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

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day