A ball is released from a great height so that it falls vertically downwards towards the surface of the Earth - AQA - A-Level Maths Mechanics - Question 17 - 2021 - Paper 2

To find the expression for v in terms of t, we start with the differential equation:

$a = \frac{dv}{dt} = g - 0.1v.$

We can rearrange this to separate variables:

$\frac{dv}{g - 0.1v} = dt.$

Integrating both sides:

$\int \frac{1}{g - 0.1v} \, dv = \int 1 \, dt.$

This leads to:

$-10 \ln|g - 0.1v| = t + C,$

where $C$ is the constant of integration. Solving for $v$, we exponentiate and rearrange:

$g - 0.1v = e^{-\frac{t + C}{10}},$

which gives:

$v = 10(g - e^{-\frac{t + C}{10}}).$

Substituting the initial condition when $t = 0$:

• $v = 0$ leads to:

$g - e^{-\frac{C}{10}} = 0 \Rightarrow e^{-\frac{C}{10}} = g.$

Thus, the expression for $v$ in terms of $t$ is:

$v(t) = 10(g - ge^{-0.1t}).$

As $t \to \infty$, in the simple model, the velocity continues to increase indefinitely, given by: $v = 2g \, \text{ms}^{-1}.$

In contrast, in Amy's refined model, the term $e^{-0.1t}$ approaches zero, leading us to:

$v(t) \to 10g \, \text{ms}^{-1} \text{ as t becomes large.}$

This indicates that while Andy's model suggests constant acceleration and an increasing velocity, Amy's model takes into account a deceleration due to air resistance, resulting in a limiting velocity.