3.7.
Visualize:
The figure shows the components
v
x
and
v
y
, and the angle
θ
.
Solve:
We have,
sin40 ,
y
v
v
= −
°
or
10 m/s
sin40 ,
v
−
= −
°
or
15.56 m/s.
v
=
Thus the
x
component is
cos40
(15.56 m/s ) cos40
11.9 m/s.
x
v
v
=
° =
° =
Assess:
Note that we had to insert the minus sign manually with
y
v
since the vector is in the fourth quadrant.
3.8.
Visualize:
Solve:
Vector
E
r
points to the left and up, so the components
x
E
and
y
E
are negative and positive, respectively,
according to the Tactics Box 3.1.
(a)
cos
x
E
E
θ
= −
and
sin .
y
E
E
θ
=
(b)
sin
x
E
E
φ
= −
and
cos .
y
E
E
φ
=
Assess:
Note that the role of sine and cosine are reversed because we are using a different angle.
θ
and
φ
are
complementary angles.
3.10.
Visualize:
We will follow the rules given in the Tactics Box 3.1.
Solve:
(a)
(2 km)sin30
1 km
x
r
= −
° = −
(2 km)cos30
1.73 km
y
r
=
° =
(b)
(5 cm/s)sin90
5 cm/s
x
v
= −
° = −
(5 cm/s)cos90
0 cm/s
y
v
=
° =
(c)
2
2
(10 m/s
)sin40
6.43 m/s
x
a
= −
° = −
2
2
(10 m/s
)cos40
7.66 m/s
y
a
= −
° = −
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(d)
(50 N)sin36.9
30.0 N
x
F
=
° =
(50 N)cos36.9
40.0 N
y
F
=
° =
Assess:
The components have the same units as the vectors. Note the minus signs we have manually inserted
according to the Tactics Box 3.1.
3.12.
Visualize:
Solve:
The magnitude of the vector is
2
2
2
2
(
)
(
)
(125 V/m)
(
250 V/m)
280 V/m.
x
y
E
E
E
=
+
=
+ −
=
In the
expression for
,
E
v
the
and
+
means that
ˆ
j
−
ˆ
i
E
r
is in quadrant IV. The angle
θ
is below the positive
x
axis. We have:


1
1
1
250 V/m
tan
tan
tan
2
63.4
125 V/m
y
x
E
E
θ
−
−
−
=
=
=
=
°
Assess:
Since 



y
x
E
E
>
, the angle
θ
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 Spring '06
 staff
 Vectors, Dot Product, Acceleration, Velocity, Displacement, vy

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