A river flows towards the East at 2 m/s. The relative velocity of a fish with the water is 4 m/s towards the north.
If the fish faces the same direction in which it swims calculate the angle of fish's body with the East direction.
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The river is flowing towards the East at 2 m/s. Relative to the water the velocity of the fish is 4 m/s towards the North. To an observer outside the water, the velocity of the fish is the sum of the two velocities.
It has a magnitude of sqrt (2^2 + 4^2) = sqrt(4 + 16) = sqrt 20 = 2*sqrt 5 m/s.
The direction of the the velocity of the fish forms an angle equal to arc tan (4/2) with the East.
arc tan (4/2) = arc tan 2 = 63.43 degrees.
As the fish is facing the direction in which it is swimming, it is at an angle of 63.43 degrees with the East.
The vector that represents the velocity of the river is added to the vector that represents the relative velocity of fish to river's velocity.
These two vectors are perpendicular and they are the legs of the right angle triangle, whose hypotenuse is the resultant of addition of the vectors. The horizontal leg is the velocity of river and the vertical leg is the velocity of fish.
The magnitude of horizontal vector is 2 and the magnitude of vertical leg is 4.
The angle that has to be calculated is made by the hypotenuse to the horizontal leg.
tan x = vertical leg/horizontal leg
tan x = 4/2
tan x = 2 => x = arctan 2 => x = 63.4 degrees.
The requested angle is x = 63.4 degrees.
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