Solution manual fundamentals of physics extended, 8th editionch04

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1. (a) The magnitude of r is

5.02 + ( −3.0) 2 + 2.02 = 6.2 m.

(b) A sketch is shown. The coordinate values are in meters.


2. Wherever the length unit is not specified (in this solution), the unit meter should be
understood.
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(a) The position vector, according to Eq. 4-1, is r = ( − 5.0 m) ˆi + (8.0 m)jˆ .

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(b) The magnitude is |r | =

x 2 + y 2 + z 2 = ( −5.0) 2 + (8.0) 2 + 0 2 = 9.4 m.

(c) Many calculators have polar ↔ rectangular conversion capabilities which make this
computation more efficient than what is shown below. Noting that the vector lies in the
xy plane, we are using Eq. 3-6:
§ 8.0 ·

θ = tan −1 ¨
¸ = − 58° or 122°
© −5.0 ¹
where we choose the latter possibility (122° measured counterclockwise from the +x
direction) since the signs of the components imply the vector is in the second quadrant.
(d) In the interest of saving space, we omit the sketch. The vector is 32° counterclockwise
from the +y direction, where the +y direction is assumed to be (as is standard) +90°
counterclockwise from +x, and the +z direction would therefore be “out of the paper.”

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&
& &
&
ˆ Therefore,
(e) The displacement is ∆r = r ' – r where r is given in part (a) and r ' = 3.0i.
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∆r = 8.0iˆ − 8.0jˆ (in meters).

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(f) The magnitude of the displacement is | ∆r | = (8.0) 2 + ( − 8.0) 2 = 11 m.

(g) The angle for the displacement, using Eq. 3-6, is found from
§ 8.0 ·
tan −1 ¨
¸ = − 45° or 135°
© −8.0 ¹

where we choose the former possibility (-45°, which means 45° measured clockwise from
+x, or 315° counterclockwise from +x) since the signs of the components imply the
vector is in the fourth quadrant.


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& &
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3. The initial position vector ro satisfies r − ro = ∆r , which results in

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ˆ − (2.0iˆ − 3.0ˆj + 6.0 k)
ˆ = − 2.0 ˆi + 6.0 ˆj − 10 kˆ
ro = r − ∆r = (3.0ˆj − 4.0k)

where the understood unit is meters.


4. We choose a coordinate system with origin at the clock center and +x rightward
(towards the “3:00” position) and +y upward (towards “12:00”).
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(a) In unit-vector notation, we have (in centimeters) r1 = 10i and r2 = − 10j. Thus, Eq.
4-2 gives
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∆ r = r2 − r1 = − 10iˆ − 10ˆj .

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Thus, the magnitude is given by | ∆ r |= (− 10) 2 + (− 10) 2 = 14 cm.

(b) The angle is
§ − 10 ·
¸ = 45 ° or − 135 ° .
© − 10 ¹

θ = tan − 1 ¨

We choose − 135 ° since the desired angle is in the third quadrant. In terms of the
& &
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magnitude-angle notation, one may write ∆ r = r2 − r1 = − 10iˆ − 10ˆj → (14 ∠ − 135 °).
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(c) In this case, r1 = − 10j and r2 = 10j, and ∆r = 20j cm. Thus, | ∆ r |= 20 cm.

(d) The angle is given by
§ 20 ·
¸ = 90 ° .
© 0 ¹

θ = tan − 1 ¨

(e) In a full-hour sweep, the hand returns to its starting position, and the displacement is
zero.
(f) The corresponding angle for a full-hour sweep is also zero.


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5. The average velocity is given by Eq. 4-8. The total displacement ∆ r is the sum of
three displacements, each result of a (constant) velocity during a given time. We use a
coordinate system with +x East and +y North.

(a) In unit-vector notation, the first displacement is given by
& §
km · § 40.0 min · ˆ
ˆ
∆r1 = ¨ 60.0
¸¨

¸ i = (40.0 km)i.
h
60
min/h
©
¹©
¹

The second displacement has a magnitude of 60.0
direction is 40° north of east. Therefore,

km
h

⋅

20.0 min
60 min/h

= 20.0 km, and its

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∆r2 = 20.0 cos(40.0°) ˆi + 20.0 sin(40.0°) ˆj = 15.3 ˆi + 12.9 ˆj

in kilometers. And the third displacement is
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km · § 50.0 min · ˆ
§
ˆ
∆r3 = − ¨ 60.0

¸¨
¸ i = ( − 50.0 km) i.
h ¹ © 60 min/h ¹
©

The total displacement is

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&
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∆r = ∆r1 + ∆r2 + ∆r3 = 40.0iˆ +15.3iˆ +12.9 ˆj − 50.0 ˆi = (5.30 km) ˆi + (12.9 km) ˆj.
The time for the trip is (40.0 + 20.0 + 50.0) = 110 min, which is equivalent to 1.83 h. Eq.
4-8 then yields
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§ 5.30 km · ˆ § 12.9 km · ˆ
ˆ
ˆ
vavg = ¨
¸i + ¨
¸ j = (2.90 km/h) i + (7.01 km/h) j.
© 1.83 h ¹
© 1.83 h ¹

The magnitude is
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| vavg |= (2.90) 2 + (7.01) 2 = 7.59 km/h.
(b) The angle is given by
§ 7.01 ·
¸ = 67.5 ° (north of east),

© 2.90 ¹

θ = tan − 1 ¨
or 22.5° east of due north.


6. To emphasize the fact that the velocity is a function of time, we adopt the notation v(t)
for dx / dt.
(a) Eq. 4-10 leads to
v (t ) =

d
ˆ = (3.00 m/s)iˆ − (8.00t m/s) ˆj
(3.00tˆi − 4.00t 2 ˆj + 2.00k)
dt

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ˆ m/s.
(b) Evaluating this result at t = 2.00 s produces v = (3.00iˆ − 16.0j)
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(c) The speed at t = 2.00 s is v = |v | = (3.00) 2 + ( − 16.0) 2 = 16.3 m/s.
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(d) And the angle of v at that moment is one of the possibilities

§ −16.0 ·
tan −1 ¨
¸ = − 79.4° or 101°
© 3.00 ¹

where we choose the first possibility (79.4° measured clockwise from the +x direction, or

281° counterclockwise from +x) since the signs of the components imply the vector is in
the fourth quadrant.


7. Using Eq. 4-3 and Eq. 4-8, we have
ˆ − (5.0iˆ − 6.0jˆ + 2.0k)
ˆ
( − 2.0iˆ + 8.0jˆ − 2.0k)
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ˆ m/s.
vavg =
= ( −0.70iˆ +1.40jˆ − 0.40k)
10


8. Our coordinate system has i pointed east and j pointed north. All distances are in
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kilometers, times in hours, and speeds in km/h. The first displacement is rAB = 483i and
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the second is r = − 966j.
BC

(a) The net displacement is
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&
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rAC = rAB + rBC = (483 km) ˆi − (966 km)jˆ
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which yields | rAC |=

(483) 2 +( − 966) 2 =1.08 × 103 km.

(b) The angle is given by
§ −966 ·
tan −1 ¨
¸ = − 63.4°.
© 483 ¹

We observe that the angle can be alternatively expressed as 63.4° south of east, or 26.6°
east of south.
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(c) Dividing the magnitude of rAC by the total time (2.25 h) gives

483 ˆi − 966jˆ
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vavg =
= 215iˆ − 429ˆj.
2.25

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with a magnitude | vavg |= (215) 2 + (− 429) 2 =480 km/h.
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(d) The direction of vavg is 26.6° east of south, same as in part (b). In magnitude-angle
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notation, we would have vavg = (480 ∠ − 63.4 °).
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(e) Assuming the AB trip was a straight one, and similarly for the BC trip, then | rAB | is the
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distance traveled during the AB trip, and | rBC | is the distance traveled during the BC trip.
Since the average speed is the total distance divided by the total time, it equals

483 + 966
= 644 km / h.
2.25


9. We apply Eq. 4-10 and Eq. 4-16.
(a) Taking the derivative of the position vector with respect to time, we have
d ˆ
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ˆ = 8t ˆj + kˆ
v=
(i + 4t 2 ˆj + t k)
dt

in SI units (m/s).
(b) Taking another derivative with respect to time leads to
d
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ˆ = 8 ˆj
a=
(8t ˆj + k)
dt

in SI units (m/s2).


10. We adopt a coordinate system with i pointed east and j pointed north; the
coordinate origin is the flagpole. With SI units understood, we “translate” the given
information into unit-vector notation as follows:

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ro = 40i
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r = 40j

and
and

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vo = − 10j
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v = 10i.

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(a) Using Eq. 4-2, the displacement ∆ r is

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∆ r = r − ro = − 40 ˆi+40 ˆj.
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with a magnitude | ∆ r |= (− 40) 2 + (40)2 = 56.6 m.
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(b) The direction of ∆ r is

§ ∆y ·
− 1 § 40 ·
¸ = tan ¨
¸ = − 45 ° or 135° .
© ∆x ¹
© − 40 ¹

θ = tan − 1 ¨

Since the desired angle is in the second quadrant, we pick 135° ( 45° north of due west).
& & &
Note that the displacement can be written as ∆ r = r − ro = ( 56.6 ∠ 135 ° ) in terms of the
magnitude-angle notation.
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(c) The magnitude of vavg is simply the magnitude of the displacement divided by the

time (∆t = 30 s). Thus, the average velocity has magnitude 56.6/30 = 1.89 m/s.
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(d) Eq. 4-8 shows that vavg points in the same direction as ∆ r , i.e, 135° ( 45° north of
due west).

(e) Using Eq. 4-15, we have
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v − vo
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aavg =
= 0.333i + 0.333j
∆t

in SI units. The magnitude of the average acceleration vector is therefore
0.333 2 = 0.471 m / s2 .
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(f) The direction of aavg is


§ 0.333 ·

¸ = 45 ° or − 135° .
© 0.333 ¹

θ = tan − 1 ¨

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Since the desired angle is now in the first quadrant, we choose 45° , and aavg points
north of due east.


11. In parts (b) and (c), we use Eq. 4-10 and Eq. 4-16. For part (d), we find the direction
of the velocity computed in part (b), since that represents the asked-for tangent line.
(a) Plugging into the given expression, we obtain
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r

t = 2.00

= [2.00(8) − 5.00(2)]iˆ + [6.00 − 7.00(16)] ˆj = 6.00 ˆi − 106 ˆj

in meters.
(b) Taking the derivative of the given expression produces
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v (t ) = (6.00t 2 − 5.00) ˆi − 28.0t 3 ˆj

where we have written v(t) to emphasize its dependence on time. This becomes, at
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t = 2.00 s, v = (19.0 ˆi − 224 ˆj) m/s.
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(c) Differentiating the v ( t ) found above, with respect to t produces 12.0t ˆi − 84.0t 2 ˆj,

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which yields a =(24.0 ˆi − 336 ˆj) m/s 2 at t = 2.00 s.
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(d) The angle of v , measured from +x, is either

§ −224 ·
tan −1 ¨
¸ = − 85.2° or 94.8°
© 19.0 ¹

where we settle on the first choice (–85.2°, which is equivalent to 275° measured
counterclockwise from the +x axis) since the signs of its components imply that it is in
the fourth quadrant.


1
12. We find t by solving ∆ x = x0 + v0 x t + ax t 2 :
2
1
12.0 = 0 + (4.00)t + (5.00)t 2
2

where ∆x = 12.0 m, vx = 4.00 m/s, and ax = 5.00 m/s2 . We use the quadratic formula and
find t = 1.53 s. Then, Eq. 2-11 (actually, its analog in two dimensions) applies with this
value of t. Therefore, its velocity (when ∆x = 12.00 m) is
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v = v0 + at = (4.00 m/s)iˆ + (5.00 m/s 2 )(1.53 s)iˆ + (7.00 m/s 2 )(1.53 s)jˆ
= (11.7 m/s) ˆi + (10.7 m/s) ˆj.
&
&

Thus, the magnitude of v is | v |= (11.7) 2 + (10.7)2 = 15.8 m/s.
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(b) The angle of v , measured from +x, is

§ 10.7 ·
tan −1 ¨
¸ = 42.6°.
© 11.7 ¹


13. We find t by applying Eq. 2-11 to motion along the y axis (with vy = 0 characterizing
y = ymax ): 0 = (12 m/s) + (−2.0 m/s2)t Ÿ t = 6.0 s. Then, Eq. 2-11 applies to motion
along the x axis to determine the answer: vx = (8.0 m/s) + (4.0 m/s2)(6.0 s) = 32 m/s.
Therefore, the velocity of the cart, when it reaches y = ymax , is (32 m/s)i^.


14. We make use of Eq. 4-16.
(a) The acceleration as a function of time is
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& dv d
a=
=
dt dt

( ( 6.0t − 4.0t ) ˆi + 8.0 ˆj) = ( 6.0 − 8.0t ) ˆi
2

in SI units. Specifically, we find the acceleration vector at t = 3.0 s to be
( 6.0 − 8.0(3.0) ) ˆi = (−18 m/s2 )i.ˆ

b

g

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(b) The equation is a = 6.0 − 8.0t i = 0 ; we find t = 0.75 s.
(c) Since the y component of the velocity, vy = 8.0 m/s, is never zero, the velocity cannot
vanish.
(d) Since speed is the magnitude of the velocity, we have
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v =|v | =

( 6.0t − 4.0t ) + (8.0 )
2 2

2

= 10

in SI units (m/s). We solve for t as follows:
squaring ( 6.0t − 4.0t 2 ) + 64 = 100
2

rearranging ( 6.0t − 4.0t 2 )

2

= 36

taking square root 6.0t − 4.0t 2 = ± 6.0

rearranging 4.0t 2 − 6.0t ± 6.0 = 0
using quadratic formula t =

6.0 ± 36 − 4 ( 4.0 )( ±6.0 )
2 ( 8.0 )

where the requirement of a real positive result leads to the unique answer: t = 2.2 s.


15. Constant acceleration in both directions (x and y) allows us to use Table 2-1 for the
motion along each direction. This can be handled individually (for ∆x and ∆y) or together
with the unit-vector notation (for ∆r). Where units are not shown, SI units are to be
understood.
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(a) The velocity of the particle at any time t is given by v = v0 + at , where v0 is the
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initial velocity and a is the (constant) acceleration. The x component is vx = v0x + axt =
3.00 – 1.00t, and the y component is vy = v0y + ayt = –0.500t since v0y = 0. When the
particle reaches its maximum x coordinate at t = tm, we must have vx = 0. Therefore, 3.00
– 1.00tm = 0 or tm = 3.00 s. The y component of the velocity at this time is

vy = 0 – 0.500(3.00) = –1.50 m/s;
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this is the only nonzero component of v at tm.

(b) Since it started at the origin, the coordinates of the particle at any time t are given by
& &
&

r = v 0 t + 21 at 2 . At t = tm this becomes

(

)

(

)

1
&
2
r = 3.00iˆ ( 3.00 ) + −1.00 ˆi − 0.50 ˆj ( 3.00 ) = (4.50 ˆi − 2.25 ˆj) m.
2


16. The acceleration is constant so that use of Table 2-1 (for both the x and y motions) is
permitted. Where units are not shown, SI units are to be understood. Collision between
particles A and B requires two things. First, the y motion of B must satisfy (using Eq. 2-15
and noting that θ is measured from the y axis)
y=

1
ayt 2 Ÿ
2

30 =

b

g

b

g

1
0.40 cosθ t 2 .
2

Second, the x motions of A and B must coincide:
vt =

1 2
1
a x t Ÿ 3.0t = 0.40 sin θ t 2 .
2
2

We eliminate a factor of t in the last relationship and formally solve for time:
t=

3.0
.
0.20 sin θ

This is then plugged into the previous equation to produce
1
30 = ( 0.40 cos θ )

2

§
·
3.0
¨
¸
© 0.20 sin θ ¹

2

which, with the use of sin2 θ = 1 – cos2 θ, simplifies to
30 =

9.0
cos θ
9.0
Ÿ 1 − cos 2 θ =
cos θ .
2
0.20 1 − cos θ
( 0.20 )( 30 )

We use the quadratic formula (choosing the positive root) to solve for cos θ :
cos θ =

−1.5 + 1.52 − 4 (1.0 )( −1.0 )
2

which yields

θ = cos−1

FG 1 IJ = 60° .
H 2K

=

1
2


17. (a) From Eq. 4-22 (with θ0 = 0), the time of flight is
t=

2h
2(45.0)
=
= 3.03 s.
g
9.80

(b) The horizontal distance traveled is given by Eq. 4-21:
∆x = v0 t = ( 250)( 3.03) = 758 m.

(c) And from Eq. 4-23, we find
v y = gt = ( 9.80)( 3.03) = 29.7 m / s.


18. We use Eq. 4-26

§ v2
·
( 9.5m/s )
v2
= ¨ 0 sin 2θ 0 ¸ = 0 =
9.80m/s 2
© g
¹max g

2

Rmax

= 9.209 m ≈ 9.21m

to compare with Powell’s long jump; the difference from Rmax is only ∆R =(9.21 – 8.95)
= 0.259 m.


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19. We designate the given velocity v = 7.6 i + 6.1 j (SI units understood) as v1 − as
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opposed to the velocity when it reaches the max height v2 or the velocity when it returns
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to the ground v3 − and take v0 as the launch velocity, as usual. The origin is at its launch
point on the ground.

(a) Different approaches are available, but since it will be useful (for the rest of the

problem) to first find the initial y velocity, that is how we will proceed. Using Eq. 2-16,
we have
v12 y = v02y − 2 g ∆y Ÿ (6.1)2 = v02 y − 2(9.8)(9.1)
which yields v0 y = 14.7 m/s. Knowing that v2 y must equal 0, we use Eq. 2-16 again but
now with ∆y = h for the maximum height:
v22 y = v02 y − 2 gh Ÿ

0 = (14.7) 2 − 2(9.8)h

which yields h = 11 m.
(b) Recalling the derivation of Eq. 4-26, but using v0 y for v0 sin θ0 and v0x for v0 cos θ0,
we have

0 = v0 y t −

1 2
gt
2

R = v0 x t
which leads to R = 2v0 x v0 y / g . Noting that v0x = v1x = 7.6 m/s, we plug in values and
obtain R = 2(7.6)(14.7)/9.8 = 23 m.
(c) Since v3x = v1x = 7.6 m/s and v3y = – v0 y = –14.7 m/s, we have
v3 = v32 x + v32 y = (7.6) 2 + (−14.7) 2 = 17 m/s.
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(d) The angle (measured from horizontal) for v3 is one of these possibilities:

§ −14.7 ·
tan −1 ¨
¸ = −63° or 117°

© 7.6 ¹

where we settle on the first choice (–63°, which is equivalent to 297°) since the signs of
its components imply that it is in the fourth quadrant.


20. We adopt the positive direction choices used in the textbook so that equations such as
Eq. 4-22 are directly applicable.
(a) With the origin at the initial point (edge of table), the y coordinate of the ball is given
by y = − 21 gt 2 . If t is the time of flight and y = –1.20 m indicates the level at which the
ball hits the floor, then
t=

2 ( −1.20 )
−9.80

= 0.495s.

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(b) The initial (horizontal) velocity of the ball is v = v0 i . Since x = 1.52 m is the
horizontal position of its impact point with the floor, we have x = v0t. Thus,

v0 =

x 152
.
=
= 3.07 m / s.
t 0.495

21. We adopt the positive direction choices used in the textbook so that equations such as
Eq. 4-22 are directly applicable. The initial velocity is horizontal so that v 0 y = 0 and
v0 x = v0 = 10 m s.
(a) With the origin at the initial point (where the dart leaves the thrower’s hand), the y
coordinate of the dart is given by y = − 21 gt 2 , so that with y = –PQ we have
PQ =

1
2

. g
b9.8gb019

2

= 018
. m.

(b) From x = v0t we obtain x = (10)(0.19) = 1.9 m.


22. (a) Using the same coordinate system assumed in Eq. 4-22, we solve for y = h:
h = y0 + v0 sin θ 0t −

1 2
gt
2

which yields h = 51.8 m for y0 = 0, v0 = 42.0 m/s, θ0 = 60.0° and t = 5.50 s.

(b) The horizontal motion is steady, so vx = v0x = v0 cos θ0, but the vertical component of
velocity varies according to Eq. 4-23. Thus, the speed at impact is
v=

( v0 cos θ 0 )

2

+ ( v0 sin θ 0 − gt ) = 27.4 m/s.
2

(c) We use Eq. 4-24 with vy = 0 and y = H:

bv sinθ g
H=
0

0

2g

2

= 67.5 m.


23. We adopt the positive direction choices used in the textbook so that equations such as
Eq. 4-22 are directly applicable. The coordinate origin is at ground level directly below
the release point. We write θ0 = –30.0° since the angle shown in the figure is measured
clockwise from horizontal. We note that the initial speed of the decoy is the plane’s speed

at the moment of release: v0 = 290 km/h, which we convert to SI units: (290)(1000/3600)
= 80.6 m/s.
(a) We use Eq. 4-12 to solve for the time:
∆x = (v0 cos θ 0 ) t

Ÿ t=

700
= 10.0 s.
(80.6) cos (−30.0°)

(b) And we use Eq. 4-22 to solve for the initial height y0:
1
1
y − y0 = (v0 sin θ 0 ) t − gt 2 Ÿ 0 − y0 = (−40.3)(10.0) − (9.80)(10.0) 2
2
2

which yields y0 = 897 m.


24. We adopt the positive direction choices used in the textbook so that equations such as
Eq. 4-22 are directly applicable. The coordinate origin is throwing point (the stone’s
initial position). The x component of its initial velocity is given by v0 x = v0 cosθ 0 and the
y component is given by v 0 y = v 0 sin θ 0 , where v0 = 20 m/s is the initial speed and θ 0 =
40.0° is the launch angle.
(a) At t = 1.10 s, its x coordinate is

b

gb

g

x = v0 t cos θ 0 = 20.0 m / s 110
. s cos 40.0° = 16.9 m

(b) Its y coordinate at that instant is
y = v0t sin θ 0 −

1 2
1
2
gt = ( 20.0m/s )(1.10s ) sin 40.0° − ( 9.80m/s 2 ) (1.10s ) = 8.21m.
2
2

(c) At t' = 1.80 s, its x coordinate is

b

gb

g

x = 20.0 m / s 180
. s cos 40.0° = 27.6 m.

(d) Its y coordinate at t' is
y = ( 20.0m/s )(1.80s ) sin 40.0° −

1
9.80m/s 2 ) (1.80s 2 ) = 7.26m.
(
2

(e) The stone hits the ground earlier than t = 5.0 s. To find the time when it hits the
ground solve y = v0 t sin θ 0 − 21 gt 2 = 0 for t. We find
t=

b

g

2 20.0 m / s
2 v0
sin 40° = 2.62 s.
sin θ 0 =
g
9.8 m / s2

Its x coordinate on landing is

b

gb

g

x = v0t cos θ 0 = 20.0 m / s 2.62 s cos 40° = 40.2 m

(or Eq. 4-26 can be used).
(f) Assuming it stays where it lands, its vertical component at t = 5.00 s is y = 0.