7 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, $v = 5 \left( 1 - \mathrm { e } ^ { - 2 t } \right)$.\\
(i) Show that this equation is consistent with the initial and terminal velocities. Calculate the velocity after 0.5 seconds as given by this model.\\
(ii) 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 $t = 0 , v = 0$.\\
(iii) 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.]\\
(iv) 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 $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(v) Which of the two models fits the data better?

\hfill \mbox{\textit{OCR MEI C4 2011 Q7 [18]}}