Drag forces on someone diving into a pool

AI Thread Summary
The discussion centers on calculating the time it takes for a diver, entering a pool at 7.0 m/s, to slow down to 2% of that speed under the influence of a drag force defined as FD = (-1.00×10^4 kg/s)v. The diver's mass is 75 kg, and the forces acting on them are analyzed using the equation ΣF = mg - FD = may. The challenge arises from the fact that the drag force is velocity-dependent, complicating the use of kinematic equations that assume constant acceleration. One participant expresses confusion over the large acceleration value obtained when using the drag force formula, prompting a re-evaluation of the drag coefficient's validity. The discussion emphasizes the need to correctly account for variable forces in motion equations.
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Homework Statement



You dive straight down into a pool of water. You hit the water with a speed of 7.0 m/s, and your mass is 75 kg.

Assuming a drag force of the form FD= (−1.00×104 kg/s)v, how long does it take you to reach 2% of your original speed? (Ignore any effects of buoyancy.)

Homework Equations



\SigmaF=ma
V=Vo+at (?)

The Attempt at a Solution



I summed the forces in the y-direction (down is positive):
\SigmaF=mg-FD=may

2% of 7 m/s = 0.14 m/s

But I don't know what to do from here. How do I solve for ay when the drag force varies with velocity? If ay varies, then how can I use the kinematic equations for constant a?

I figured this out the hard way after plugging in 7.0 m/s for v and solving for a, which is an absurdly large number. (-923 m/s2)
 
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Are you sure that FD=(-1.00x104)v?
This coefficient seems kind of large to be for drag, recheck the numbers.
 
Last edited:
Lazyshot,

I should've placed the negative sign outside the parenthesis, but yes, according to my book and MasteringPhysics, that is the coefficient.

FD=-(1.00x104)v
 
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