis up, the three displacements can be written
in terms of their components and the unit vectors: r1 = 21.0‘1‘ m, r2 = (8.50 m)(icos 225° +jsin 225°):
—(6.01)(i+j) m, and r3 = —1.10k m. (Note that
southwest is 180° + 45° CCW from east.) The total
displacement is the vector surn r; + r2 + r3 = {(21.0 — 6.01)‘i — 6.013 — 1.10k] m, and its magnitude is
the square root of the sum of its components, «(15.0)? + (—6.01)2 + (—1.10)2 m = 16.2 m. Problem 52. A cosmic ray plows into Earth’s upper
atmosphere, liberating a shower of subatomic
particles One of these particles moves downward
3.2 km, then undergoes a collision, after which it
moves 1.6 km at 27° northward of the vertical. It then undergoes another collision that sends it
moving horizontally eastward for 2.1 km before it
annihilates with another particle. What is the
magnitude of its overall displacement? Solution Take a standard coordinate system with zaxis east,
yaxis north, and zaxis up. The particle begins
moving downward (say along the z—axis) through a
displacement r1 = ——3.2f£ km. It is then deﬂected in
the y—z plane (northward of the vertical) by 27°, so r2 + (1.5 km)(jsin 27° — Ecos27°), and ﬁnally eastward through 113 = 2.1i km. Thus, r1 + r2 + r3 =
(2.1)2 + (1.6sin 27°)2 + (—3.2 — 1.6cos 27°)2 km = 5.13 km.
1} q
'2 Problem 52 Solution. , Problem Cg A car is heading into a turn at 85 km/h. It enters
the turn, slows to 55 km/h, and emerges 28 3 later
at 35° to its original direction, still moving at 55 km/h. What are (a) the magnitude and (b) the direction of its average acceleration, the
latter measured with respect to the car’s original
direction?» 'y Solution The initial and ﬁnal velocities have magnitudes of V 85 km/h and 55 km/h, respectively, and make an
angle of 35? as shown, where we chose the :raxis parallel to v. and the y—axis in the direction of the turn. The change in velocity is _ Av=vf — v; = (55 km/h)(icos35° +jsin35°)  _ (85 km/h)i = (—39.9‘1‘ + 31.55) km/h: (a) The
magnitude of the average acceleration is , a = IAvl /At = ‘,/(—39.9)2 + (31.5)2 (km/h)/28 s =
. 1.82 km/h/s = 0.505 m/s’. (b) The direction of “ 5 is the same as the direction of Av, or 0 = CHAPTER 3 43 tan‘1(31.5/  39.9) = 142°. (This problem could also
have been solved using the laws of cosines and Sines;
see the solution to the next problem.) 4'»?
‘U .u
A399,. Problem 53 Solution. Problem 54. The Galileo space probe was originally to be
launched directly toward its destination, Jupiter.
But the 1986 explosion of the space shuttle
Challenger led to a decision that Galileo’s
liquidfueled booster rocket was unsafe to ﬂy on
the shuttle. As a result, Galileo’s trajectory
became a complicated path through the inner
solar system, picking up speed through a socalled
“gravity assist” maneuver involving close
encounters with the planets Venus and Earth. The
ﬁrst Earth encounter occurred on December 8,
1990, when Galileo, outbound from Venus, passed
200 miles from Earth. Thirty days before the
encounter, Galileo was approaching Earth at
2.99x104 m/s. Thirty days after the encounter,
Galileo was moving at 54° to its preencounter
direction, at a speed of 3.51x104 m/s. What were
(a) the magnitude and (b) the direction of
Galileo’s average acceleration duringthis interval,
the latter measured with respect to its
preencounter direction? 351mg]? A7 6 2:29.? Ian/~— Problem 54 Solution. Solution The initial and final velocities for the 60 d interval are
given, as indicated in the diagram. Instead of using 44 CHAPTER 3 unit vectors, as in the previous problem, let’s use the laws of casinos and Sines. (3.) Av =
«(29.9)i + (35.1)2 — 2(29.9)(35.1) cos 54° kin/s =
29.9 km/s. Therefore a = Av /At (29.9 km/s)—:
(60x86,400 s) = 5.76 Him/$2. (b) o = 180° —
sin—1(35.1 sin54°/29.9) = 180° — 71.9° = 108°. Problem The sweepsecond hand of a clock is 3.1 cm long.
What are the magnitude of (a) the average
velocity and (b) the average acceleration of the
band’s tip over a 506 interval? (c) What is the
angle between the average velocity and
acceleration vectors? Solution There will be numerous occasions to use vector
components to analyze circular motion in later
chapters (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 s
interval is 0 = (5/60)(360°) = 30°. The position
vectors (from the center hub) of the tip at the
beginning and end of the interval, r1 and r2, form the
sides of an isosceles triangle whose base is the
magnitude of the displacement, l Ar l = 2 lrl sin %9 =
2(3.1 cm) sin(30°/2) = 1.60 cm, and whose base angle
is %(180° — 30°) = 75°. Thus, the average velocity has
magnitude I Ar l/At = 1.60 cm/5 s = 0.321 cm/s and
direction 180° — 75° 2 105°CW from n. (b) The
instantaneous speed of the tip of the secondth is a
constant and equal to the circumference divided by 60
s, or v = 21r(3.1 cm)/60 s = 0.325 cm/s. The direction
of the velocity of the tip is tangent to the
circumference, or perpendicular to the radius, in the Problem 55 Solution. direction of motion (CW). The angle between two
tangents is the same as the angle between the two
corresponding radii, so v1, V2 and Av form an ,
isoscch triangle similar to the one in part (a). Thus
[Av] = 2 M sin $0 = 2(0.325 cm/s) sin(% x 30°) =
0.168 cm/s. The magnitude of the average
acceleration is lAvl /At = (0.168 cm/s)/5 s =
3.36x10‘2 cm/sz, and its direction is 105° cw from
the direction of v1, or 195° CW from the direction of
n. (c) The angle between a and 17, from parts (a) and
(b), is 195°  105° = 90°. (Note: This is the geometry
used in Section 4.4 to discuss centripetal acceleration.) Problem 56. A proton in a cyclotron follows a circular path
23 cm in diameter, completing one revolution in
0.17 ,us. What are the magnitude of (a) the
average velocity and (b) the average acceleration
as the proton sweeps through onetwelfth of the
full circle? (c) What is the angle between the
average velocity and acceleration vectors? Solution Onetwelfth of a revolution is 30°, so the geometry of
this problem is the same as that for the previous one,
and the directions of 17, a and the angle between them
are the same as in Problem 55. The magnitudes,
however, are different. (a) [Ar] = 2 r sin %0 = (23 cm) sin15° = 5.95 cm and At = 0.17 [1.8/12 =
14.2 ns. Thus, M = [Ar] /At = 4.20 Min/s. (b) The
instantaneous constant speed is 7r(23 cm) / (0.17 us) =
4.25 Mm/s, so lAvI =2(4.25 Mm/s) sin 15° = 2.20 Mm/s. Then [a] = [Av] /At = 1.55x1014 m/sz. Problem 57. A ferryboat sails between two towns directly
opposite one another on a river. If the boat sails
at 15 km/h relative to the water, and if the
current ﬂows at 6.3 km/h, at what angle should
the boat head? Problem 57 Solution. Solution The velocity of the boat relative to the ground, v, is
perpendicular to the velocity of the water relative to ...
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 Spring '08
 WORMER
 Physics

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