Drag Force in pool of water

[Cmur]

Homework Statement

You dive straight down into a pool of water. You hit the water with a speed of 7.0m/s, and your mass is 75kg. Assuming a drag force of the form F_D = (−1.00×10^4kg/s)*v, how long does it take you to reach 2% of your original speed? (Ignore any effects of buoyancy.)

Homework Equations

F_D = (-1.00*10^4kg/s)*v
F_g = m*g
F_net = F_D - F_g
(V_f - V_i) / a = t

The Attempt at a Solution

I have tried various attempts at this solution, all of them giving the wrong answer, and I have one chance left at the question before it gives me 0%. I assumed the general approach would be as follows:

V_i = 7m/s
V_f = 0.14m/s
m = 75kg
F_g = 735N
F_D = (-10000*v) - this part is confusing me, as I'm unsure which velocity value to use, but I have come up with 68600N using the change in initial velocity and final velocity.

F_D = (-10000kg/s * (0.14m/s - 7m/s) = 68600N

Given that F_Net = 68600N - 735N,

m*a = 67865N

a=904.8667m/s^2 (this value seems incredibly large, suggesting I have done something wrong at this point)

Anyways, plugging into my kinematics equation supplied, I get: (0.14m/s - 7m/s)/-904.8667m/s^2 = 0.00758s.

This answer, obviously, is quite wrong. Which value of velocity should I be using in the F_D formula?

 

[Cmur]

I suppose I may as well point out all the answers in time that I have tried, all of which were wrong:

0.0074 seconds
0.37 seconds
0.014 seconds
0.0076 seconds
0.77 seconds

 

[Cmur]

Well, I decided to give one last try at it after reading a similar thread, and it turns out the answer is actually 0.035 seconds to slow down. Seems very absurd, but anyone else with the same Mastering Physics question may as well try this answer.