Two ships A and B simultaneously leave a port with velocity of 20 kmph each. A goes in the direction 30 north of west while B goes in 30 north of east. Find the magnitude of the velocity of ship B relative to that of A
Answers
Answer:
Given that θ
A
=45° , and θ
B
=40° ,
as defined in the figure, the velocity vectors (relative to the shore) for ships A and B are given by
v
A
=−(V
A
cos45
o
)
i
^
+(V
A
sin45
o
)
j
^
v
B
=−(V
B
sin40
o
)
i
^
+(V
B
cos40
o
)
j
^
,
with v
A
=24knots and v
B
=28knots.
We take east as +
i
^
and north as
j
^
.
The velocity of ship A relative to ship B is simply given by
v
AB
=
V
A
−
V
B
(a) The relative velocity is
v
AB
=
V
A
−
V
B
=(V
B
sin40
o
−V
A
cos45
o
)
i
^
+(V
B
cos40
o
+V
A
sin45
o
)
j
^
=(1.03knots)
i
^
+(38.4knots)
j
^
the magnitude of which is
∣
v
AB
∣=
(1.03knots)
2
+(38.4knots)
2
≈38.4knots,or38knots
(b) The angle θ
AB
that
v
AB
makes with north is given by
θ
AB
=tan
−1
(
v
AB,y
v
AB,x
)=tan
−1
(
38.4knots
1.03knots
)=1.5
o
which is to say that
v
AB
points 1.5° east of north.
(c) Since the two ships started at the same time, their relative velocity describes at what
rate the distance between them is increasing. Because the rate is steady, we have
t=
∣
v
AB
∣
∣Δr
AB
∣
=
38.4knots
160nauticalmiles
=4.2h.
(d) The velocity
v
AB
does not change with time in this problem, and
r
AB
is in the same
direction as
v
AB
since they started at the same time. Reversing the points of view,
we have
v
AB
=−
v
BA
so that
r
AB
=−
r
BA
(i.e., they are 180° opposite to each other).
Hence, we conclude that B stays at a bearing of 1.5° west of south relative to A during the
journey (neglecting the curvature of Earth).