4 A particle is moving vertically downwards in a liquid. Initially its velocity is zero, and after \(t\) seconds it is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Its terminal (long-term) velocity is \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
A model of the particle's motion is proposed. In this model, \(v = 5 \left( 1 - \mathrm { e } ^ { - 2 t } \right)\).
- Show that this equation is consistent with the initial and terminal velocities. Calculate the velocity after 0.5 seconds as given by this model.
- Verify that \(v\) satisfies the differential equation \(\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 2 v\).
In a second model, \(v\) satisfies the differential equation
$$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 0.4 v ^ { 2 }$$
As before, when \(t = 0 , v = 0\).
- Show that this differential equation may be written as
$$\frac { 10 } { ( 5 - v ) ( 5 + v ) } \frac { \mathrm { d } v } { \mathrm {~d} t } = 4$$
Using partial fractions, solve this differential equation to show that
$$t = \frac { 1 } { 4 } \ln \left( \frac { 5 + v } { 5 - v } \right)$$
This can be re-arranged to give \(v = \frac { 5 \left( 1 - \mathrm { e } ^ { - 4 t } \right) } { 1 + \mathrm { e } ^ { - 4 t } }\). [You are not required to show this result.]
- Verify that this model also gives a terminal velocity of \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
Calculate the velocity after 0.5 seconds as given by this model.
The velocity of the particle after 0.5 seconds is measured as \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
- Which of the two models fits the data better?