Resisted motion in HSC Mathematics Extension 2: rectilinear motion under gravity with resistance proportional to velocity or velocity squared, and terminal velocity

What this dot point is asking

NESA wants you to model a particle moving in a straight line under gravity while a resistive force opposes its motion. The resistance is taken proportional to the speed or to the square of the speed. You set up Newton's second law as a differential equation, separate variables to find velocity as a function of time or position, and identify the terminal velocity where acceleration vanishes.

Setting up Newton's second law

For a particle of mass $m$, the net force equals mass times acceleration:

The forces are gravity, of magnitude $mg$, and resistance, which always opposes the direction of motion. Choosing a positive direction is essential and the sign of the resistance term must oppose velocity. Taking downwards as positive for a falling body with resistance proportional to speed:

where $k > 0$ is the resistance constant. For resistance proportional to the square of the speed, replace $kv$ by $kv^2$, again ensuring the sign opposes motion.

The forces and the velocity curve, stage by stage

A falling body with resistance is best understood through the force balance, because the balance is what bends the velocity-time graph into its characteristic shape. The panels below track the forces from release to terminal velocity, then show the curve they produce.

Stage 1, the instant of release. At $t = 0$ the body is at rest, so $v = 0$ and the resistance $kv = 0$. The only force is gravity $mg$ pulling down, so the net force is at its largest and the acceleration is the full $g$. The body starts to speed up exactly as it would in free fall.

Stage 2, falling and speeding up. Once the body moves, the resistance $kv$ acts upward, opposing the downward motion. The net downward force is now $mg - kv$, smaller than gravity alone, so the acceleration has dropped below $g$. The faster the body falls, the larger $kv$ becomes and the smaller the acceleration: the body keeps speeding up, but by less and less.

Stage 3, terminal velocity. Eventually $kv$ grows to equal $mg$. Now the two forces cancel, the net force is zero, and the acceleration is zero. The body cannot speed up any further, so its velocity stays fixed at the terminal velocity $v_T = \dfrac{mg}{k}$. This is the steady state the whole motion is heading towards.

Stage 4, the velocity-time curve. Putting the three force pictures together gives the velocity-time graph. It starts steep (acceleration near $g$), curves over as the resistance builds, and levels off towards the horizontal asymptote $v = v_T$. The body gets arbitrarily close to $v_T$ but never quite reaches it, which is why terminal velocity is always stated as a limit.

Terminal velocity

As a falling body speeds up, the resistance grows until it balances gravity and the acceleration drops to zero. Setting $\dfrac{dv}{dt} = 0$ in $m\dfrac{dv}{dt} = mg - kv$ gives

For the square law $mg - kv^2 = 0$ gives $v_T = \sqrt{\dfrac{mg}{k}}$. The velocity approaches $v_T$ but never reaches it in finite time; it is a horizontal asymptote of the velocity-time graph. A useful sense-check: terminal velocity does not depend on how the body started, only on the balance $mg = kv_T$, so a dropped body and a body thrown downward both tend to the same $v_T$. A body thrown downward faster than $v_T$ would actually slow down, because then the resistance exceeds gravity and the net force points up.

Velocity as a function of time

To find $v(t)$, use $a = \dfrac{dv}{dt}$ and separate variables. From $m\dfrac{dv}{dt} = mg - kv$,

The left integral is a logarithm; solving and applying the initial condition (often $v = 0$ at $t = 0$ for a dropped body) gives an exponential approach to $v_T$, of the form $v = v_T(1 - e^{-kt/m})$.

Velocity as a function of position

When the question asks for $v$ in terms of distance fallen, use $a = v\dfrac{dv}{dx}$ instead:

Separate as $\displaystyle\int \frac{mv\,dv}{mg - kv} = \int dx$ and integrate. The same care with signs and the same use of initial conditions apply. A rising body decelerates under both gravity and resistance, so for upward motion (taking up as positive) the equation is $m\dfrac{dv}{dt} = -mg - kv$, with both forces acting downwards.

Up-and-down problems

A projectile thrown upward and then falling back is two separate problems joined at the highest point. On the way up, both gravity and resistance oppose the upward motion, so the deceleration is large and the equation carries $-mg - kv$ (or $-mg - kv^2$). At the top the velocity is zero. On the way down, gravity drives the motion while resistance opposes it, so the equation becomes $mg - kv$ (or $mg - kv^2$). Because the resistance always opposes the current direction of travel, the sign of the resistance term flips between the two phases. A consequence worth quoting is that the body returns to its starting level with a speed less than its launch speed, since energy is lost to resistance throughout.

Choosing the integral form

As with all rectilinear motion, the choice between $\frac{dv}{dt}$ and $v\frac{dv}{dx}$ is dictated by what the question asks. If you need velocity as a function of time, or the time to reach a height, use $\frac{dv}{dt}$. If you need velocity as a function of distance, or the distance to come to rest, use $v\frac{dv}{dx}$. The resulting integrals are standard: a velocity-only resistance gives a logarithm, while a velocity-squared resistance typically gives a logarithm or an arctangent depending on the sign of the constant term, which is exactly why the inverse-trig and logarithm integrals from the calculus strand are prerequisites here.

How exam questions ask about resisted motion

The wording flags both the model and the tool:

• "A body falls / is dropped and meets a resistance proportional to its speed (or to the square of its speed)." Write $m\dot{v} = mg - kv$ (or $mg - kv^2$), taking down as positive.
• "Show that the terminal velocity is ..." Set $\dot{v} = 0$ and solve $mg = kv_T$ (or $mg = kv_T^2$). One line, one mark.
• "Find $v$ as a function of time" / "how long to reach a given speed." Use $\dfrac{dv}{dt}$, separate, integrate to a logarithm (linear law) or use $\int \frac{dv}{a^2 - v^2}$ / $\int \frac{dv}{a^2 + v^2}$ (square law). Apply the initial condition.
• "Find $v$ as a function of distance fallen" / "the distance to reach a given speed." Use $v\dfrac{dv}{dx}$ and separate instead; time never appears.
• "A body is projected vertically upward against gravity and resistance. Find the time to the top / the greatest height." This is the upward phase: both forces act downward, so $\dot{v} = -(g + kv^2/m)$ and you integrate, often to an arctangent. The top is where $v = 0$.
• "Show the body returns with a smaller speed than it was launched with." An energy or up-then-down argument: resistance removes energy on both legs, so the return speed is below the launch speed (and below terminal velocity).
• "Two bodies move towards each other in a resisting medium." Subtract the two equations of motion to cancel gravity and reduce to a single first-order equation in the separation, as in the 2024 question above.