In military exercises, parachutists jump from stationary helicopters and their motion is tracked by sensors tagged to their bodies. One parachutist jumps vertically from a helicopter. At time t seconds, the parachutist is x metres below the helicopter with velocity v ms^{-1}. It experiences a downward gravitational acceleration of 9.8 ms^{-2} and an upward acceleration R due to air resistance, where R < 9.8. R is directly proportional to v, with a constant of proportionality k > 0.

1. Write down the differential equation from the given model.

Given that $R$ is directly proportional to $v$ with constant $k > 0$, we have $R = kv$.
The rate of change of $v$ is the difference between gravitational acceleration and $R$:
$\frac{\mathrm{d}v}{\mathrm{d}t} = 9.8 - kv$

2. Separate the variables and integrate to find the general solution.

$\int \frac{1}{9.8 - kv} \, \mathrm{d}v = \int 1 \, \mathrm{d}t$
$-\frac{1}{k} \ln |9.8 - kv| = t + C$
Since $R < 9.8$, we have $kv < 9.8$, so $9.8 - kv > 0$. We can drop the absolute value:
$-\frac{1}{k} \ln(9.8 - kv) = t + C$

3. Apply the initial condition to find the constant of integration.

Assuming the parachutist jumps from rest from a stationary helicopter, $v = 0$ when $t = 0$.
$-\frac{1}{k} \ln(9.8 - k(0)) = 0 + C$
$C = -\frac{1}{k} \ln 9.8$

4. Rearrange the equation to make $v$ the subject.

Substitute $C$ back into the equation:
$-\frac{1}{k} \ln(9.8 - kv) = t - \frac{1}{k} \ln 9.8$
$\ln(9.8 - kv) - \ln 9.8 = -kt$
$\ln \left( \frac{9.8 - kv}{9.8} \right) = -kt$
$\frac{9.8 - kv}{9.8} = e^{-kt}$
$9.8 - kv = 9.8e^{-kt}$
$kv = 9.8(1 - e^{-kt})$
$v = \frac{9.8}{k}(1 - e^{-kt})$

1. Determine the terminal velocity by considering the behaviour of $v$ as time $t$ approaches infinity.

Using the expression for $v$ derived in the previous part, $v = \frac{9.8}{k}\left(1 - e^{-kt}\right)$.

As $t \to \infty$, since $k > 0$, the term $e^{-kt} \to 0$.

Alternatively, terminal velocity occurs when the acceleration is zero (i.e. $\frac{dv}{dt} = 0$), giving $9.8 - kv = 0$.

In both cases, the terminal velocity $v = \frac{9.8}{k}$.