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hscscience Ext 2 · Y12
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Module 16 · L13 of 16 ~45 min ⚡ +90 XP available

Resisted Motion, Vertical Falling

When a body falls through air or fluid, gravity pulls it down while resistance pushes back. The faster it falls, the harder resistance fights, until the two forces balance and acceleration vanishes. That balance point is the terminal velocity. In this lesson you'll set up the equation of motion $\dot v = g - kv$ (and the quadratic form $\dot v = g - kv^2$), solve it by separation of variables, and answer the canonical HSC question: how long does it take to reach a given fraction of terminal velocity?

Today's hook, A skydiver of mass $m$ falls under gravity with linear air resistance of magnitude $mkv$ (so $k$ has units $\text{s}^{-1}$). Before reading on, predict: (a) what is the terminal velocity in terms of $g$ and $k$? (b) Will the diver reach exactly the terminal velocity in finite time? Compare your answers after card 05.
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You are here

Recall, your gut answer first

+5 XP warm-up A body of mass falls under gravity with linear resistance proportional to its speed.

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Recall, your gut answer first
+5 XP warm-up

A body of mass $m$ falls under gravity with linear resistance proportional to its speed. Take down as positive. Before checking write the equation of motion using Newton's second law, then solve $\dot v = 0$ to find the terminal speed in terms of $g$ and the resistance constant.

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The two moves for resisted-falling problems

+5 XP to read Every vertical resisted-motion problem rewards two habits: choose a sign convention (down positive is standard...

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The two moves for resisted-falling problems
+5 XP to read

Every vertical resisted-motion problem rewards two habits: choose a sign convention (down positive is standard for a falling object), then match the calculus form to the question, use $\dfrac{dv}{dt}$ when time is wanted, $v\dfrac{dv}{dx}$ when distance is wanted. Picking the wrong form turns a one-line separation into a tangled mess.

The convention-equation-form reading: (1) fix downward as positive, (2) write $m\dot v = mg - R(v)$ where $R(v)$ is the resistance, (3) pick $\dfrac{dv}{dt}$ (time question) or $v\dfrac{dv}{dx}$ (displacement question).

Linear: $\dot v = g - kv$  ·  Quadratic: $\dot v = g - kv^2$  ·  Terminal: $\dot v = 0$

Sequence flow for setting up and solving a resisted-motion differential equation.
$v_T = \dfrac{g}{k}$ (linear)$\;\;$ or $\;\; v_T = \sqrt{\dfrac{g}{k}}$ (quadratic)
Sign matters
Take down positive for a falling body so $g > 0$ and resistance acts in the $-$ direction: $\dot v = g - kv$. Get the sign wrong and the terminal velocity comes out negative.
Choose the right form
Question asks for time $\to$ use $\dfrac{dv}{dt}$. Question asks for distance fallen $\to$ use $v\dfrac{dv}{dx}$. Both forms come from the same Newton's law.
Terminal is an asymptote
The body approaches $v_T$ but never reaches it in finite time. Questions ask for time to a fraction (e.g., 90%) of $v_T$, not to $v_T$ itself.
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What you'll master
Know

Key facts

  • Equation of motion for vertical fall with linear resistance: $\dot v = g - kv$
  • Equation of motion with quadratic resistance: $\dot v = g - kv^2$
  • Terminal velocity $v_T$ found by setting $\dot v = 0$
  • $v_T = g/k$ (linear) and $v_T = \sqrt{g/k}$ (quadratic)
Understand

Concepts

  • Why resistance opposes velocity and changes sign if the body reverses
  • Why $v \to v_T$ asymptotically but is never reached in finite time
  • Why $v\dfrac{dv}{dx}$ is the right form for displacement questions
Can do

Skills

  • Set up Newton's second law for a falling body with resistance
  • Solve $\dot v = g - kv$ by separation of variables, applying $v(0)=0$
  • Find the time to reach a given fraction $\alpha v_T$ of terminal velocity
  • Use $v\dfrac{dv}{dx}$ to find distance fallen to reach a target speed
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Key terms
ResistanceA force opposing motion through a medium. Modelled as $kv$ (linear, low speed) or $kv^2$ (quadratic, higher speed). Always opposes velocity.
Terminal velocity ($v_T$)The constant limiting speed reached when gravity exactly balances resistance, so $\dot v = 0$. Approached asymptotically but never attained in finite time.
$\dfrac{dv}{dt}$ formAcceleration as a function of time. Use when the question asks for time elapsed or velocity at a given time. Separate as $\dfrac{dv}{g - kv} = dt$.
$v\dfrac{dv}{dx}$ formAcceleration as a function of displacement. Use when the question asks for distance fallen. Separate as $\dfrac{v\,dv}{g - kv^2} = dx$ (quadratic case integrates cleanly).
Linear resistance ($kv$)$\dot v = g - kv$ with $k > 0$. Terminal $v_T = g/k$. Solution: $v(t) = v_T(1 - e^{-kt})$ starting from rest.
Quadratic resistance ($kv^2$)$\dot v = g - kv^2$. Terminal $v_T = \sqrt{g/k}$. Solution uses partial fractions or $\tanh$: $v(t) = v_T \tanh(\sqrt{gk}\, t)$.
MEX-M1NESA outcome (Applications of Calculus to Mechanics): solves problems involving rectilinear motion with resistance, including motion under gravity and uniform circular motion.
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Setting up Newton's second law and solving for $v(t)$

Take downward as positive.

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Setting up Newton's second law and solving for $v(t)$
core concept

Take downward as positive. Two forces act on a body of mass $m$ falling from rest: gravity $mg$ (down, +) and resistance $R(v)$ (up, $-$, because the body moves down). Newton's second law gives

$$m\dfrac{dv}{dt} = mg - R(v).$$

For the linear case $R(v) = mkv$ (so the constant $k$ has units $\text{s}^{-1}$):

$$\dfrac{dv}{dt} = g - kv.$$

Terminal velocity. Set $\dot v = 0$: $g - kv_T = 0$, so $v_T = g/k$.

Separation of variables (starting from rest, $v(0)=0$):

  1. $\dfrac{dv}{g - kv} = dt$.
  2. Integrate: $-\dfrac{1}{k}\ln|g - kv| = t + C$.
  3. At $t = 0$, $v = 0$: $C = -\dfrac{1}{k}\ln g$.
  4. Combine: $\ln\!\left(\dfrac{g}{g - kv}\right) = kt$, so $g - kv = g\,e^{-kt}$.
  5. Solve: $v(t) = \dfrac{g}{k}\bigl(1 - e^{-kt}\bigr) = v_T\bigl(1 - e^{-kt}\bigr)$.

Worked through the hook. With the diver of mass $m$ and resistance $mkv$, the terminal velocity is $v_T = g/k$. Because $v(t) = v_T(1 - e^{-kt}) < v_T$ for every finite $t$, the diver never reaches the terminal velocity, the curve approaches it asymptotically. HSC questions therefore ask for the time to reach a fraction such as $0.9\,v_T$.

Connecting to proof. The exponential form $v = v_T(1 - e^{-kt})$ shows that as $t \to \infty$, $e^{-kt} \to 0$, so $v \to v_T$. This is a one-line justification you can cite in any HSC working, write "as $t \to \infty$, $e^{-kt} \to 0$, so $v \to v_T = g/k$" rather than re-deriving the limit.

Down positive. $m\dot v = mg - R(v)$. · Linear case: $\dot v = g - kv$; $v_T = g/k$; $v(t) = v_T(1 - e^{-kt})$. · Set $\dot v = 0$ to find $v_T$. · $v \to v_T$ as $t \to \infty$ but never equals it in finite time.

Pause, copy the linear-falling ODE $m\dot v = mg-kv$, $v_T = g/k$, $v(t) = v_T(1-e^{-kt})$, and the rule to set $\dot v = 0$ for $v_T$ into your book.

Quick check: A body falls from rest under gravity with linear resistance, satisfying $\dot v = g - kv$ with $g, k > 0$. What is the terminal velocity?

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Quadratic resistance and the displacement form

We just saw the linear-drag falling ODE $m\dot v = mg-kv$, giving $v_T = g/k$ (linear ratio) and $v(t) = v_T(1-e^{-kt})$, terminal approached but never reached in finite time.

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Quadratic resistance and the displacement form
core concept

We just saw the linear-drag falling ODE $m\dot v = mg-kv$, giving $v_T = g/k$ (linear ratio) and $v(t) = v_T(1-e^{-kt})$, terminal approached but never reached in finite time. That raises a question: for quadratic resistance $R = kv^2$, how do we find the displacement as a function of velocity? This card answers it → use $v\,dv/dx = g-kv^2$, factor as $k(v_T-v)(v_T+v)$, integrate by partial fractions to get $x = (1/2k)\ln(v_T^2/(v_T^2-v^2))$.

At higher speeds resistance is better modelled by $R(v) = mkv^2$, giving

$$\dfrac{dv}{dt} = g - kv^2.$$

Setting $\dot v = 0$: $v_T^2 = g/k$, so $v_T = \sqrt{g/k}$. The factored form makes integration tractable:

$$g - kv^2 = k\bigl(v_T^2 - v^2\bigr) = k(v_T - v)(v_T + v).$$

For distance fallen, swap to the chain-rule form $\dot v = v\dfrac{dv}{dx}$:

$$v\dfrac{dv}{dx} = g - kv^2 \;\;\Longrightarrow\;\; \dfrac{v\,dv}{g - kv^2} = dx.$$

The left side integrates by inspection: let $u = g - kv^2$, then $du = -2kv\,dv$, so $\displaystyle\int \dfrac{v\,dv}{g-kv^2} = -\dfrac{1}{2k}\ln|g - kv^2|$. Cleanly applying $v(0) = 0$ gives

$$x = \dfrac{1}{2k}\ln\!\left(\dfrac{g}{g - kv^2}\right) = \dfrac{1}{2k}\ln\!\left(\dfrac{v_T^2}{v_T^2 - v^2}\right).$$
Common mistake. Students reach for $\dfrac{dv}{dt}$ even when the question asks for the distance fallen. The chain-rule swap to $v\dfrac{dv}{dx}$ avoids having to find $v(t)$ explicitly and then integrate again, it is one integration instead of two.

Quadratic: $\dot v = g - kv^2$; terminal $v_T = \sqrt{g/k}$. · Use $v\dfrac{dv}{dx}$ when distance is wanted, $\dfrac{dv}{dt}$ when time is wanted. · $g - kv^2 = k(v_T - v)(v_T + v)$, factored form helps partial fractions. · $x = \dfrac{1}{2k}\ln\!\left(\dfrac{v_T^2}{v_T^2 - v^2}\right)$ (distance fallen, quadratic case, from rest).

Pause, copy the quadratic-falling ODE $m\dot v = mg-kv^2$, terminal $v_T = \sqrt{g/k}$, the factored form $k(v_T-v)(v_T+v)$, the partial-fraction split, and the displacement formula $x = (1/2k)\ln(v_T^2/(v_T^2-v^2))$ into your book.

Did you get this? True or false: a body falling from rest with linear resistance $\dot v = g - kv$ reaches the terminal velocity $v_T = g/k$ in finite time.

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Worked examples · 3 in a row, reveal as you go

Worked examples · 3 in a row, reveal as you go

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · TIME TO REACH 90% OF TERMINAL VELOCITY

A particle falls from rest with linear resistance $\dot v = g - kv$. Find the time taken to reach $90\%$ of the terminal velocity, expressing the answer in terms of $k$.

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Terminal velocity: $\dot v = 0 \Rightarrow v_T = g/k$. Solution from rest (derived earlier): $v(t) = v_T(1 - e^{-kt})$.
Start by writing $v_T$, the question is framed as a fraction of it. The general solution converts the problem to a one-equation question.
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PROBLEM 2 · QUADRATIC RESISTANCE, DISTANCE FALLEN

A particle falls from rest with $\dot v = g - kv^2$. Find the distance fallen by the time the speed reaches $\tfrac{1}{2} v_T$, where $v_T = \sqrt{g/k}$.

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Distance is wanted, so use $\dot v = v\dfrac{dv}{dx}$: $\;v\dfrac{dv}{dx} = g - kv^2 \Rightarrow \dfrac{v\,dv}{g - kv^2} = dx$.
Choose the form that matches the question. Using $\dot v = dv/dt$ would force you to solve for $v(t)$ then integrate again, two steps instead of one.
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PROBLEM 3 · NON-ZERO INITIAL VELOCITY

A body of mass $m$ is projected downward at speed $u$ where $u < v_T = g/k$, with linear resistance $\dot v = g - kv$. Show that $v(t) = v_T - (v_T - u)e^{-kt}$, and find the value of $u$ for which the body falls with constant velocity from the outset.

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Separate: $\dfrac{dv}{g - kv} = dt \Rightarrow -\dfrac{1}{k}\ln|g - kv| = t + C$. At $t=0$, $v=u$: $C = -\dfrac{1}{k}\ln(g - ku)$.
Same separation as the from-rest case, but the initial condition $v(0) = u$ changes the constant. Be careful $g - ku > 0$ since $u < v_T$.
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Fill the gap: For a body falling from rest with $\dot v = g - kv$, the velocity is $v(t) = v_T(1 - e^{p})$, where the terminal velocity is $v_T = q$. Give $p =$ and $q =$ .

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Misconceptions to fix · the 3 traps that cost marks

Misconceptions to fix · the 3 traps that cost marks

Misconceptions to fix · the 3 traps that cost marks

Trap 01
Wrong sign on resistance
Resistance always opposes velocity. If down is positive and the body falls, resistance is $-kv$ (or $-kv^2$). Writing $\dot v = g + kv$ gives an exponential blow-up and a negative terminal velocity, both unphysical and easy to spot.
Trap 02
"Time to reach terminal velocity"
There is no finite time at which $v = v_T$, the approach is asymptotic. If a question asks for "time to reach terminal velocity", it really means a fraction (e.g., $0.99 v_T$). Read carefully; answer in terms of that fraction.
Trap 03
Wrong calculus form for the question
"Distance fallen" $\Rightarrow$ use $v\dfrac{dv}{dx}$. "Time taken" $\Rightarrow$ use $\dfrac{dv}{dt}$. Picking $dv/dt$ for a distance question forces you to solve $v(t)$ then integrate, possible but slow, and a sign error is more likely.

Did you get this? True or false: for $\dot v = g - kv^2$ the terminal velocity is $v_T = g/k$.

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Revisit your thinking

Activities · practice with the ideas

Activities · practice with the ideas

1

A particle of mass $m$ falls from rest with resistance $mkv$. Write the equation of motion (down positive) and find the terminal velocity in terms of $g$ and $k$.

2

Starting from $v(t) = v_T(1 - e^{-kt})$, find the time taken to reach $\tfrac{1}{2} v_T$ in terms of $k$.

3

For $\dot v = g - kv^2$ falling from rest, derive an expression for the distance fallen $x$ in terms of $v$, $g$ and $k$.

4

A body is projected downwards at $u = 2 v_T$ in a linear-resistance medium with $\dot v = g - kv$. Will the body speed up or slow down? Write the equation for $v(t)$.

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Show that the distance fallen as $v \to v_T$ for the quadratic-resistance case becomes infinite, and explain physically.

Odd one out: Three of these are correct facts about a body falling from rest with $\dot v = g - kv$. Which is NOT?

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Revisit your thinking

Earlier you set up Newton's second law for a falling body with resistance, and you predicted whether the diver could actually reach $v_T$.

The equation $\dot v = g - kv$ gives $v(t) = v_T(1 - e^{-kt})$, so the diver's speed climbs exponentially toward $v_T = g/k$ but never quite gets there: $v < v_T$ for every finite $t$. The HSC question is therefore framed as "time to reach $0.9\,v_T$" or similar, never "time to reach $v_T$". Recognising this distinction is the single most useful sanity check in Module 16.

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Show what you have learned

Multiple choice, then short answer under exam conditions.

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer
ApplyBand 32 marks

Q1. A particle of mass $m$ falls from rest in a medium where the resistance is $mkv$ ($k > 0$). Write the equation of motion using $\dfrac{dv}{dt}$ and state the terminal velocity. (2 marks)

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ApplyBand 43 marks

Q2. For a body falling from rest with $\dot v = g - kv$, show that $v(t) = \dfrac{g}{k}(1 - e^{-kt})$ and hence find the time taken to reach $80\%$ of the terminal velocity. (3 marks)

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AnalyseBand 53 marks

Q3. A particle falls from rest under gravity in a medium where the resistance per unit mass is $kv^2$, so $\dot v = g - kv^2$. (a) Find the terminal velocity $v_T$. (b) Show that the distance fallen as a function of speed is $x = \dfrac{1}{2k}\ln\!\left(\dfrac{v_T^2}{v_T^2 - v^2}\right)$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. Down positive: $m\dot v = mg - mkv \Rightarrow \dot v = g - kv$. Setting $\dot v = 0$: terminal velocity $v_T = g/k$.

2. $v = \tfrac{1}{2}v_T$: $\tfrac{1}{2} = 1 - e^{-kt} \Rightarrow e^{-kt} = \tfrac{1}{2} \Rightarrow t = \dfrac{\ln 2}{k} \approx \dfrac{0.693}{k}$.

3. $v\dfrac{dv}{dx} = g - kv^2$. Separate, $u = g - kv^2$, $du = -2kv\,dv$: $x = -\dfrac{1}{2k}\ln(g - kv^2) + C$. From $v(0)=0,\;x=0$: $C = \dfrac{1}{2k}\ln g$. So $x = \dfrac{1}{2k}\ln\!\left(\dfrac{g}{g - kv^2}\right)$.

4. At $v = 2v_T = 2g/k$: $\dot v = g - k(2g/k) = -g < 0$, so the body slows. General solution with $v(0) = 2v_T$: $v(t) = v_T - (v_T - 2v_T)e^{-kt} = v_T + v_T e^{-kt} = v_T(1 + e^{-kt})$. As $t \to \infty$, $v \to v_T$ from above.

5. From Q3, $x = \dfrac{1}{2k}\ln\!\left(\dfrac{v_T^2}{v_T^2 - v^2}\right)$. As $v \to v_T^-$, $v_T^2 - v^2 \to 0^+$, so the logarithm $\to +\infty$ and $x \to \infty$. Physically the body never reaches $v_T$ in finite time, so it must fall an unbounded distance to approach it.

Q1 (2 marks): Equation $\dot v = g - kv$ [1]. Terminal velocity: $\dot v = 0 \Rightarrow v_T = g/k$ [1].

Q2 (3 marks): Separation: $\dfrac{dv}{g - kv} = dt$; integration: $-\tfrac{1}{k}\ln|g - kv| = t + C$; IC $v(0)=0 \Rightarrow C = -\tfrac{1}{k}\ln g$; rearrange to $v = \dfrac{g}{k}(1 - e^{-kt})$ [2 marks]. For $v = 0.8\,v_T$: $0.8 = 1 - e^{-kt} \Rightarrow e^{-kt} = 0.2 \Rightarrow t = \dfrac{\ln 5}{k}$ [1].

Q3 (3 marks): (a) $\dot v = 0 \Rightarrow kv_T^2 = g \Rightarrow v_T = \sqrt{g/k}$ [1]. (b) Use $\dot v = v\,dv/dx$: $\dfrac{v\,dv}{g - kv^2} = dx$. Sub $u = g - kv^2$: $x = -\dfrac{1}{2k}\ln(g - kv^2) + C$ [1]. From rest: $C = \dfrac{1}{2k}\ln g$; therefore $x = \dfrac{1}{2k}\ln\!\left(\dfrac{g}{g - kv^2}\right) = \dfrac{1}{2k}\ln\!\left(\dfrac{v_T^2}{v_T^2 - v^2}\right)$ [1].