NSWMaths Extension 2Syllabus dot point

How does a resistance proportional to speed change rectilinear motion, and what is the terminal velocity of a body falling under gravity with resistance?

Model rectilinear motion under gravity with a resistive force proportional to velocity or to the square of velocity, and determine terminal velocity

A focused answer to the HSC Maths Extension 2 dot point on resisted motion. Newton's second law with resistance proportional to speed, separating variables for velocity, the concept of terminal velocity, and motion of a rising and falling body, with verified worked examples.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Setting up Newton's second law
Terminal velocity
Velocity as a function of time
Velocity as a function of position

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:

m\,a = \sum F.

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:

m\frac{dv}{dt} = mg - kv,

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.

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

mg - kv_T = 0 \quad\Longrightarrow\quad v_T = \frac{mg}{k}.

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.

Velocity as a function of time

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

\int \frac{m\,dv}{mg - kv} = \int dt.

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:

mv\frac{dv}{dx} = mg - kv.

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.

Worked example

A body of mass $m$ falls from rest with resistance $kv$ . Find $v$ as a function of $t$ , and confirm the terminal velocity

Taking down as positive, Newton's second law gives

m\frac{dv}{dt} = mg - kv.

Separate variables:

\int \frac{dv}{mg - kv} = \int \frac{1}{m}\,dt.

Integrate. Let $u = mg - kv$ , so $du = -k\,dv$ :

-\frac{1}{k}\ln|mg - kv| = \frac{t}{m} + C.

At $t = 0$ , $v = 0$ , so $-\tfrac{1}{k}\ln(mg) = C$ . Substituting back:

-\frac{1}{k}\ln\frac{mg - kv}{mg} = \frac{t}{m}.

Solve for $v$ : $\ln\dfrac{mg - kv}{mg} = -\dfrac{kt}{m}$ , so $mg - kv = mg\,e^{-kt/m}$ , giving

v = \frac{mg}{k}\left(1 - e^{-kt/m}\right).

As $t \to \infty$ , $e^{-kt/m} \to 0$ , so $v \to \dfrac{mg}{k}$ , the terminal velocity $v_T = \dfrac{mg}{k}$ . This matches setting $\dfrac{dv}{dt} = 0$ in the original equation.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2021 HSC3 marksAn object of mass 1 kg is projected vertically upwards with an initial velocity of u m/s. It experiences air resistance of magnitude k v^2 newtons, where v is the velocity in m/s and k is a positive constant. Acceleration due to gravity is g m/s^2. Show that the time for the object to reach its maximum height is (1/sqrt(gk)) arctan(u sqrt(k/g)) seconds.

Show worked answer →

Take upwards as positive. While the object rises, both gravity and resistance act downwards, so for unit mass

dv/dt = -g - k v^2 = -(g + k v^2).

Separate variables and integrate from launch (t = 0, v = u) to the top (t = T, v = 0):

dt = -dv/(g + k v^2), so T = integral from 0 to u of dv/(g + k v^2).

Factor out k: 1/(g + k v^2) = (1/k) . 1/((g/k) + v^2). Using integral of dv/(a^2 + v^2) = (1/a) arctan(v/a) with a = sqrt(g/k):

integral of dv/(g + k v^2) = (1/k) . sqrt(k/g) arctan(v sqrt(k/g)) = (1/sqrt(gk)) arctan(v sqrt(k/g)).

Evaluating from 0 to u (the arctan is 0 at v = 0):

T = (1/sqrt(gk)) arctan(u sqrt(k/g)) seconds, as required.

Mark notes: 1 mark for the equation of motion dv/dt = -(g + k v^2), 1 mark for separating and integrating to an arctan form, 1 mark for applying the limits v = u to v = 0 to reach the printed expression.

2024 HSC4 marksTwo particles, A and B, each of mass 1 kg, are in a medium that exerts a resistance to motion equal to k v, where k > 0 and v is the velocity. Both maintain vertical trajectories. The particles are simultaneously projected towards each other with the same speed v0 m s^-1, where 0 < v0 < g/k. Particle A is initially d metres directly above particle B, where d < 2 v0/k. Find the time taken for the particles to meet.

Show worked answer →

Let s = x_A - x_B be the separation (A above B). Subtracting the two equations of motion removes gravity, because both particles feel the same -g.

For each particle, acceleration = -g - k v (the medium resists motion). Differencing:
d^2 s/dt^2 = a_A - a_B = (-g - k v_A) - (-g - k v_B) = -k (v_A - v_B) = -k ds/dt.

Let p = ds/dt. Then dp/dt = -k p, so p = L e^(-k t). Initially A moves down (towards B) at v0 and B moves up at v0, so v_A - v_B = -2 v0 at t = 0, giving L = -2 v0:

ds/dt = -2 v0 e^(-k t).

Integrate: s = (2 v0/k) e^(-k t) + C. At t = 0, s = d, so C = d - 2 v0/k:

s = (2 v0/k) e^(-k t) + d - 2 v0/k.

The particles meet when s = 0:
(2 v0/k) e^(-k t) = 2 v0/k - d, so e^(-k t) = (2 v0 - k d)/(2 v0), giving e^(k t) = 2 v0/(2 v0 - k d).

Therefore t = (1/k) ln( 2 v0/(2 v0 - k d) ).

(The condition d < 2 v0/k makes 2 v0 - k d > 0, so the logarithm is defined and t > 0.)

Mark notes: 1 mark for the equations of motion, 1 mark for differencing to dp/dt = -k p, 1 mark for finding the separation s(t) with correct constants, 1 mark for solving s = 0 to give t = (1/k) ln(2 v0/(2 v0 - k d)).