Graph of acceleration against time?

Question

Considering no air resistance, the acceleration time graph for a free falling object would be a horizontal line as the acceleration remains unchanged. When we do take into account air resistance, the graph becomes a curve with negative gradient. However, to me, this is unclear. I understand why this must be a negative relationship, but why can it not be a negative sloping line.

When we graph acceleration against air resistance, we see this is a negative sloping line. Why must this be the case? I am looking for both an intuitive and mathematical answer.

Answer 1

Air resistance (or drag) is a force proportional to the velocity (or velocity squared) of the body: $F_\text{D}=\beta v$ (or $F_\text{D} = \beta v^2$) for some constant $\beta$. You might consider a proof by contradiction. An acceleration that decreases linearly with time takes the form $a(t) = At + B$ for constants $A\neq 0$ and $B$. Then the velocity must be $v(t) = \frac{1}{2}At^2 + Bt + C$ for another constant $C$. Newton's second law then reads

\begin{align} F = ma &= m(At + B) \\ &= -mg + F_\text{D} = -mg + \beta v \\ &= -mg + \beta\left(\frac{1}{2}At^2 + Bt + C\right) \end{align}

which can't be true! The top line is linear in time, and the last line is quadratic in time. You can check for a drag force proportional to the square of the velocity that you again reach a contradiction. Hence the acceleration cannot increase/decrease linearly in time.

Answer 2

The two examples often cited are with the air resistance proportional to the velocity, $m\frac{dv}{dt} = mg - \alpha v \,\,\bf (1) $, or velocity squared, $m\frac{dv}{dt} = mg - \beta v^2\,\,\bf (2)$.

In both of these cases the velocity $v$ is not linearly dependent on time and thus graphs of acceleration against time will have a negative gradient (decreasing acceleration with time) and non-linear.

When we graph acceleration against air resistance, we see this is a negative sloping line.

Looking at the two equations of motion quoted above this is only true if the air resistance is proportional to the velocity, equation $\bf (1) $.