Question : 3
Speed of current is 10% of speed of ship B. At 9:00
AM, ship A was 800 meter ahead of ship B, and ship
B completely crosses the ship A at 9:05 AM.
Calculate in what time ship B crosses ship C when
distance between them is 2300 meters and both are
travelling in same direction, given that sum of length
of ship B and ship A is 100 meters and speed of ship
C is same as ship A and length of ship C and ship A
is also same.
Step-by-step explanation:
It is clear from the diagram that the shortest distance between the ship A and B is PQ.
Here, Sin45
0
=
OQ
PQ
⟹PQ=100×
2
1
=50
2
m
Also,
V
AB
=
V
A
2
+V
B
2
=
10
2
+10
2
=10
2
km/h
So, the time taken to reach shortest path is
t=
V
AB
PQ
=
10
2
50
2
=5h