Two ships A and B are 10km apart on a line
running south to north. Ship A farther north is
streaming west at 20 km/h and ship B is
streaming north at 20 km/h. Their distance of
closest approach and how long do they take
to reach it?
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Ships A and B are moving with same speed 20 km/h in the directions shown in figure. It is two dimensional, two body problem with zero acceleration.
Let us find
v
BA
v
BA
=
v
B
−
v
A
Here, ∣
v
BA
∣=
(20)
2
+(20)
2
=20
2
km/h
i.e.,
v
BA
is20
2
km/h at an angle of 45
∘
from east towards north.
Thus, the given problem can be simplified as:
A is at rest and B is moving with
v
BA
in the direction shown in figure.
Therefore, the minimum distance between the two is
s
min
=AC=ABsin45
∘
=10(
2
1
)km
=5
2
km
And the desired time is:
t=
∣
v
BA
∣
BC
=
20
2
5
2
(BC=AC=5
2
km)
=
4
1
h
=15 mins
Open in answr app
Open_in_app
Ships A and B are moving with same speed 20 km/h in the directions shown in figure. It is two dimensional, two body problem with zero acceleration.
Let us find
v
BA
v
BA
=
v
B
−
v
A
Here, ∣
v
BA
∣=
(20)
2
+(20)
2
=20
2
km/h
i.e.,
v
BA
is20
2
km/h at an angle of 45
∘
from east towards north.
Thus, the given problem can be simplified as:
A is at rest and B is moving with
v
BA
in the direction shown in figure.
Therefore, the minimum distance between the two is
s
min
=AC=ABsin45
∘
=10(
2
1
)km
=5
2
km
And the desired time is:
t=
∣
v
BA
∣
BC
=
20
2
5
2
(BC=AC=5
2
km)
=
4
1
h
=15 mins
Answered by
0
Answer:
5Km
15 minutes
Explanation:
5km ( because they having equal speed the mid point of the line must they meet)
20 km ⇒ 1 h
∴ 5km ⇒ 1/4 hours
= 15 minutes
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