Drag

For swimmers, drag can be a blessing and a curse. In training, a swimmer want's as much excess drag as she can get in order to become stronger. In a race, however, a swimmer want's to minimize drag at all costs. Even the smallest amount of drag created by arm and leg hair is minimized in order to try to shave off a hundredth of second.
When training, swimmers will sometimes swim with a small parachute tied around their waist in order to create a large amount of drag. Not surprisingly, these small parachutes are typically called drag chutes and look a lot like this. 

http://www.finisinc.com/Swim-Parachute

To calculate the drag force that this chute creates is fairly easy using the following equation

F=.5PAC(v)^2

the force of drag is equal to half the density of water multiplied by the cross sectional surface area of the object multiplied by the drag coefficient multiplied by the velocity squared.

Using this equation and the fact that the density of water is 1000 kg/m^3$\frac{kg}{{m}^{3}$, the drag coefficient is .42, and the radius of the circle is 3in, we can determine that

F = .5(1000 kg/m^3)(0.0103*pi m^2)(0.42)(v)^2

For this problem, we are going to assume that with the drag chute, the swimmer is moving at 1.5 m/s$\frac{}{}$

Keeping this in mind we can say that the force of drag the swimmer is overcoming is

F = 4.877 N

After a swimmer has been swimming for a while with this added force working against them they start to not notice it as much. Their bodies begin to tolerate this extra force and start to build more muscle to overcome it.

Now we must figure out the force of thrust with the drag chute. ${F}_{thrust} = mass\left(acceleration\right)$

$$F = Fd + m * a

Force of thrust is mass times acceleration plus the drag force

F = 4.877 N + 59.65 kg(a)

$$ This would be the equation for the force of thrust with our drag chute on.

$$x = (xo) + (vo)t + 0.5at^2

The final position of an object is equal to the initial position plus the initial velocity multiplied by time plus half the acceleration multiplied by time squared

Substituting in that x is equal to 22.88m and t is equal to 15.25s we get

$$
22.88m = 0.5a(15.25s)^2
a = 0.197 m/s^2

Plugging that into our force of thrust equation we get that
F = 16.63 N${F}_{thrust} = 16.63N$

Now if we remove the extra drag force and tell the swimmer to swim as if that force was still there, how fast would the swimmer go now?
Well by putting our force of thrust while overcoming the chute in our force equation we get

16.63 N = 59.65kg(a)

a = 0.279 m/s^2

$$
Putting that acceleration back into our position equation we get

22.88 m = 0.5(0.279 m/s^2)t^2

t = 12.81s

$$$t = 12.81s$
In order to get from one end of the pool to the other in that amount of time a swimmer would have to be going at an average velocity of 1.79 m/s$$

An almost .3 $$m/s difference between the two velocities is huge and could be the difference between first and last in a race.