CHAPTER 4 PROBLEM SOLUTIONS
Problem
4. An airliner is flying at a velocity of
260
î
m/s
, when a wind gust gives it an acceleration of
0
.
38
î
+
0
.
72
ˆ
j
m
/
s
2
for a
period of 24 s. (a) What is its velocity at the end of that time? (b) By what angle has it been deflected from its original
course?
Solution
(a) Equation 4-3 gives
v
=
260
î
m
/
s
+
(0.38
î
+
0.72
ˆ
j
)(
m
/
s
2
24 s
)
=
(
269
î
+
17
.
3
ˆ
j
)
m/s.
(b) Since
v
0
is along the
x
-axis, the
angular deflection is just
tan
!
1
(
v
y
=v
x
)
=
tan
!
1
(17.3
=
269)
=
3.67
°
.
Problem
6. An airplane heads northeastward down a runway, accelerating from rest at the rate of 2.1 m/s
2
. Express the plane’s
velocity and position at
t
=
30 s
in unit vector notation, using a coordinate system with
x
-axis eastward and
y
-axis
northward, and with origin at the start of the plane’s takeoff roll.
Solution
Since the acceleration is constant and the airplane starts from rest
(
v
0
=
0)
at the origin
(
r
0
=
0),
Equations 4-3 and 4-4
give
v
(
t
)
=
a
t
and
r
(
t
)
=
1
2
a
t
2
.
Both vectors are in the NE direction, parallel to
a
=
(
2
.
1 m
/
s
2
î
cos
45
° +
ˆ
j
sin
45
°
)
=
(
1
.
48 m
/
s
2
î +
ˆ
j
).
Thus,
v
(
30 s)
=
(
1
.
48 m
/
s
2
30 s
î
+
ˆ
j
)
=
(
44
.
5 m
/
s
)(
î
+
ˆ
j
),
and
r
(
30 s
)
=
1
2
(
1
.
48 m/s)(30 s)
2
(
î
+
ˆ
j
)
=
(
668 m)
(
î
+
ˆ
j
).
(Note:
(
î
+
ˆ
j
)
=
is a unit vector in the NE direction.)
Problem
13. Figure 4-27 shows a cathode-ray tube, used to display electrical signals in oscilloscopes and other scientific instruments.
Electrons are accelerated by the electron gun, then move down the center of the tube at
2.0
!
10
9
cm/s.
In the 4.2-cm-
long deflecting region they undergo an acceleration directed perpendicular to the long axis of the tube. The acceleration
“steers” them to a particular spot on the screen, where they produce a visible glow. (a) What acceleration is needed to
deflect the electrons through 15
°
, as shown in the figure? (b) What is the shape of an electron’s path in the deflecting
region?
Solution
(a) With
x-y
axes as drawn on Fig. 4-27, the electrons emerge from the deflecting region with velocity
v
=
v
0
î
+
at
ˆ
j
,
after a time
t
=
x
0
,
where
x
=
4.2 cm
and
v
0
=
2
!
10
9
cm/s.
The angle of deflection (direction of
v
) is 15
°
, so
tan15
° =
v
y
x
=
at
0
=
ax
0
2
.
Thus,
a
=
v
0
2
tan15
°
=
x
=
2.55
!
10
17
cm/s
2
(when values are substituted). (b) Since the
acceleration is assumed constant, the electron trajectory is parabolic in the deflecting region.

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