Aswimmer heading directly across a river 200 m wide reaches the opposite bank in 6 min 40 s, during which time she is swept downstream 480m. how fast can she swim in still water

Short method:

Assuming the direction that she is swimming is directly across the river (y-direction), and she is swept downstream (x-direction) by the current in the river. We can directly calculate her y-velocity:

y = v*t

v = y/t = 200/(6*60 + 40) = 0.5 m/s

Long method:

The river’s speed does not contribute to her progress across the river. It only provides an x-direction component. If we want to calculate her component of velocity over the resultant path, we would calculate the resultant distance and velocity.

v_x = 480m/400s = 1.2 m/s

v_x is the speed of the river current

The distance covered is calculated using hypotenuse formula:

d = sqrt(200^2 + 480^2) = 520 m

The resultant velocity is:

v = d/t = 520/400 = 1.3 m/s

However this is still the resultant speed of the swimmer and the river current together. Therefore using the hypotenuse formula again to find for v_x:

v^2 = v_x)^2 + (v_y)^2

v^2 - (v_x)^2 = (v_x)^2

v_y = sqrt [ v^2 - (v_x)^2 ] = sqrt( 1.3^2 - 1.2^2) = 0.5 m/s

This is the swimmers velocity without the current.

0.5 m/s