HW03SolWalker4e

7/26/2019 HW03SolWalker4e

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Physics 1401Homework Solutions - Walker, Chapter 3

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Conceptual Questions

CQ3. (a.) B

must be the same as A

. (b.) Direction of B

must be oppposite the direction of A

. So

graphically A

and B

might look like:

To add these two vectors graphically, we could pick up B

and move it so that its tail is at the tip of A

, as follows:

Then the vector BA

+ is the vector drawn from the tail of A

to the tip of B

. This is just the single pointshown

above. This vector, which consists of just a single point, is called the zero vector, 0

. In component form, the zero

vector is written yx 000 +=

. Suppose A

were expressed in component form as yAxAA yx +=

. Then in order for

BA

+ to give the zero vector, B

must be yAxAB yx =

. In other words, B

must be the negative of A

.

CQ4. No. Consider a vector A

given in component form by yAxAA yx +=

. The magnitude of A

is

22yx AAA +=

, which is always at least as big as xA or yA (actually, the absolute valueof xA or of yA ). To put

it another way, cosAAx

= and sinAAy

= , but neither cos nor sin is ever greater than 1.

Problems6. The problem is pictured below. Doing a quick unit conversion, I find that 530 ft = 0.100 mi.

(a.) 10.100

sin 4.81.2

= =

(b.) When Ive gained an additional 150 ft of elevation, my elevation will be 680 ft (0.129 mi), so Ill have:

0.129 misin4.8

d= ,

A

B

A

B

BA

+

1.2 mi

530 ft (0.100 mi)


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in which dis the total distance I will have driven. So0.129 mi

1.54 misin4.8

d= =

. So I have to drive 0.34 mi

farther (after driving 1.2 mi) to gain the additional 150 ft of elevation.

8. (a.)

(b.) ( ) ( )75 m cos 65.0 32 mxr = =

( ) ( )75 m sin 65.0 68 myr = =

16. The situation is as shown below. The first displacement from the starting point is 1d

(15.0 m north... the red

vector). There are two choices for the second displacement 2d

, shown in blue: 22.0 m east or 22.0 m west.

For each choice of 2d

, there are two choices for the third displacement 3d

: 5.00 m north or south (green

vectors). The purple vectors show the four possible netdisplacements from the starting point, 1R

through 4R

.

For the net displacement 1R

, we have, in component form:

( )1 15.0 md y=

( )2 22.0 md x=

35.0( )75 m sin 35.0 43 myr = =

( )75 m cos 35.0 61 mxr = =

x

m75=r

y

N (y)

E (x)

15.0 m

W

S

22.0 m22.0 m5.00 m

5.00 m

5.00 m

5.00 m

1d

2d

2d

3d

3d

3d

3d

1R

2R

3R

4R


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( )yd m00.53 =

Adding all these up, I get: ( ) ( )1 1 2 3 22.0 m 10.0 mR d d d x y= + + = +

.

In magnitude and direction, I get:

( ) ( )2 2

1 22.0 m 10.0 m 24.2 mR = + =

11 1

1

10.0 mtan tan 24.4

22.0 m

y

x

R

R

= = =

above +xaxis.

But the book asked us for the direction relative to north, so should give the angle as 65.6

east of north.

By appealing to symmetry, we can see that 4R

is the same in magnitude as 1R

, but its direction will be 74.4to

the westof north. To see this more formally, write down 1d

, 2d

, and 3d

. (Note that only 2d

changes.):

( )1 15.0 md y=

( )2 22.0 md x=

( )yd m00.53 =

So we have: ( ) ( )4 1 2 3 22.0 m 10.0 mR d d d x y= + + = +

.

In magnitude and direction: ( ) ( )2 2

4 22.0 m 10.0 m 24.2 mR = + =

41 1

4

10.0 mtan tan 24.4

22.0 m

y

x

R

R

= = =

.

Notice that the angle my calculator gives is 24.4. With reference to the picture above, its clear that the angle

that the calculator found is the angle between 4R

and the xaxis. This angle is 24.4in magnitude. The minus

sign in the result from the calculator occurs for the following reason: It is customary, for reasons that have todo with ensuring that the tangent is a continuous function over some restricted domain to refer to angles

below thexaxis as negative. Now, in addition to the angle that 4R

makes with the xaxis, there is another

angle this one 24.4below the +xaxis, in the 4thquadrant for which the tangent is 10.0 m 22.0 m .

Following convention in the assigning the algebraic sign of angles below thex axis, we would refer to this angle

as 24.4. When you type in 110.0 m

tan22.0 m

on your calculator, your calculator cannot tell whether you

mean the angle in the 4thquadrant or the angle in the 2ndquadrant after all, they bothhave tangents of (-10/22)

so it assumes you mean the angle in the 4thquadrant. This is why the calculator gives the answer 24.4. Insituations like this, you simply need to draw a picture and keep it clearly in mind when interpreting the resultthat your calculator gives you.

So the final result, then, is that 4R

is 24.2 m in magnitude and is directed at an angle of 24.4 to the north of

west, or 65.6 to the west of north.


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For 2R

, we have: ( )1 15.0 md y=

( )2 22.0 md x=

( )yd m00.53 =

So: ( ) ( )2 1 2 3 22.0 m 20.0 mR d d d x y= + + = +

.

In magnitude and direction: ( ) ( )2 2

2 22.0 m 20.0 m 29.7 mR = + =

21 1

2

20.0tan tan 42.3

22.0

y

x

R

R

= = =

above +xaxis.

So 2R

is 29.7 m in magnitude and its direction is 47.7 east of north.

By symmetry, 3R

will be 29.7 m in magnitude and 47.7 to the westof north.

27. ( ) ( ) ( ) ( ) 54 m cos 42 54 m sin 42 40 m 36 mr x y x y = =

34. ( ) ( ) ( ) ( )yxyxA m96.0m1.140sinm5.140cosm5.1 +=+=

( ) ( ) ( ) ( )yxyxB m65.0m.9119sinm.0219cosm.02 ==

( ) ( ) ( ) ( )yxyxC m.420m.91025sinm.0125cosm0.1 +=+=

( )yD m5.1=


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38. (a.) Take the cats starting place to be at the origin. Then the cats first two displacements are 1d

and 2d

, as

shown in the picture below:

We want to find the magnitude and direction of 3d

. Well, clearly, 0321

=++ ddd if the cat returns home to

his starting point. So lets express 1d

and 2d

in component form:

yd m1201=

xd m722 =

.

So we want the vector ydxdd yx 333 +=

such that 0321

=++ ddd , or:

( ) yxydxd yx 00m120m72 33 +=++ .

In order for this to be satisfied, thexandycomponents of the vectors on the LHS and RHS must be separatelyequal. So we must have:

( ) 0m723 =xd

and ) 0m1203 =+yd .

So: yxd m120m723 =

.

In magnitude and direction: ( ) ( ) m140m120m72 2223233 =+=+= yx ddd

5972

120tantan 1

3

31 =

=

=

x

y

d

d

N (y)

E (x)

120 m

W

S

72 m

1d

2d

3d


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If we imagine extending the direction of 3d

down into the 4thquadrant, as shown by the dashed line in the

above picture, then -59 is the angle shown as ... So 3d

is 59 south of east.

(b.) This would not affect the displacement needed to bring it home at all, because the cat would be at the same

place after the first two displacements, no matter whether they are done in the order 1d

then 2d

or 2d

then

1d

. To put it another way, vector addition is commutative 1 2 2 1d d d d + = +

.

41. The picture is shown below. Note that Ive called the netdisplacement from the starting point r

, so

21 ddr

+= . Note also that avv

always points in the same direction as r

t is always positive, remember

so it looks something like what Ive shown below.

So the directionof avv

will just be the direction of r

, and the magnitudeof avv

is given by:

t

rvav

=

.

Lets first find the magnitude of r

. According to Pythagoras:

ft291525001500 222

2

2

1 =+=+= ddr

.

So the magnitude of avv is: ft/min107.9min0.3

ft2915 2==avv , keeping 2 sig figs. To find the direction

of avv

, find the angle :

04.591500

2500tan 1 =

=

So avv

is directed at an angle of 59.04north of east.

N (y)

E (x)1500 ft

W

S

2500 ft

1d

2d

r

t

rvav


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45. Here is a diagram showing the situation. Note that Ive taken the positivexaxis to be parallel to the

inclined surface, as shown.

The problem is almost tooeasy, once you realize that the acceleration is constant. (After all, neither gnor ischanging.) So you just refer to the four facts for one-dimensional motion with constant acceleration. In

particular, I find:

atvv += 0

So at t = 3.25 s: ( )( )22.95 m/s 3.25 s 9.59 m/sv= =

54. Let the positivexdirection be east and the positiveydirection be north, as usual. Then we can write thevelocity of the passenger relative to the ferry as:

( )yvPF m/s5.1=

Also, if the velocity of the passenger relative to the wateris 4.5 m/s at an angle of 30.0west of north, then we

can write this as:

( ) ( ) yxv PW 30cosm/s5.430sinm/s5.4

+=

So the velocity of theferryrelative to the wateris:

PWFPFW vvv

+=

Note the order of the subscripts: This equation says that the velocity of theferryrelative to the wateris thevelocity of theferryrelative to thepassengerPLUS the velocity of thepassengerrelative to the water. Now,

theres one hitch in all of this: We arent given the velocity of theferryrelative to thepassenger, FPv

. But we

can figure out what it is very easily. We know that the passenger is moving relative to the ferry with a velocityof 1.5 m/s due north. So suppose for a moment that you are the passenger. If you move northat 1.5 m/srelative to the ferry, what does it look like theferryis doing, from your perspective? Answer: It looks like the

ferry is moving at 1.5 m/s south! So the velocity of the ferry relative to the passenger is:

( )yvFP m/s5.1=

So the velocity of theferryrelative to the wateris:

( ) ( )yxvFW m/s40.2m/s25.2 +=

In magnitude and direction:

0=v

17.5=

? at t 3.25 sv= =

x

0=x

2sin 9.81sin 17.5 2.95 m/sa g = = =


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( ) ( ) m/s29.340.225.2 22 =+=FWv

8.4625.2

40.2tan 1 =

=

So the direction is 46.8north of west, or 43.2west of north.

55. This problem is similar to Example 3-2. Choosing the positiveydirection to be upstream and the positivexdirection to be toward the right, we know the velocity of the water relative to the ground is:

( )yvWG m/s.82=

.

Furthermore, we are given that the velocity of the jet ski relative to the ground is (in component form):

( ) ( ) yxvJG 0.20sinm/s5.90.20cosm/s5.9

+=

We want the velocity of the jet ski relative to the water, JWv

. We can find it from:

GWJGJW vvv

+= .

Now, by the same reasoning used in the previous problem, the velocity of the groundrelative to the wateris thenegativeof the velocity of the water relative to the ground. That is:

( )yvv WGGW m/s8.2==

So: ( ) ( )yxvJW m/s0.6m/s9.8 +=

The speedof the jet ski relative to the water is the magnitude of this velocity:

( ) ( ) m/s110.69.8 22 =+=JWv

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HW03SolWalker4e