chap3

CHAPTER 3

THE VECTOR DESCRIPTION OF MOTION
Problem
3. A migrating whale follows the west coast of Mexico and North America toward its summer home in Alaska. It first travels 360 km due northwest to just off the coast of Northern California and then turns due north and travels 400 km toward its destination. Determine graphically the magnitude and direction of its displacement vector.

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 are the magnitude and direction of your displacement vector?

Solution
We can find the magnitude and direction of the vector sum of the two displacements either using geometry and a diagram, or by adding vector components. From the law of cosines:
C = = A2 + B 2 − 2AB cos γ (360 km)2 + (400 km)2 − 2(360 km)(400 km) cos 135◦

Solution
The triangle formed by the two displacement vectors and their sum is a right triangle, so√ Pythagorean the Theorem gives the magnitude C = A2 + B 2 = (220 m)2 + (150 m)2 = 266 m, and the basic defi nition of the tangent gives β = tan−1 (150 m/220 m) = 34.3◦ . The direction of C can be specified as 34.3◦ N of W, or 55.7◦ W of N, or by the azimuth 304.3◦ (CW from N), etc.

= 702 km.

From the law of sines: C/ sin γ = B/ sin β, or β= sin−1 B sin γ = sin−1 C 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 the x-axis (east) in the illustration.

Problem 1 Solution.

Problem
2. An ion in a mass spectrometer (a device that sorts atomic-size 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
= πr = (a) The length of the semicircle is π(15.2 cm) = 47.8 cm. (b) The magnitude of the displacement vector, from the start of the semicircle to its end, is just a diameter, or 2(15.2 cm) = 30.4 cm.
1 2 (2πr)

Problem 3 Solution. In a coordinate system with x-axis east and y-axis north, the first displacement is 360 km (ˆ cos 135◦ + ı ˆ sin 135◦ ) and the second simply 400 km ˆ Their sum  . is (−255ˆ+ 255ˆ+ 400ˆ km = (−255ˆ+ 655ˆ km, ı  ) ı )

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CHAPTER 3 A = (3.0 m)ˆ B = (4.0 m)ˆ and C = −(A + B) = ı, , (−3.0 m)ˆ + (−4.0 m)ˆ = Cxˆ + Cyˆ The magnitude of ı  ı . C is C2 + C2 = (−3.0 m)2 + (−4.0 m)2 , and the x y 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 also be specified as −127◦, or CW from the x-axis.)

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 since Cy > 0 and Cx < 0, θx is in the second quadrant.)

Problem
4. A city’s streets are laid out with its north-south blocks twice as long as its east-west blocks. You walk 8 blocks east and 3 blocks north. Determine (a) the total distance you’ve walked and (b) the magnitude of your displacement vector. Express in units of east-west blocks.

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

Solution
(a) Eight blocks east is 8 units, but three blocks north is 3 × 2 = 6 units, so the total distance walked is 14 units. (b) The magnitude of your displacement vector is the hypotenuse of a right triangle, with sides of 8 units and 6 units; its length is (8 u)2 + (6 u)2 = 10 units.

Solution
Any two vectors (V1 and V2 ) and their sum (V3 = V1 + V2 ) form a triangle. If V1 is perpendicular to V2 , the triangle is a right triangle, and the magnitudes are related by the Pythagorean Theorem, 2 2 V3 = V√+ V2 . (a) For V1 = A and V2 = |2B| = 1 2A, V3 = 5A. (b) For V1 = 3A and V2 = | − B| = A, √ V3 = 10A.

Problem
5. Vector A has magnitude 3.0 m and points to the right; vector B has magnitude 4.0 m and points vertically upward. Find the magnitude and direction of a vector C such that A + B + C = 0.

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

figure 3-22 Problem 7 Solution.

Solution
Problem 5 Solution. The vectors A and B in Fig. 3-22 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 also true; see Problem 60). Then A + B is along the perpendicular bisector of the base A − B of an isosceles triangle, and vice versa. Using the given angles, 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 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

Solution
The vectors A, B, and C form a 3-4-5 right triangle, as shown in the sketch. Therefore, C = 5 m, and the direction of C, measured CCW from the direction of A, is 180◦ + tan−1 (B/A) = 180◦ + 53.1◦ = 233◦ (other angles could have been chosen). C could also be determined algebraically from components, with x-axis parallel to A and y-axis parallel to B. Then

CHAPTER 3 components and unit vectors. Figure 3-22 suggests a coordinate system with x-axis to the right and y-axis up, as shown. Then A = A(ˆ cos 20◦ + ˆ sin 20◦ ) and ı  B = A(ˆ cos 20◦ − ˆ sin 20◦ ), from which A ± B are ı  easily obtained.) (c) A + B + C, (d) A + B − C.

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Problem
8. Vector A has magnitude 1.0 m and points at 35◦ clock-wise from the x-axis. Vector B has magnitude 1.8 m. What angle should B make with the x-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 = 1.8A and A + B are perpendicular to the x-axis, two are possible. In one, the angle opposite side B is β = 125◦ , in the other β = 55◦ , as sketched. For either, the law of sines gives ◦ 125◦ α = sin−1 ( A sin β ) = sin−1 ( sin1.8 ) = sin−1 ( sin 55 ) = B 1.8 27.1◦. Therefore, B makes an angle of ± 62.9◦ with the negative x-axis or ± 117◦ with the x-axis (look at suitable right triangles in the sketch). (This result can also be obtained from components: since A + B is vertical, 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 be positive or negative.) 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 vector equation 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, 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 |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-axis parallel to C and y-axis perpendicular, as shown superposed on Fig. 3-23. Then A = |A|(ˆ√ 120◦ + ı cos √ ˆ sin 120◦ ) = L(−ˆ + 3ˆ  ı )/2, B =L(−ˆ − 3ˆ ı )/2, and C = Lˆ It is a simple matter to find ı. √ , (a) A + B = −Lˆ (b) A − B = 3Lˆ ı, (c) A + B + C = 0, and (d) A + B − C = − 2Lˆ ı. 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 Charleston-to-Atlanta leg of your flight.

Problem 8 Solution.

Problem
9. Three vectors A, B, and C have the same magnitude L and form an equilateral triangle, as shown in Fig. 3-23. Find the magnitude and direction of the vectors (a) A + B, (b) A − B,

Solution
The displacements between the three cities are as shown in the diagram. Graphical techniques can be

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CHAPTER 3

confirmed by calculations using components and unit vectors, or trigonometry. Evidently, the displacement from Charleston to Atlanta is C = A − B = (660 km)(ˆ cos(90◦ + 29◦ ) +ˆ sin 119◦) − ı  (510 km)(ˆ cos(90◦ − 9.3◦ ) +ˆ sin 80.7◦ ) = ı  (−320ˆ+ 577ˆ− 82.4ˆ− 503ˆ km = (−420ˆ+ 74.0ˆ km. ı  ı ) ı ) Its magnitude is (−402)2 + (74.0)2 km = 409 km and its direction is θ = tan−1 (74.0/(−402)) = 170◦ CCW from the x-axis, or θ − 90◦ = 79.6◦ CCW from the y-axis. For those solving this problem from the laws of cosines and sines (660)2 + (510)2 − 2(660)(510) cos 38.3◦ = 409, sin−1 (660 sin 38.3◦ /409) = 91.1◦ , and 90◦ − (91.1◦ − 80.7◦ ) = 79.6◦ .

Solution
Equation 3-1 gives the magnitude of a vector in terms of its Cartesian components: A = A2 + A2 = x y (34 m)2 + (13 m)2 = 36.4 m. Equation 3-3 gives the angle 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 used here, 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 unit vector notation, with the x-axis horizontally to the right and the y-axis vertically upward.

Problem 10 Solution.

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

Section 3-3: Coordinate Systems, Vector Components, and Unit Vectors Problem
11. Vector V represents a displacement of 120 km at 29◦ counterclockwise from the x-axis. Write V in unit vector notation.

Solution
Take the x-axis to the right and the y-axis 90◦ counterclockwise from it. Then A = 10(ˆcos 35◦ +ˆ sin 35◦ ) = 8.19ˆ+ 5.74ˆ ı  ı  B = 6(ˆ cos 235◦ +ˆ sin 235◦ ) = − 3.44ˆ − 4.91ˆ ı  ı  C = 8(ˆ cos 115◦ +ˆ sin 115◦ ) = − 338ˆ+ 7.25ˆ ı  ı 

Solution
Take the y-axis 90? CCW from the x-axis, as in Figs. 3-10 and 11. Then V = Vxˆ + Vyˆ = V (ˆ cos θx + ı  ı ˆ cos θy ) = V (ˆ cos θx +ˆ sin θx ) = (120 km)×  ı  (ˆ cos 29◦ +ˆ sin 29◦ ) = (105ˆ+ 58.2ˆ km. (Note: The ı  ı ) 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, θy = |θx − 90◦ |, and cos θy = sin θx .)

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

Solution
The same reasoning as in Problem 12 shows that √ √ (a) the magnitude of ˆ +ˆ is 12 + 12 = 2, and that ı  (b) the angle it makes CCW from the x-axis√ is √ ˆ ı )/ 2 is a θx = cos−1 (1/ 2) = 45◦ . (Thus, n = (ˆ + ˆ unit vector midway between the x- and y-directions; see Problem 63.)

Problem
12. Find the magnitude of the vector 34ˆ+ 13ˆ m, and ı  determine the angle it makes with the x axis.

Problem
15. Repeat Problem 3, using unit vector notation.

CHAPTER 3

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Solution
See solution to Problem 3.

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

√ √ 1 ˆ sin 30◦ ) = 2 ( 3ˆ + ˆ km; (c) r = (ˆ +ˆ km/ 2;  ı ) ı ) √ 1 (d) r = 2 (ˆ + 3ˆ km; (e) r =ˆ km; and ı )  (f) r = −ˆ km . ı

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
See the solution to Problem 9.

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 x-component of A + 1.5B − 2.2C is Ax + 1.5Bx − 2.2Cx . Therefore (a) A + B +C = 1.37ˆ+ 8.07ˆ = ı  8.19 (ˆcos 80.4◦ +ˆ sin 80.4◦ ) (b) A − B + C = ı  8.25ˆ+ 17.9ˆ = 19.7(ˆcos 65.3◦ + ˆ sin 65.3◦ ) (c) A + ı  ı  1.5B − 2.2C = 10.5ˆ− 17.6ˆ = 20.5(ˆcos 301◦ + ı  ı ˆ sin 301◦ ) The first form is the vector in components,  the second gives the magnitude ( x2 + y 2 ) and direction θx = tan−1 (y/x).

Problem
ˆ 17. Let A = 15ˆ− 40ˆ and B = 31ˆ + 18k. Find a ı   vector C such that A + B + C = 0.

Solution
ˆ C = − A − B = −(15ˆ− 40ˆ − (31ˆ + 18k) = ı )  ˆ (Since A and B are specified in −15ˆ+ 9ˆ − 18k. ı  terms of unit vectors, this form is also appropriate for C.)

Problem
18. A proton travels in a circular path around the 2.0-km-diameter 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◦ ; (d) 60◦ ; (e) 90◦ ; (f) 180◦ , as measured counterclockwise from the x-axis.

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 = − 1.37ˆ− 8.07ˆ= 8.19(ˆ cos 260◦ +ˆ sin 260◦ ). ı  ı  (Note that 180◦ + θ is the opposite direction to θ, 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, (a) in unit vector notation; and (b) as a magnitude and compass direction.

Problem 18 Solution.

Solution
The position of any point on the circumference of a circle, relative to the center, is r = xˆ + yˆ = r(ˆ cos θ + ı  ı ˆ sin θ), where θ is the angle measured CCW from the  x-axis. At Fermilab, r = 1 km (half the diameter), so for the values of θ given, (a) r = (1 km)(ˆ cos 0◦ + ı ˆ sin 0◦ ) =ˆ km; (b) r = (1 km)(ˆ cos 30◦ +  ı ı

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 x-axis E and y-axis N, R1 = 0.80km, R2 = (2.1 km)(ˆ cos 45◦ + ˆ sin 45◦ ) = 1.48ˆ+ 1.48ˆ km, ı  ı  and R = 2.47ˆ+ 2.47ˆ km. Therefore R3 = ı 

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CHAPTER 3 θ′ = 54◦ − 21◦ = 33◦ so A′ = 10 cos 33◦ = 8.39 and x A′ = 10 sin 33◦ = 5.45 (b) Direct calculation shows y that (5.88)2 + (8.09)2 = (8.39)2 + (5.45)2 = 10, which reflects the fact that sin2 + cos2 = 1. (The mathematical definition of a two-dimensional vector is a pair of numbers (Vx , Vy ) which transform like the position vector (x, y) when the coordinate axes are rotated.)

(2.47 − 0.80 − 1.48)ˆ+ (2.47 − 1.48)ˆ km = ı  0.190ˆ+ 0.990ˆ km = (1.01 km)(ˆ cos 79.1◦ + ı  ı ˆ sin 79.1◦ ). (See Equations 3-1 and 3-2 or the solution  to Problem 19 for the last step.)

Problem
22. For the vectors of Fig. 3-23, 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 |A| = L, |cB| = |c|L, and |A + cB| = 2.18L, are related by the law of 1 cosines. Since |c|L cos 60◦ = 2 cL for c > 0 is the 1 ◦ same as |c|L cos 120 = (−c)L(− 2 ) for c < 0, the law of cosines for both possibilities is (2.18)2 L2 = L2 + c2 L2 − cL2 , or c2 − c − 3.75 = 0. The quadratic formula gives the two solutions as c = [1 ± 1 + 4(3.75)]/2 = 2.50 or −1.50, respectively.

Problem
24. Vector A is 10 units long and points 30◦ counterclockwise (CCW) from horizontal. What are the x and y components on a coordinate system (a) with the x-axis horizontal and the y-axis vertical; (b) with the x-axis at 45◦ CCW from horizontal and the y-axis 45◦ CCW from vertical; and (c) with the x-axis 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 (≤ 180◦). Thus, (a) Ax = 10 cos 30◦ = 8.66, Ay = 10 cos 60◦ = 5.00 (b) A′ = 10 cos 15◦ = 9.66, A′ = 10 cos 105◦ = −2.59 x y (c) A′′ = 10 cos 0◦ = 10, A′′ = 10 cos 90◦ = 0. y x

Problem 24 Solution.

Problem 22 Solution.

Problem
ˆ 25. Express the sum of the unit vectors ˆ ˆ and k in ı, , unit vector notation, and determine its magnitude.

Problem
23. In Fig. 3-14 the angle between x- and x′ -axes is 21◦ , the angle between the vector A and the x-axis 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
r =ˆ +ˆ + k; |r| = ı  ˆ 12 + 12 + 12 = √ 3.

Problem
26. A mountain expedition starts a base camp at an altitude of 5500 m. Four climbers then establish an advance camp at an altitude of 7400 m; 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 8900-m summit, a horizontal distance of 2.1 km.

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

CHAPTER 3 Using a coordinate system with the x-axis 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 straight-line distance from base camp to summit.

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Section 3-4: Velocity and Acceleration Vectors Problem
28. An object is moving at 18 m/s at an angle of counterclockwise from the x-axis. What are the x and y components of its velocity?

Solution
The vector from base to advance camps is A = (8.2 km)(ˆ cos 45◦ −ˆ sin 45◦ ) + (7.4 km − ı  ˆ ˆ 5.5 km)k = (5.80ˆ− 5.80ˆ + 1.90k) km, and that ı  ˆ from advance camp to summit B = (2.1ˆ + 1.5k) km.  Therefore, the vector from base camp to summit is ˆ C = A + B = (5.80ˆ − 5.80ˆ+ 1.90k) km + ı  ˆ km + (5.80ˆ − 3.70ˆ + 3.40k) km. The ˆ (2.1ˆ + 1.5k)  ı  straight-line distance is C = (5.8)2 + (−3.7)2 + (3.4)2 km = 7.67 km.

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, 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 x-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)ˆ mi. For the second  segment, r2 = 5ˆ mi. The time spent on this segment is ı t2 = 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 ˆ cos 225◦ +ˆ sin 225◦ = ı  √ − (ˆ + ˆ ı )/ 2, so r 3 = − (3/ 2)(ˆ +ˆ mi. These ı ) displacements and their sum are shown in the sketch. (a) The total displacement is rtot = r1 + r2 + r3 = √ ı )] ı ) [(20/3)ˆ + 5ˆ − (3/ 2)(ˆ +ˆ mi = (2.88ˆ + 4.55ˆ mi.  ı (b) The total time is 10 min + 5 min +6 min = 21 min, so the average velocity for the trip is v = rtot /ttot = ¯ (2.88ˆ + 4.55ˆ mi/(21/60) h = (8.22ˆ+ 13.0ˆ mi/h. ı ) ı ) (Note: Instead of unit vector notation, rtot and v ¯ 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, θ = tan−1 (4.55/2.88) = 57.7◦ N of E.)

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. (a) Express the displacement vector ∆r from start to summit in each 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(ˆ cos θA + ı ˆ sin θA ) = (2.3 km)(ˆ cos 30◦ +ˆ sin 30◦ ) =  ı  (1.99ˆ+ 1.15ˆ ı )km, B = (1 km)(ˆ cos 60◦ +ˆ sin 60◦ ) = ı  (0.50ˆ+ 0.87ˆ km, and C = (2.9 km)(ˆ cos 30◦ + ı ) ı ˆ sin 30◦ ) = (2.51ˆ+ 1.45ˆ  ı )km. Thus, ∆r = A + B + C = (5.00ˆ+ 3.47ˆ km, and ı ) ∆r = (5.00)2 + (3.47)2 km = 6.09 km. A similar ′ ′ calculation in the x′ -y ′ system, with θA = θC = 0 and ′ ◦ ′ θB = 30 , yields A = 2.3ˆ km, B = (1 km)× ı (ˆ′ cos 30◦ +ˆ′ sin 30◦ ) = (0.87ˆ′ + 0.50ˆ′ ) km, C = ı  ı  2.9ˆ′ km, ∆r = (6.07ˆ′ + 0.50ˆ′ ) km, and ı ı  (6.07)2 + (0.50)2 = 6.09, of course.

Problem
30. A biologist studying the motion of bacteria notes a ˆ bacterium at position r1 = 2.2ˆ+ 3.7ˆ − 1.2k µ m ı  −6 (1 µ m = 10 m). After 6.2 s the bacterium is at ˆ r2 = 4.6ˆ + 1.9k µ m. What is its average velocity? ı Express in unit vector notation, and calculate the magnitude.

Solution
v = (r2 − r1 )/(t2 − t1 )

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CHAPTER 3 ¯ (4.76ˆ − 2.75ˆ cm. The average velocity is v = ı ) (r2 − r1 )/20 min = (4.76ˆ − 8.25ˆ cm/20 min = ı ) (0.238ˆ− 0.413ˆ cm/min. ı )

Problem 29 Solution. Problem 32 Solution. ˆ ˆ = [(4.6ˆ + 1.9k) − (2.2ˆ+ 3.7ˆ − 1.2k)]µ m/6.2 s ı ı  ˆ = (0.387ˆ− 0.597ˆ+ 0.500k)µ m/s ı  |v| = (0.387)2 + (−0.597)2 + (0.5)2 µ m/s = 0.869 µ m/s.

Problem
33. A hot-air 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 x-axis eastward, the y-axis northward, and the z-axis upward.

Problem
31. The Orlando-to-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 x-axis east and the y-axis north.

Solution
(b) The displacement from Orlando to Atlanta calculated in Problem 10 was A = (−320ˆ+ 577ˆ km ı ) in a coordinate system with x-axis east and y-axis north. If this trip took 2.5 h, the average velocity was v = A/2.5 h = (−128ˆ+ 231ˆ km/h. (a) This has ˆ ı ) magnitude (−128)2 + (231)2 km/h = 264 km/h and direction θ = tan−1 (231/−128) = 119◦ (which was given).

Solution
The displacement for the first segment of the balloon’s ˆ excursion is r1 = 0.8k km, in the coordinate system specified, and for the second segment r2 = 14ˆ km. ı The third segment has length (24 km/h)(15 min) = 6 km in the northeast √ direction ˆ cos 45◦ +ˆ sin 45◦ , ı  ı ) so r3 = (6 km)(ˆ +ˆ ı )/ 2 = (4.24ˆ + 4.24ˆ km. Finally, the last segment’s displacement is r4 = ˆ ˆ (250 m − 800 m)k = − 0.55k km (this is a drop of 550 m from the preceding altitude). The total displacement ∆r = r1 + r2 + r3 + r4 = ˆ [(14 + 4.24)ˆ+ 4.24ˆ+ (0.8 − 0.55)k] km = (18.2ˆ + ı  ı ˆ 4.24ˆ + 0.25k) km is accomplished in total time  ∆t = (10 + 27 + 15 + 5) min = 0.950 h, so the average ˆ velocity is ∆r/∆t = (19.2ˆ+ 4.47ˆ + 0.263k) km/h. ı 

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 y-axis toward noon and x-axis toward 3 o’clock?

Problem
34. Figure 3-25 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.

Solution
At the hour, the tip of the minute hand has position r1 = (5.5 cm)ˆ while at 20 min past the hour, it has , position r2 = (5.5 cm)(ˆ cos ( − 30◦ ) +ˆ sin ( − 30◦ )) = ı 

CHAPTER 3
y (mm) 4 3 6.0 s 2 0s 1 2 3 4 1.0 s 5.0 s

9

turn?

Solution
In a coordinate system with x-axis east and y-axis north, the initial velocity of the airplane at the beginning of its turn is v1 = 2100ˆ km/h, and the final ı velocity at the end is v2 = −1800ˆ km/h. The average  ¯ acceleration (Equation 3-8) is a = (v2 − v1 )÷ (t2 − t1 ) = (−1800ˆ − 2100ˆ  ı)(km/h)/2.5 min = 2 −(3.89ˆ + 3.33ˆ m/s , with magnitude ı ) 2 (−3.89)2 + (−3.33)2 m = 5.12 m/s and direction −1 ◦ θ = tan (−3.33/3.89) = 221 (in the third quadrant, nearly southwest).

4.0 s 1 –4 –3 –2 –1 3.0 s –1 –2 –3 –4

x (mm)

2.0 s

figure 3-25 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 = ∆r/∆t. (a) va = [r(2 s) − r(1 s)]/(2 s − 1 s) = [(−3ˆ mm) − (3ˆ − ˆ) mm]/1 s = −(3ˆ+ 2ˆ mm/s.  ı  ı ) ¯ (b) vb = [r(4 s) − r(2 s)]/(4 s − 2 s) =[( −ˆ + ı 0.5ˆ mm −(−3ˆ mm)]/2 s = (−0.5ˆ + 1.75ˆ mm/s. )  ı ) ¯ (c) vc = [r(6 s) − r(0)]/6 s = [( −ˆ + 2ˆ mm − 0]/6 s = ı ) (−0.167ˆ+ 0.333ˆ mm/s. ı )

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 ∆v = −vˆ −(vˆ = −v(ˆ +ˆ where ˆ is ı) ı ), ı east and ˆ is north. The direction of the average  acceleration is the same as that of ∆v, which is parallel to −(ˆ +ˆ or southwest. ı ) (θ = tan−1 (−1/−1) = 225◦.)

Problem
35. An object’s position as a function of time is given by r = 12tˆ + (15t − 5.0t2 )ˆ m, where t is time in s. ı  (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?

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

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)ˆ + ı 2 [(15 m/s)(2 s)− (5.0 m/s )(2 s)2 ]ˆ = 24ˆ+ 10ˆ m,  ı  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 ¯ v = [r(2 s) − r(0)]/(2 s − 0) = 12ˆ+ 5ˆ m/s. (c) The ı  instantaneous velocity at any time is dr/dt = (12 m/s)l + [(15 m/s) − (5.0 m/s2 )2t]ˆ = v(t) (see  Appendix A-2 for the derivative of tn ), so when t = 2 s, v(2s) = 12ˆ − 5ˆ m/s. ı 

Solution
(a) In six months (half a year or 1 ×3.156×107 s) the 2 Earth has traveled half way around its orbit, so its velocity has merely reversed direction, vJul =−vJan . The magnitude of the average acceleration is 1 aa = |vJul − vJan | /∆t = 2 |vJul | / 2 y. The Earth’s ¯ orbital speed is nearly constant at 30km/s; therefore 2 aa = 4(30 km/s)/3.156 × 107 s = 3.80 mm/s . The ¯ ¯ direction of aa is parallel to vJul . (b) In just three months, the Earth covers one fourth of its orbit, so its velocity changes by almost 90◦ , i.e., vApr ⊥vJan . Then ∆v = vApr − vJan forms the hypotenuse of an isosceles right triangle, as shown in the sketch, with magnitude √ 2|v|. Therefore, the magnitude of the average √ √ acceleration is ab = 2|v|/ 1 y = 4 2(30 km/s)÷ ¯ 4 3.156×107 s = 5.38 mm/s2 . (c) From the sketch, one

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

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CHAPTER 3

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

Problem
40. Attempting to stop on a slippery road, a car moving at 80 km/h skids across the road at a 30◦ angle to its initial motion, coming to a stop in 3.9 s. Determine the average acceleration in m/s2 , using a coordinate system with the x-axis in the direction of the car’s original motion and the y-axis toward the side of the road to which the car skids.

Problem 38 Solution. Problem 40 Solution.

Problem
39. What are (a) the average velocity and (b) the average acceleration of the tip of the 2.4-cm-long hour hand of a clock in the interval from 12 p.m. to 6 p.m.? Express in unit vector notation, with the x-axis pointing toward 3 p.m. and the y-axis toward 12 p.m.

Solution
The car’s acceleration is opposite to the direction of the 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 = 2 5.70 m/s . In relation to the coordinate system 2 specified, a = (5.70 m/s )(ˆ cos 210◦ +ˆ sin 210◦ ) = ı  √ −(5.70 m/s2 )( 3ˆ + ˆ ı )/2 = −(4.93ˆ+ 2.85ˆ m/s2 . ı ) (Note that v0 , the velocity at the start of the skid, is not in the direction of the initial motion before the skid.)

Solution
The position and velocity of the tip of the hour hand are shown in the diagram for 12 p.m. and 6 p.m., in the coordinate system specified. The magnitude of the position is a constant, namely, the 2.4-cm radius. The magnitude of the velocity is also constant, namely, the circumference divided by 12 h, or 2π(2.4 cm)/12 h = ¯ 1.26 cm/h. (a) v = (r2 − r1 )/6 h = ¯ (−2.4ˆ − 2.4ˆ cm/6 h = − 0.8ˆ cm/h. (b) a =  )  (v2 − v1 )/6 h = (−1.26ˆ− 1.26ˆ ı ı)(cm/h)/6 h. = −0.419ˆ cm/h2 . ı

Problem
41. An object undergoes acceleration of 2.3ˆ + ı 2 3.6ˆ m/s over a 10-s interval. At the end of this  time, its velocity is 33ˆ + 15ˆ m/s. (a) What was ı  its velocity at the beginning of the 10-s interval? (b) By how much did its speed change? (c) 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 = v0 + at, so v0 = v − at, or v0 = (33ˆ + 15ˆ m/s − (2.3ˆ + 3.6ˆ m/s2 (10 s) = ı ) ı ) (10ˆ− 21ˆ m/s. (b) v0 = (10)2 + (−21)2 = 23.3 m/s, ı ) and v = (33)2 + (15)2 = 36.2 m/s, so the change in speed is ∆v = v − v0 = 13.0 m/s (we did not round off before subtracting). (c) θ = tan−1 (15/33) = 24.4◦ and θ0 = tan−1 (−21/10) = 295◦ = −64.5◦ (positive angles CCW, negative angles CW, from x-axis) so the direction changed by ∆θ = θ − θ0 = 89.0◦ . (d) at =

Problem 39 Solution.

CHAPTER 3 (2.3)2 + (3.6)2 m/s (10 s) = 42.7 m/s = ∆v. The difference between at = |v − v0 | and ∆v = v − v0 can be seen from the triangle inequality: | v − v0 | ≤ | v − v0 | ≤ v + v0 .
2

11

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 s.

Solution
(a) The average acceleration during the interval used ¯ ¯ in Problem 35 is a = [¯ (2 s) − v(0)]/(2 s − 0) = v 2 [(12ˆ − 5ˆ m/s− (12ˆ + 15ˆ m/s]/(2 s) = − 10ˆ m/s ı ) ı )  ˆ (see part (c) of the solution to Problem 35 for v(t)). (b) The instantaneous acceleration at any time is 2 ¯ a(t) = d¯ /dt = − (10 m/s )ˆ which is constant, v , including when t = 2 s. Problem 41 Solution.

Section 3-5: 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?

Problem
42. An object’s position as a function of time is given ˆ by r = (bt3 + ct)ˆ + dt2ˆ + (et + f )k, where b, c, d, ı  e, and f are constants. Determine the velocity and acceleration as functions of time.

Solution
Differentiating each term using Equation 2-3, we find ˆ v = dr = dt = (3bt2 + c)ˆ + 2dtˆ + ek, and ı  a = dv/dt = 6btˆ = 2dˆ ı .

Solution
(a) Let S be a frame of reference fixed on the shore, with x-axis upstream, and let S ′ be a frame attached to the barge. The velocity of S ′ relative to S is V = (14 km/h)ˆ The velocity of the dog relative to ı. the shore is (from Equation 3-25) v = v′ + V, and its speed is v = |v′ + V|, where v ′ = 4 km/h. When the dog is walking upstream, v′ V, v = (4 + 14) km/h = 18 km/h, and when walking downstream, −v′ V, and v = (14 −√ km/h = 10 km/h. When 4) v′ ⊥ V, v = 142 + 42 = 14.7 km/h. (In general, v 2 = v ′2 + V 2 + 2v ′ V cos θ′ , where θ′ is the angle between v′ and V in S ′ .) (b) Since the current flows downstream, according to Equation 3-25: vel. of barge rel. to water = vel. of barge rel. to shore − vel. of water rel. to shore

Problem
43. The position of an object is given by r = (ct − bt3 )ˆ + dt2ˆ with constants c = 6.7 m/s, ı , 3 b = 0.81 m/s , and d = 4.5 m/s2 . (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 )ˆ + 2dtˆ so ı , v(0) = (6.7 m/s)ˆ (b) The direction of v(t) is ı. θ(t) = tan−1 (2dt/(c − 3bt2 )), measured CCW from the x-axis, so θ(0) = 0◦ . θ(t) = 90◦ when the argument of the arctan is ∞, or t = c/3b. Thus, t = (6.7 m/s)/3(0.81 m/s ) = 1.66 s. (The direction of motion was −90◦ at t = −1.66 s.) (c) Since vx = 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.
3

= 14ˆ − (−3ˆ km/h. ı ı) Going through the same steps as in part (a), for a new frame S moving with the water, with a new relative velocity V = (17 km/h)ˆ we find the speed of the dog ı, relative 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 −y direction at its orbital speed of 24 km/s. Find the spacecraft’s velocity relative to Mars.

Solution
Equation 3-10 says that the velocity of the spacecraft relative to Mars, vCM , equals the difference of the velocities of each relative to the Sun, vCS , −vMS . (Our notation vCM means the velocity of C relative to M.) vMS is given as −(24 km/s)ˆ in the Sun’s reference  frame, the x-y coordinates in the problem. The Earth has velocity vES = (30 km/s)ˆ in the Sun’s frame, and ı vCE = (40 km/s)ˆ where the y-axes in the Earth’s and , Sun’s frames are assumed to be parallel. A second application of Equation 3-10 gives vCS = vCE + vES = (40ˆ + 30ˆ km/s; therefore vCM = vCS − vMS =  ı) (30ˆ = 40ˆ km/s − (−24ˆ km/s = (30ˆ + 64ˆ km/s. ı ) ) ı )

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 3-7, we are given that v = (1500 km/100 min)(−ˆ = −900ˆ km/h, and )  v′ = −(1000 km/h)(ˆ sin 15◦ +ˆ cos 15◦ ). The wind ı  velocity is V = v − v′ = (−900ˆ + 259ˆ+ 966ˆ =  ı ) (259ˆ + 65.9ˆ km/h. Therefore, the wind speed is ı√ ) V = 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 the direction facing the wind, in this case 14.3◦ S of W.)

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 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 3-10. (a) Evidently, sin θ = |V| / |v′ |, or θ = sin−1 (0.57/1.3) = 26.0◦ , which is your heading upstream. (b) |v| = |v′ | cos θ = (1.3 m/s) cos 26.0◦ = 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.

Problem 46 Solution. Problem 48 Solution.

Problem
47. A spacecraft is launched toward Mars at the instant Earth is moving in the +x 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 +y direction. At the launch time, Mars is moving in the

Problem
49. 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

CHAPTER 3 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?

13

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 speed”), and the velocity of the airplane relative to the air, v′ (the “air speed”) form the sides of a triangle given by Equation 3-10, v′ + V = v, as shown. We are given that the triangle is a right triangle (v ⊥ V), that the angle between the airspeed and groundspeed is 32◦ , and the hypotenuse (magnitude of airspeed) is 370 km/h. From trigonometry, the magnitude of the windspeed is |V| = (370 km/h) sin 32◦ = 196 km/h.

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 S ′ relative to S is a constant, 2.2 m/s in the x-x′ 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 ∆y ′ = −1.6 m, and its average velocity 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 button follows a projectile trajectory to be described in Chapter 4. Here, we observe that while the button falls vertically through a displacement ∆y = ∆y ′ = −1.6 m, it also moves horizontally (in the direction of the conveyor belt) through a displacement ∆x = (2.2 m/s)(0.57 s) = 1.25 m. Its net displacement in S is ∆r = ∆xˆ + ∆yˆ = ı  (1.25ˆ− 1.60ˆ m, so its average velocity is ı ) ¯ v = ∆r/∆t = (1.25ˆ − 1.60ˆ m/0.57 s = ı ) (2.20ˆ− 2.8ˆ m/s. These have magnitudes |∆r| = ı ) (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 the same direction θ = tan−1 (−1.60/1.25) = −51.9◦ from the horizontal, or 38.1◦ from the downward vertical.

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 x-axis east, y-axis north, and z-axis up, the three displacements can be written in terms of their components and the unit vectors: r1 = 21.0ˆ m, r2 = (8.50 m)(ˆ cos 225◦ +ˆ sin 225◦ ) = ı ı  ˆ −(6.01)(ˆ +ˆ m, and r3 = −1.10k m. (Note that ı ) southwest is 180◦ + 45◦ CCW from east.) The total displacement is the vector sum r1 + r2 + r3 = ˆ [(21.0 − 6.01)ˆ− 6.01ˆ − 1.10k] m, and its magnitude is ı  the square root of the sum of its components, (15.0)2 + (−6.01)2 + (−1.10)2 m = 16.2 m.

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 jet stream?

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
Problem 50 Solution. Take a standard coordinate system with x-axis east, y-axis north, and z-axis up. The particle begins moving downward (say along the z-axis) through a ˆ displacement r1 = −3.2k km. It is then deflected in the y-z plane (northward of the vertical) by 27◦ , ˆ so r2 + (1.6 km)(ˆ sin 27◦ − k cos 27◦ ), and  finally eastward through r3 = 2.1ˆ km. ı

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 Solution.

Problem 52 Solution.

Problem
53. A car is heading into a turn at 85 km/h. It enters the turn, slows to 55 km/h, and emerges 28 s 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?

solar system, picking up speed through a so-called “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 2.99×104 m/s. Thirty days after the encounter, Galileo was moving at 54◦ to its pre-encounter direction, at a speed of 3.51×104 m/s. What were (a) the magnitude and (b) the direction of Galileo’s average acceleration during this interval, the latter measured with respect to its pre-encounter direction?

Solution
The initial and final velocities have magnitudes of 85 km/h and 55 km/h, respectively, and make an angle of 35? as shown, where we chose the x-axis parallel to vi and the y-axis in the direction of the turn. The change in velocity is ∆v = vf − vi = (55 km/h)(ˆ cos 35◦ + ˆ sin 35◦ ) − ı  (85 km/h)ˆ = (−39.9ˆ + 31.5ˆ km/h: (a) The ı ı ) magnitude of the average acceleration is a = |∆v| /∆t = (−39.9)2 + (31.5)2 (km/h)/28 s = ¯ 2 1.82 km/h/s = 0.505 m/s . (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 also have been solved using the laws of cosines and sines; see the solution to the next problem.)

Problem 54 Solution.

Solution
The initial and final velocities for the 60 d interval are given, as indicated in the diagram. Instead of using unit vectors, as in the previous problem, let’s use 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)÷ ¯ 2 (60×86,400 s) = 5.76 mm/s . (b) θ = 180◦ − sin−1 (35.1 sin 54◦ /29.9) = 180◦ − 71.9◦ = 108◦ .

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 liquid-fueled booster rocket was unsafe to fly on the shuttle. As a result, Galileo’s trajectory became a complicated path through the inner

Problem
55. The sweep-second 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 hand’s tip over a 5.0-s interval? (c) What is the

CHAPTER 3 angle between the average velocity and acceleration vectors?

15

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 θ = (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, | ∆r | = 2 |r| sin 1 θ = 2 2(3.1 cm) sin(30◦ /2) = 1.60 cm, and whose base angle 1 is 2 (180◦ − 30◦ ) = 75◦ . Thus, the average velocity has magnitude | ∆r | /∆t = 1.60 cm/5 s = 0.321 cm/s and direction 180◦ − 75◦ = 105◦CW from r1 . (b) The instantaneous speed of the tip of the second-hand is a constant and equal to the circumference divided by 60 s, or v = 2π(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

direction of motion (CW). The angle between two tangents is the same as the angle between the two corresponding radii, so v1 , v2 and ∆v form an isosceles triangle similar to the one in part (a). Thus 1 |∆v| = 2 |v| sin 2 θ = 2(0.325 cm/s) sin( 1 × 30◦ ) = 2 0.168 cm/s. The magnitude of the average acceleration is |∆v| /∆t = (0.168 cm/s)/5 s = 2 3.36×10−2 cm/s , and its direction is 105◦ CW from the direction of v1 , or 195◦ CW from the direction of ¯ ¯ r1 . (c) The angle between a and v, 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 µs. What are the magnitude of (a) the average velocity and (b) the average acceleration as the proton sweeps through one-twelfth of the full circle? (c) What is the angle between the average velocity and acceleration vectors?

Solution
One-twelfth of a revolution is 30◦ , so the geometry of this problem is the same as that for the previous one, ¯ ¯ and the directions of v, a and the angle between them are the same as in Problem 55. The magnitudes, however, are different. (a) |∆r| = 2 |r| sin 1 θ = 2 (23 cm) sin 15◦ = 5.95 cm and ∆t = 0.17 µs/12 = 14.2 ns. Thus, |¯ | = |∆r| /∆t = 4.20 Mm/s. (b) The v instantaneous constant speed is π(23 cm)/(0.17 µs) = 4.25 Mm/s, so |∆v| = 2(4.25 Mm/s) sin 15◦ = 2 2.20 Mm/s. Then |¯| = |∆v| /∆t = 1.55×1014 m/s . a

Problem 55 Solution.

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.

16

CHAPTER 3

Solution
The velocity of the boat relative to the ground, v, is perpendicular to the velocity of the water relative to the ground, the current velocity V, which form a right triangle with hypotenuse v′ equal to the velocity of the boat relative to the water, as shown in the diagram and as required by Equation 3-10. The heading upstream is θ = sin−1 (|V| / |v′ |) = sin−1 (6.3/15) = 24.8◦.

Problem
58. A flock of geese is attempting to migrate due south, but the wind is blowing from the west at 5.1 m/s. If the birds can fly at 7.5 m/s relative to the air, in what direction should they head? figure 3-26 Problem 59 (figure is not to scale).

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

of the displacement from its initial position P1 to its final position P2 , and its average velocity and acceleration, during the same tracking interval, by using the geometrical analysis in the solution to Problem 55, to which the reader is referred. In the diagram based on Fig. 3-26, O is the center of the Earth, RE = 6370 km is the average radius of the Earth, A is the tracking station, and r = RE + h = 6370 km + 240 km = 6610 km is the radius of the orbit. We apply the law of cosines to triangle OAP1 to find AP1 , and then the law of sines to 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◦ + = 1295 km. (6370 km)2 cos2 95◦ + (6610 km)2 − (6370 km)2

Problem 58 Solution.

Supplementary Problems Problem
59. A satellite is in a circular orbit 240 km above Earth’s surface, moving at a constant speed of 7.80 km/s. A tracking station picks up the satellite when it is 5.0◦ above the horizon, as shown in Fig. 3-26. The satellite is tracked until it is directly overhead. What are the magnitudes of (a) its displacement, (b) its average velocity, and (c) its average acceleration during the tracking interval? Is the value of the average acceleration approximately familiar?

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

(We are keeping four significant figures in the intermediate results, but will round off to three at the end.) Then sin ∆θ/AP1 = sin 95◦ /r gives ∆θ = sin−1 (1295 sin 95◦ /6610) = 11.26◦ . (a) Now that we know ∆θ, the magnitude of the displacement can be found from the isosceles triangle OP1 P2 as in Problem 55. P1 P2 = 2r sin 1 ∆θ = 2 1 2(6610 km) sin 2 (11.26◦) = 1296 km ≈ 1300 km. (b) To find the magnitude of the average velocity, we first need to find the tracking interval ∆t . The time for a complete orbit (called the period) is the orbital circumference divided by the speed, or T = 2πr/v = 2π(6610 km)/(7.8 km/s) = 5.325×103 s. During the tracking interval, the satellite completes only a fraction ∆θ/360◦ of a complete orbit, so ∆t = (11.26◦ /360◦)(5.325×103 s) = 166.5 s. Thus, |¯ | = P1 P2 /∆t = 1296 km/166.5 s = 7.78 km/s. v

CHAPTER 3 (c) As shown in the solution to Problem 55, the velocity vectors at P1 and P2 form an isosceles triangle similar to OP1 P2 . Thus | ∆v| = 2 |v| sin 1 ∆θ = 2 2(7.8 km/s) sin 1 (11.26◦ ) = 1.530 km/s, and the 2 magnitude of the average acceleration is 2 |¯| = |∆v| /∆t = (1.530 km/s)/166.5 s = 9.19 m/s . a When we recall that the Earth’s gravity holds the satellite in its orbit, it is not surprising that this magnitude is close to g. Of course, at an altitude of 240 km, the Earth’s gravitational field is only 9.11 m/s2 (compared to 9.81 m/s2 at the surface). We must also remember that there is a discrepancy between the average and instantaneous accelerations due to the finite size of ∆t .

17

Problem 60 Solution.

|B| = |A|, but pointing at an angle 90◦ more than the angle of A with the x-axis, as shown in the sketch, i.e., θB = 90◦ + θA . The components of A are Ax = |A| cos θA = a and Ay = |A| sin θA = b, while the components of B are Bx = |B| cos θB = |A| cos(90◦ + θA ) = − |A| sin θA = −b and By = |B| sin(90◦ + θA ) = |A| cos θA = a. Thus, B = −bˆ + aˆ is perpendicular to A = aˆ + bˆ Clearly, ı  ı . −B = bˆ − aˆ is also perpendicular to A, as is any ı  scalar multiple of B. (Another way to do this problem is to use the scalar product defined in Section 7.2.)

Problem 61 Solution. Problem 59 Solution

Problem Problem
60. The sum, A + B, of two vectors is perpendicular to the difference, A − B. How do the magnitudes of the two vectors compare? 62. Find the angle between the vectors 3.0ˆ + 1.7ˆ and ı  6.1ˆ − 4.2ˆ ı .

Solution
The angle a vector makes with the x-axis can be found from its components, θ = tan−1 (Ay /Ax ) (see Equation 3-2 and Fig. 3-11(a)). For the first vector, θ1 = tan−1 (1.7/3.0) = 29.5◦ (in the first quadrant) and for the second, θ2 = tan−1 (−4.2/6.1) = 325◦ or −34.6◦ (in the fourth quadrant). The angle between them is the difference θ1 − θ2 = 29.5◦ − (−34.6◦ ) = 64.1◦ . (Note: 29.5◦ − 325◦ = −296◦ + 360◦ = 64.1◦ . There are two possible angles between any two vectors, one less than, the other greater than 180◦. By convention, the smaller one is always used.)

Solution
The vectors A + B and A − B are the two diagonals of the parallelogram formed by sides A and B. If the diagonals are perpendicular, the parallelogram is a rhombus; hence |A| = |B|. (Students are advised to redo this problem after the “dot” product is defined in Section 7.2.)

Problem
61. Find two vectors in the x-y plane that are perpendicular to the vector aˆ + bˆ ı .

Problem
63. Write an expression for a unit vector that lies at 45◦ between the positive x- and y-axes.

Solution
One vector perpendicular to the given vector, A = aˆ + bˆ is a vector B of the same magnitude, ı ,

18

CHAPTER 3
y
◦

Solution
A vector of unit magnitude, making a 45 angle CCW with the x-axis, can √ expressed as 1· cos 45◦ˆ + be ı 1· sin 45◦ˆ = (ˆ +ˆ  ı )/ 2. (A unit vector in any direction in the x-y plane is therefore n = ˆ ˆ cos θ +ˆ sin θ.) ı 

C

B

θB

Problem
64. A vector A has components Ax and Ay in a coordinate system with axes x and y. Find its components A′ and A′ in a coordinate system x y whose axes x′ and y ′ are rotated counterclockwise through an angle θ from the x- and y-axes. Test your result for the cases θ = 0 and θ = 90◦ .
A

θA

x

figure 3-27 Problem 65.

Problem
y y′ Ay Aysin θ Ay′ Axcos θ ˆ J ˆ′ J
θ

Aycos θ

x′ AZ′

î′ î

θ

Axsin θ x

AZ

66. You wish to paddle a canoe perpendicularly across a river of width w and back. If the river’s flow speed is c, and you can paddle at speed v relative to the water, show that your round-trip travel √ time is given by ∆t = 2w/ v 2 − c2 .

figure 3-14 Problem 64.

Solution
Crossing perpendicular to the current, your velocity relative to the ground, u, and the current velocity, c, form a right triangle with hypotenuse equal to your velocity relative to the water, v. (See Equation 3-10 √ and the sketch.) Thus, u = v 2 − c2 , and the time for a round-trip of length 2w is ∆t = 2w/u.

Solution
By drawing an extra line on Fig. 3-14 (through Ax on the x-axis perpendicular to the dotted line through A′ y on the y ′ -axis), one sees that A′ = Ax cos θ + Ay sin θ x and A′ = −Ax sin θ + Ay cos θ. For θ = 0, y A′ = Ax and A′ = Ay (no change), while for θ = 90◦ , x y A′ = Ay and A′ = −Ax (a 90◦ CW rotation of A). x y

Problem
65. Figure 3-27 shows two arbitrary vectors A and B that sum to a third vector C. By working with components, prove the law of cosines: C 2 = A2 + B 2 − 2AB cos γ.

Problem 66 Solution.

Solution
Adding some lines parallel to the axes and labeling the angles of the vectors in Fig. 3-27, one sees that Cx = Ax + Bx = A cos θA + B cos θB and Cy = A sin θA + B sin θB . Squaring, adding, and using two trigonometric identities from Appendix A, one 2 2 gets C 2 = Cx + Cy = A2 (cos2 θA + sin2 θA )+ 2 2 2 B (cos θB + sin θB ) + 2AB(cos θA cos θB + sin θA sin θB ) = A2 + B 2 + 2AB cos(θB − θA ). This is the law of cosines when we replace θB − θA by 180◦ − γ, where γ is the angle between sides A and B in the vector triangle.

Problem
67. Town B is located across the river from town A and at a 40.0◦ angle upstream from A, as shown in Fig. 3-28. A ferryboat travels from A to B; it sails at 18.0 km/h relative to the water. If the current in the river flows at 5.60 km/h, at what angle should the boat head? What will be its speed relative to the ground? Hint: Set up Equation 3-10 for this situation. Each component of Equation 3-10 yields two equations in the

CHAPTER 3 unknowns v, the magnitude of the boat’s velocity relative to the ground, and φ, the unknown angle. Solve the x equation for cos φ and substitute into the second equation, using the relation sin φ = 1 − cos2 φ. You can then solve for v, then go back and get φ from your first equation.

19

cos−1 (13.9 cos 40◦ /18.0) = 53.8◦. (Equations similar to these will be solved when collisions in two-dimensions and the conservation of momentum are discussed in Chapter 11.)

V V B

u 40° A

figure 3-28 Problem 67.

Solution
The diagram shows the velocities and coordinate axes added to Fig. 3-28. v′ (the velocity of the boat relative to the water), v (the velocity of the boat relative to the ground), and V (the velocity of the water relative to the ground) are related by Equation 3-10, v′ = v − V. This vector equation is equivalent to two scalar equations, one for each ′ ′ component, vx = vx − Vx and vy = vy − Vy . The components of each vector, in terms of its ′ magnitude and angle, are vx = v ′ cos θ′ × ′ ′ ′ ◦ vy = v sin θ , vx = v cos 40 , vy = v sin 40◦ , and Vx = 0, Vy = −V. Therefore, the x and y component equations are v ′ cos θ′ = v cos 40◦ and v ′ sin θ′ = v sin 40◦ + V. We can eliminate θ′ by squaring and adding (since sin2 + cos2 = 1): v ′2 (cos2 θ′ + sin2 θ′ ) = v 2 (cos2 40◦ + sin2 40◦ )+ 2vV sin 40◦ + V 2 , or v 2 + 2vV sin 40◦ − v ′2 + V 2 = 0. The positive root of this quadratic for v (appropriate for 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 find the speed relative to the ground is v = 13.9 km/h. Going back to the x component equation, we find the heading θ′ = cos−1 (v cos 40◦ /v ′ ) =

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 at 25◦ to the equator. What is the shuttle’s velocity relative to scientists tracking it from the equator?

Solution
Let v and V be the velocities of the space shuttle and the tracking station, respectively, relative to a reference frame fixed at the center of the Earth. The shuttle’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 the shuttle and the station. Suppose that the shuttle is passing directly over the tracking station (as at point A). Then the angle between v and V is 25◦ , while magnitudes are given, so v′ can be found from the law of 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◦ .

Problem 68 Solution.