# Ball dropped in viscous oil

1. Jan 13, 2012

### getty102

1. The problem statement, all variables and given/known data
In an experiment to determine the viscosity of some oil, a ball is dropped into some oil. The position of the ball is given by the formula y=Y+At+Be-Ct. At t=0, y=0.070 m, the velocity dy/dt is 0 and the acceleration is 0.0983 m/s2. As t→∞ the velocity approaches 0.480 m/s. What is the value of Y in MKS units?

2. Relevant equations
y=Y+At+Be-Ct
vy(t)=A+-CBe-Ct
ay(t)=C2Be-Ct

3. The attempt at a solution
With four unknowns (Y,A,B,C), I came up with three equations:
B+Y=0.070m
A=CB
BC2=0.0983

I wasn't sure what to do as t→∞

2. Jan 13, 2012

### CanIExplore

Looking at your velocity equation, what happens to the exponential term as t approaches infinity? What does the exponential function look like as its exponent approaches infinity? What if the exponential is in the denominator? Try using a graphing program or wolfram alpha to plot it.

3. Jan 13, 2012

### getty102

as t→∞ the exponent would turn to negative, therefore it would go to 0. But that would give me
vy(∞)=A-CBe-C∞
0.480=A-CB(1)

if A=CB

0.480=CB-CB

and

0.480≠0

4. Jan 13, 2012

### CanIExplore

You said the exponential term would go to zero but you plugged a 1 into your equation. Yes $e^{-t}$ goes to zero as t approaches $\infty$.

5. Jan 13, 2012

### getty102

I skipped a step there, sorry. Yea I figured e^-Ct would go to 0 as t goes to infinity, but the math did not work out. Any more suggestions?

6. Jan 13, 2012

### CanIExplore

If the exponential term goes to 0 in your velocity equation, then you end up with 0.480=A correct? This gives you a 4th equation to solve for your 4 unknowns. Now you just have to do some algebra and solve the system of equations for Y.