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CHAPTER 4 MOTION IN MORE THAN ONE
DIMENSION
ActivPhysics
can help with these problems:
All Activities in Section 3, Projectile Motion
Section 41:
Velocity and Acceleration
Problem
1. A skater is gliding along the ice at 2.4 m/s, when
she undergoes an acceleration of magnitude
1.1 m/s
2
for 3.0 s. At the end of that time she is
moving at 5.7 m/s. What must be the angle
between the acceleration vector and the initial
velocity vector?
Solution
For constant acceleration, Equation 43 shows that the
vectors
v
0
,
a
Δ
t
and
v
form a triangle as shown. The
law of cosines gives
v
2
=
v
2
0
+ (
a
Δ
t
)
2
−
2
v
0
a
Δ
t
×
cos(180
◦
−
θ
0
). When the given magnitudes are
substituted, one can solve for
θ
0
: (5
.
7 m/s)
2
=
(2
.
4 m/s)
2
+ (1
.
1 m/s
2
)
2
(3
.
0 s)
2
+ 2(2
.
4 m/s)
×
(1
.
1 m/s
2
)(3
.
0 s) cos
θ
0
,
or cos
θ
0
= 1
.
00 (exactly), and
θ
0
= 0
◦
.
Since
v
0
and
a
are colinear, the change in
speed is maximal.
Problem 1 Solution.
Problem
2. In the preceding problem, what would have been
the magnitude of the skater’s ±nal velocity if the
acceleration had been perpendicular to her initial
velocity?
Solution
When
v
0
and
a
Δ
t
are perpendicular, Equation 43 and
the Phythagorean theorem imply
v
=
r
v
2
0
+ (
a
Δ
t
)
2
=
R
(2
.
4 m/s)
2
+ (1
.
1 m/s
2
)(3
.
0 s)
2
= 4
.
08 m/s
.
Problem
3. An object is moving in the
x
direction at 1.3 m/s
when it is subjected to an acceleration given by
a=
0
.
52
ˆ
m/s
2
.
What is its velocity vector after
4.4 s of acceleration?
Solution
From Equation 43,
v =v
0
+a
Δ
t
= (1
.
30 m/s)
ˆ
ı +
(0
.
52
ˆ
m/s
2
)(4
.
4s) = (1
.
30
ˆ
ı+
2
.
29
ˆ
) m/s
.
Problem
4. An airliner is ²ying at a velocity of 260
ˆ
ı
m/s, when
a wind gust gives it an acceleration of 0
.
38
ˆ
ı+
0
.
72
ˆ
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 de²ected from its original course?
Solution
(a) Equation 43 gives
v
= 260
ˆ
ı
m/s
+
(0
.
38
ˆ
ı+
0
.
72
ˆ
)(m
/
s
2
)(24 s) = (269
ˆ
ı+
17
.
3
ˆ
) m/s
.
(b) Since
v
0
is along the
x
axis, the angular de²ection is just
tan
−
1
(
v
y
/v
x
) = tan
−
1
(17
.
3
/
269) = 3
.
67
◦
.
Section 42:
Constant Acceleration
Problem
5. The position of an object as a function of time is
given by
r =
(3
.
2
t
+ 1
.
8
t
2
)
ˆ
ı+
(1
.
7
t
−
2
.
4
t
2
)
ˆ
m,
where
t
is the time in seconds. What are the
magnitude and direction of the acceleration?
Solution
One can always ±nd the acceleration by taking the
second derivative of the position,
a
(
t
) =
d
2
r
(
t
)
/dt
2
.
However, collecting terms with the same power of
t,
one can write the position in meters as
r
(
t
) =
(3
.
2
ˆ
ı+
1
.
7
ˆ
)
t
+ (1
.
8
ˆ
ı

2
.
4
ˆ
)
t
2
.
Comparison with
Equation 44 shows that this represents motion with
constant acceleration equal to twice the coe³cient of
the
t
2
term, or
a
= (3
.
6
ˆ
ı

4
.
8
ˆ
) m/s
2
.
The magnitude
and direction of
a
are
r
(3
.
6)
2
+ (
−
4
.
8)
2
m
/
s
2
=
4
.
49 m/s
2
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