→
r
=
13.5
mi
, at a bearing angle of
50.4
o
Step-by-step explanation:
We're asked to find the total displacement, both the magnitude and direction, of the ship after it leaves the port with the given conditions.
First, I'll explain what a bearing is.
A bearing is NOT a regular angle measure; normally, angles are measured anticlockwise from the positive
x
-axis, but bearing angles are measured clockwise from the positive
y
-axis.
Therefore, a bearing of
34.0
o
indicates that this is an angle
90.0
o
−
34.0
o
=
56.0
o
measured normally. We'll use this angle in our calculations.
We're given that the first displacement is
10.4
mi
at an angle of
56.0
o
(as calculated earlier). Let's split this up into its components:
x
1
=
10.4
cos
56.0
o
=
5.82
m
y
1
=
10.4
sin
56.0
o
=
8.62
m
Our second displacement is a simple
4.6
mi
due east, that is, the positive
x
-direction. The components are thus
x
2
=
4.6
mi
y
2
=
0
mi
To find the total displacement from the port, we'll add these two vectors' components and use the distance formula:
Δ
x
=
x
1
+
x
2
=
5.82
mi
+
4.6
mi
=
10.42
mi
Δ
y
=
y
1
+
y
2
=
8.62
mi
+
0
mi
=
8.62
mi
r
=
√
(
x
total
)
2
+
(
y
total
)
2
=
√
(
10.42
mi
)
2
+
(
8.62
mi
)
2
=
13.5
mi
The direction of the displacement vector is given by
tan
θ
=
Δ
y
Δ
x
so the angle is then
θ
=
arctan
(
Δ
y
Δ
x
)
=
arctan
(
8.62
mi
10.42
mi
)
=
39.6
o
The question asked for the bearing angle, which is just this angle subtracted from
90
o
:
Bearing angle
=
90
o
−
39.6
o
=
50.4
o