A student swims at 4m/s. She is swimming from west to east across a river that's is flowing 2.2m/s [S].

A) if the river is 100 m wide, what is the quickest time that she can swim the entire width?
B) how far downstream will she be?
C) if she decides to go directly across the river, what direction should she swim?
D) how long will it take her to cross the river going this direction?

To answer these questions, we need to consider the relative velocities and use vector addition.

A) The quickest time to swim the entire width of the river can be achieved by swimming perpendicular to the current. In this case, the effective velocity of the swimmer will be the vector sum of the swimmer's velocity and the river's current velocity.

To find the effective velocity, we can use the Pythagorean theorem as follows:

Effective velocity = sqrt((swimmer's velocity)^2 + (river's velocity)^2)

= sqrt((4 m/s)^2 + (2.2 m/s)^2)

= sqrt(16 + 4.84)

= sqrt(20.84)

= 4.57 m/s (approx.)

Now, we can calculate the time it takes to cross the river by dividing the width of the river by the effective velocity:

Time = distance / velocity

= 100 m / 4.57 m/s

= 21.88 s (approx.)

Therefore, the quickest time that she can swim the entire width of the river is approximately 21.88 seconds.

B) To find how far downstream she will be, we need to calculate the displacement caused by the river's flow during the time it takes her to cross the river.

Displacement = velocity x time

= 2.2 m/s x 21.88 s

= 48.14 m (approx.)

Therefore, she will be displaced approximately 48.14 meters downstream.

C) If she decides to go directly across the river, she should swim in a direction opposite to the direction of the river's flow. In this case, she should swim from east to west.

D) If she chooses to go directly across the river, the time it takes her to cross can be calculated using the width of the river and her swimming speed. Time = distance / velocity = 100 m / 4 m/s = 25 s.

Therefore, it will take her approximately 25 seconds to cross the river going in the opposite direction of the river's flow.