Diver decelerating in water with drag force F_D = cv

[Symstar]

Homework Statement

You dive straight down into a pool of water. You hit the water with a speed of 6.5 m/s, and your mass is 65 kg.

Assuming a drag force of the form $$F_D=cv=(-1.2*10^4 \tfrac{kg}{s})*v$$, how long does it take you to reach 2% of your original speed? (Ignore any effects of buoyancy.)

Homework Equations

$$ma=F_D-mg$$
$$v(t)=v_0+at$$

The Attempt at a Solution

$$ma=F_D-mg$$
$$a=\frac{F_D}{m}-g$$
$$a=\frac{(-1.2*10^4)*-6.5}{65}-9.8=1190.2$$

$$v(t)=v_0+at$$
$$-6.5*.02=-6.5+1190.2t$$
$$t=5.4*10^{-3}$$

Logically, the answer does not make sense, nor is it correct. Where did I go wrong?

 

[Rake-MC]

You have assumed acceleration is constant. In your equation, $$a = \frac{cv}{m} - g$$

And then you used constant acceleration formulae. However, think about what happens as v changes.

 

[Symstar]

You have assumed acceleration is constant. In your equation, $$a = \frac{cv}{m} - g$$

And then you used constant acceleration formulae. However, think about what happens as v changes.

Ah, yes. As velocity decreases, the acceleration does as well. This is problematic for me, however, as I have not worked with non-constant accelerations. How should I go about solving this problem?

 

[Rake-MC]

Try this:
solve for F_net first, before acceleration.

 

[Symstar]

If you mean:
$$F_{net}=cv-mg$$

I can solve for it, but do I use 6.5 as my value for v? I still don't know what to do with the resulting value? Doesn't that still have the same problem of assuming constant acceleration?

 

[Rake-MC]

well think about this:

$$F = m \frac{v}{t}$$so $$m \frac{v_s}{t} = cv - mg$$

where v_s is the one which is at 2%
re arrange to get t.

I hope I didn't make any careless mistakes I was up all night

 

[Symstar]

$$F_{net}=cv-mg=(-1.2*10^4)(-6.5)-65(9.8)=77363$$

$$F_{net}=m\tfrac{v_s}{t}$$
$$77363=65(\tfrac{0.13}{t}$$
$$t=1.09*10^{-4}$$

This is also an incorrect answer =/

 

[Symstar]

Shameless bump - still haven't resolved this question :(

 

[catamara]

You have to use partial derivatives.

FD-mg = ma
cv-mg = m dv/dt
cv/m-g=dv/dt
integral(dt)|0,t = (dv/ (cv/m)-g)

 

[Rake-MC]

Ahh catamara is totally right, we were on the right track though