Relative velocity in inertial frame

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Satvik Pandey
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The swimmer will move in the direction of resultant of his velocity and velocity of water.
Draw the diagram and the try to solve it.
 
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slaw155 said:

The Attempt at a Solution


So here's what I've got so far:
(a) v(swimmer relative to water) = v(swimmer relative to current) + v(current relative to water)
so v^2(swimmer relative to current) = 1.98^2 -0.508^2 -> this gives you v, and then you divide 2230m/v to get time, however, I end up getting the incorrect answer using this working (according to the textbook answer). And I need the correct answer to this to be able to answer (b).

You are using distance = width of the river , but dividing by the net speed of the swimmer .

The velocity of swimmer has two perpendicular components , one across the width of the river ,other along the flow of water .

What is the component of velocity across the width of the river ?
 
Vibhor said:
You are using distance = width of the river , but dividing by the net speed of the swimmer .

The velocity of swimmer has two perpendicular components , one across the width of the river ,other along the flow of water .

What is the component of velocity across the width of the river ?


So the velocity across the width of the river is 1.98m/s. So I would go 2230/1.98 to get the time? And then to get horizontal distance traveled I would multiply this time by 0.508 (the downstream speed)?