PHYS 1111 - Summer 2007 - Professor Caillault

Homework Solutions

Chapter 3

13. Picture the Problem: The whale dives along a straight line tilted 20.0°below horizontal for 150 m as shown in the figure.

Strategy: Resolve the whale’s displacement vector into horizontal andvertical components in order to find its depth ry and its horizontal traveldistance rx.

Solution: 1. (a) The depth is given by ry: ry = r sinθ = 150 m( )sin 20.0°( ) = 51 m

2. (b) The horizontal travel distance is given by rx: rx = r cosθ = 150 m( )cos 20.0°( ) = 140 m = 0.14 km

Insight: Note that both answers are limited to two significant figures, because although “20.0°” has three, “150 m” hasonly two significant figures.

19. Picture the Problem: The vectors involved in the problem are depicted atright.

Strategy: Use the vector component method of addition and subtractionto determine the components of each combination of

rA and

rB . Once

the components are known, the length and direction of each combinationcan be determined fairly easily.

Solution: 1. (a) Determine the components of rA +

rB :

rA +

rB = –5( ) y + 10( ) x = 10x – 5y

2. Find the magnitude of rA +

rB :

rA +

rB = 10( )2 + −5( )2 = 11 units

3. Determine the direction of rA +

rB , measured

counterclockwise from the positive x axis. θ r

A+rB = tan−1 −5

10

= −27° or 333°

4. (b) Determine the components of rA −

rB :

rA −

rB = –5( ) y − 10( ) x = −10x – 5y

5. Find the magnitude of rA −

rB :

rA −

rB = −10( )2 + −5( )2 = 11 units

6. Determine the direction of rA −

rB , measured

counterclockwise from the positive x axis. θ r

A−rB = tan−1 −5

−10

= 27° +180° = 207°

7. (c) Determine the components of rB −

rA :

rB −

rA = 10( ) x − –5( ) y− = 10x + 5y

8. Find the magnitude of rB −

rA :

rB −

rA = 10( )2 + 5( )2 = 11 units

9. Determine the direction of rB −

rA , measured

counterclockwise from the positive x axis. θ r

B−rA = tan−1 5

10

= 27°

Insight: This problem is simplified by the fact that rA and

rB have only one component each, but a similar approach

will work even with more complicated vectors. Notice that you must have a picture of the vectors in your head (or onpaper) in order to correctly interpret the directions in steps 3, 6, and 9.

29. Picture the Problem: The vectors involved in the problemare depicted at right.Strategy: Use the information given in the figure todetermine the components of vectors

rA,

rB, and

rC . Then add

the components.Solution: 1. Addthe x component ofeach vector:

Ax = 1.5 m( )cos 40°( ) = 1.1 m

Bx = 2.0 m( )cos −19°( ) = 1.9 m

Cx = 1.0 m( )cos 180 − 25°( ) = − 0.91 mrA +

rB +

rC( )

x= 2.1 m

2. Add the ycomponent of eachvector:

Ay = 1.5 m( )sin 40°( ) = 0.96 m

By = 2.0 m( )sin −19°( ) = − 0.65 m

Cy = 1.0 m( )sin 180 − 25°( ) = 0.42 mrA +

rB +

rC( )

y= 0.74 m

3. Express the sum in unit vectornotation:

rA +

rB +

rC = 2.1 m( ) x + 0.74 m( ) y

Insight: In this problem the vector component method of addition is much quicker than the graphicalmethod.

31. Picture the Problem: The displacement vectors are depicted atright. North is in the y direction and east is in the x direction.

Strategy: Sum the components of the vectors in order to determine

rA +

rB . Multiply that vector by −1 in order to reverse its direction.

Then find the magnitude and direction of the reversed vector.

Solution: 1. (a) Add the twodisplacement vectors:

rA +

rB = −72 m( ) x + 120 m( ) y

2. Multiply by −1 in order toreverse thedirection of the net displacementand bring the cat back home:

−

rA +

rB( ) = 72 m( ) x + −120 m( ) y

3. Find the magnitude of thedesired displacement:

−rA +

rB( ) = 72 m( )2 + 120 m( )2 = 140 m

4. Find the direction of thedesired displacement:

θ = tan−1 −120 m72 m

= −59° = 59° south of east

5. (b) Vector addition is independent of the order in which the addition is accomplished. The initialdisplacement is the same, so there is no change in the displacement for the homeward part of the trip.

Insight: In this problem we could claim the cat’s initial displacement is a single vector with the givencomponents. The answers wouldn’t change, but it would simplify the solution a little bit.

43. Picture the Problem: The vectors involved in this problem aredepicted at right.

Strategy: Let rvyw = your velocity with respect to the walkway,

rvwg = walkway’s velocity with respect to the ground, and add the

vectors according to equation 3-8 to find rvyg = your velocity with

respect to the ground. Then find the time it takes you to travel the 85-m distance.

Solution: 1. Findyour velocity withrespect to thewalkway:

rvyw =

ΔxΔt

x =

85 m68 s

x = 1.25 m/s( ) x

2. Apply equation 3-8 to find yourvelocity with respectto the ground:

rvyg =

rvyw +

rvwg = 1.25 m/s( ) x + 2.2 m/s( ) x = 3.45 m/s( ) x

3. Now find the time oftravel:

t =Δxvyg

=85 m

3.45 m/s= 25 s

Insight: The moving walkway slashed your time of travel from 68 s to 25 s, a factor of 2.7! Note thatwe bent the significant figures rules a little bit by not rounding

rvyw to 1.3 m/s. This helped us avoid

rounding error.

55. Picture the Problem: The three-dimensional vector is depicted in the diagram atright.Strategy: Determine the z component of

rA by applying the cosine function to the

right triangle formed in the z direction. Then find the projection of rA onto the xy

plane (A sin 55°) in order to find the x and y components of rA .

Solution: 1. Find the z component of rA :

Az = 65 m( )cos55° = 37 m

2. Find the projection onto the xy plane: Axy = Asin55° = 65 m( )sin55°

3. Find the x component of rA :

Ax = 65 m( )sin55° cos 35° = 44 m

4. Find the y component of rA :

Ay = 65 m( )sin55° sin 35° = 31 m

Insight: A knowledge of right triangles can help you find the components of even a three dimensional vector. Once thecomponents are known, then addition and subtraction of vectors become straightforward procedures.