Your weak spots

Insights load after your first practice round.

Module 16 · L03 of 16 ~45 min ⚡ +90 XP available

Acceleration as a Function of v

When the resisting force depends on speed (drag, friction proportional to $v$ or $v^2$), acceleration becomes a function of velocity. The trick is choosing the right form of $a$: use $a = \dfrac{dv}{dt}$ when you want $v(t)$, and $a = v\dfrac{dv}{dx}$ when you want $v(x)$. Both are separable first-order ODEs that the Extension 2 student can solve cleanly. Aligned to NESA outcome MEX-M1.

Today's hook — A boat moving at $10\text{ m/s}$ has its engine cut. Water resistance gives a deceleration of $a = -0.2v$ m/s$^2$. Before reading on, predict: (i) does it ever come to rest? (ii) what is the form of $v(t)$? Sketch your guess on $v$ vs $t$ axes and check after Worked Example 1.

0/5QUESTS

Recall — your gut answer first

+5 XP warm-up

If $\dfrac{dv}{dt} = -kv$ with $v(0) = v_0$, what is $v(t)$? Before checking — separate variables, integrate both sides, and apply the initial condition.

auto-saved

The two forms of acceleration

+5 XP to read

When $a$ depends only on $v$, two equivalent expressions are available: $a = \dfrac{dv}{dt}$ gives $v(t)$, and $a = v\dfrac{dv}{dx}$ gives $v(x)$. Choosing the wrong form forces an extra integration; choosing the right one makes the problem one separation away from finished.

The question decides the form: if the unknown is time, use $\dfrac{dv}{dt} = f(v)$. If the unknown is position, use $v\dfrac{dv}{dx} = f(v)$. Both are separable.

$\displaystyle \int \frac{dv}{f(v)} = \int dt \;\;\text{or}\;\; \int \frac{v\,dv}{f(v)} = \int dx$

$\displaystyle a = \frac{dv}{dt} = v\frac{dv}{dx}$

Pick the form that matches the unknown

If the question asks "how long until ...", use $dv/dt$. If it asks "how far until ...", use $v\,dv/dx$. This single decision saves a whole integration.

Both forms are separable

$\dfrac{dv}{f(v)} = dt$ or $\dfrac{v\,dv}{f(v)} = dx$. Once separated, integrate each side and apply the initial condition.

Watch the sign of $f(v)$

Resistive forces oppose motion: $a = -kv$ or $a = -kv^2$. Keep the negative; it determines whether the body slows, stops or just approaches rest asymptotically.

What you'll master

Know

Key facts

$a = \dfrac{dv}{dt} = v\dfrac{dv}{dx}$ — two interchangeable forms
When $a = f(v)$, both forms give separable ODEs
Initial conditions are required to determine the constant of integration
NESA outcome MEX-M1 covers this content

Understand

Concepts

Why "what is unknown?" decides whether to use $dv/dt$ or $v\,dv/dx$
Why resistive motion problems often produce exponential decay
Why some bodies reach a terminal velocity instead of stopping

Can do

Skills

Set up and solve $\dfrac{dv}{dt} = f(v)$ to find $v(t)$
Set up and solve $v\dfrac{dv}{dx} = f(v)$ to find $v(x)$
Apply initial conditions cleanly to evaluate constants

Key terms

AccelerationRate of change of velocity. Expressible as $\dfrac{dv}{dt}$, $\dfrac{d^2x}{dt^2}$ or $v\dfrac{dv}{dx}$ depending on the variable of interest.

Separable equationA first-order ODE that can be rewritten so each variable sits on its own side: $g(v)\,dv = h(t)\,dt$ or $g(v)\,dv = h(x)\,dx$.

Resistive forceA force opposing motion, often modelled as proportional to $v$ (low speed) or $v^2$ (higher speed). Produces $a = -kv$ or $a = -kv^2$.

Terminal velocityThe limiting velocity at which gravity balances resistance, so $a = 0$. Found by solving $f(v_T) = 0$.

Initial conditionA specified value such as $v(0) = v_0$ or $x(0) = 0$, used to determine the constant of integration.

Chain rule (for $a$)$\dfrac{dv}{dt} = \dfrac{dv}{dx}\cdot\dfrac{dx}{dt} = v\dfrac{dv}{dx}$. The bridge between time-dependent and position-dependent forms.

MEX-M1NESA outcome (Applications of Calculus to Mechanics): solves problems involving simple harmonic motion and motion with resistance using $a = dv/dt$ and $a = v\,dv/dx$.

Form 1: $\dfrac{dv}{dt} = f(v)$ → $v(t)$

core concept

When the question asks for velocity (or time) explicitly as a function of time, write $a$ as $\dfrac{dv}{dt}$ and separate variables:

$$\frac{dv}{dt} = f(v) \;\Longrightarrow\; \frac{dv}{f(v)} = dt \;\Longrightarrow\; \int \frac{dv}{f(v)} = \int dt = t + C.$$

The left integral produces some function of $v$, the right gives $t$. Rearrange for $v$ (if possible) and apply the initial condition to find $C$.

Works whenever $f(v) \neq 0$ on the interval of motion.
If $f(v_T) = 0$ at some $v_T$, that value is a terminal velocity — the body approaches it but never crosses it.
Pay attention to the sign of $v$: usually $v > 0$, so $|v| = v$ in any $\ln|v|$ that appears.

Worked through the hook. $\dfrac{dv}{dt} = -0.2v$ ⇒ $\dfrac{dv}{v} = -0.2\,dt$ ⇒ $\ln v = -0.2t + C$. At $t=0$, $v = 10$, so $C = \ln 10$. Hence $v(t) = 10e^{-0.2t}$ m/s — the boat slows exponentially but never quite stops.

Form 1: $\dfrac{dv}{dt} = f(v)$ ⇒ $\displaystyle\int\frac{dv}{f(v)} = t + C$ · Use this when the unknown involves $t$ · Always state the initial condition before evaluating $C$ · $a = -kv$ gives exponential decay; the body never stops

Pause — copy Form 1 $dv/dt = f(v) \Rightarrow \int dv/f(v) = t+C$, the exponential-decay result for $a = -kv$, and the rule that the body never stops under this model into your book.

Quick check: A particle has $\dfrac{dv}{dt} = -kv$ with $v(0) = v_0 > 0$. Which expression correctly gives $v(t)$?

Form 2: $v\dfrac{dv}{dx} = f(v)$ → $v(x)$

core concept

We just saw Form 1: $dv/dt = f(v)$ solved by separating $\int dv/f(v) = t+C$, giving $v(t)$ — used when time is the target; $a = -kv$ gives exponential decay in $t$. That raises a question: what if we want $v$ as a function of position $x$ instead? This card answers it → Form 2: $v\,dv/dx = f(v)$ solved by $\int v\,dv/f(v) = x+C$, giving $v(x)$; for $a = -kv$, this yields $v$ linear in $x$ — the body stops after finite distance.

When the question asks for velocity as a function of position (or distance travelled), use the chain-rule form $a = v\dfrac{dv}{dx}$. Separate variables differently:

$$v\frac{dv}{dx} = f(v) \;\Longrightarrow\; \frac{v\,dv}{f(v)} = dx \;\Longrightarrow\; \int \frac{v\,dv}{f(v)} = x + C.$$

Note the $v$ in the numerator on the left — this is what makes the integral different. Common simplifications:

If $f(v) = -kv$, then $\dfrac{v\,dv}{-kv} = -\dfrac{1}{k}\,dv$, an easy integral giving $v$ linear in $x$.
If $f(v) = -kv^2$, then $\dfrac{v\,dv}{-kv^2} = -\dfrac{dv}{kv}$, giving $\ln v$ linear in $x$.
If $f(v) = g - kv^2$ (gravity with quadratic drag), use the substitution $u = g - kv^2$ ⇒ $du = -2kv\,dv$.

Common mistake. Students often drop the $v$ in the numerator and write $\int dv/f(v) = x + C$. That ignores the chain rule and gives the wrong answer — effectively solving a different ODE. Keep $v\,dv$ together when finding $v(x)$.

Form 2: $v\dfrac{dv}{dx} = f(v)$ ⇒ $\displaystyle\int\frac{v\,dv}{f(v)} = x + C$ · Use this when the unknown involves $x$ · $a = -kv$ here gives $v$ linear in $x$ — the body does stop after finite distance · For $a = g - kv^2$, substitute $u = g - kv^2$

Pause — copy Form 2 $v\,dv/dx = f(v) \Rightarrow \int v\,dv/f(v) = x+C$, the linear-$v$-in-$x$ result for $a = -kv$ (body stops at finite distance), and the $a = g-kv^2$ substitution $u = g-kv^2$ into your book.

Did you get this? True or false: if $v\dfrac{dv}{dx} = -kv$ with $v(0) = v_0$, then $v(x) = v_0 - kx$ until the body comes to rest.

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

PROBLEM 1 · $v(t)$ FROM $a = -kv$

A boat moving at $10\text{ m/s}$ has its engine cut, with subsequent acceleration $a = -0.2v$ (m/s$^2$). Find $v(t)$ and the time for the speed to fall to $1\text{ m/s}$.

Choose the time-form: $\dfrac{dv}{dt} = -0.2v$. Separate: $\dfrac{dv}{v} = -0.2\,dt$.

The unknown involves time, so $\dfrac{dv}{dt}$ is the right form. Resistive sign $-0.2$ is kept exactly.

PROBLEM 2 · $v(x)$ FROM $a = -kv^2$

A particle moves with $a = -0.5v^2$ (m/s$^2$) and $v(0) = 8$ m/s at $x = 0$. Find $v$ as a function of $x$.

Choose the position-form: $v\dfrac{dv}{dx} = -0.5v^2$. Divide both sides by $v$ (valid since $v > 0$): $\dfrac{dv}{dx} = -0.5v$.

The unknown involves $x$, so $v\,dv/dx$ is the natural form. The $v$ cancels neatly because $f(v)$ has $v^2$.

PROBLEM 3 · TERMINAL VELOCITY

A body falls under gravity with linear air resistance: $a = g - kv$ where $g, k > 0$ and $v(0) = 0$. Find $v(t)$ and the terminal velocity.

Terminal velocity satisfies $a = 0$: $g - kv_T = 0 \Rightarrow v_T = g/k$. Use $\dfrac{dv}{dt} = g - kv$ ⇒ $\dfrac{dv}{g - kv} = dt$.

Find the equilibrium speed first — it gives the limiting behaviour. Then separate variables in the time-form because we want $v(t)$.

Fill the gap: To find velocity as a function of position when $a = f(v)$, write $a$ as and separate to get $\displaystyle\int \dfrac{v\,dv}{f(v)} = \int $.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Using the wrong form of $a$

If you want $v(x)$ but use $\dfrac{dv}{dt} = f(v)$, you get $v(t)$ first and must integrate again to find $x(t)$, then eliminate $t$. Picking $v\dfrac{dv}{dx} = f(v)$ from the start skips that work.

Trap 02

Dropping the $v$ in $v\,dv/dx$

Some students write $\int dv/f(v) = x + C$ when they should write $\int v\,dv/f(v) = x + C$. The missing $v$ effectively solves a different ODE. Always keep $v\,dv$ as a unit.

Trap 03

Forgetting the initial condition

A general solution with $+C$ is not a final answer. Always state the initial condition explicitly (e.g. $v(0) = v_0$, $x(0) = 0$), substitute, and write down $C$. Markers deduct for missing this step.

Did you get this? True or false: a body with $a = -kv$ ($k > 0$) and positive initial velocity comes to rest after a finite time.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

A particle has $\dfrac{dv}{dt} = -3v$ with $v(0) = 6$. Find $v(t)$ and the time for $v$ to fall to $2$.

For $a = -0.1v^2$ with $v(0) = 5$ m/s at $x = 0$, find $v$ as a function of $x$.

A body falls under gravity with $a = 10 - 0.5v$ (m/s$^2$) and $v(0) = 0$. Find the terminal velocity and write $v(t)$.

A particle is decelerated by $a = -kv$ ($k > 0$), $v(0) = v_0$. Find the total distance travelled before "effective rest" by integrating $v\dfrac{dv}{dx} = -kv$.

Compare the stopping distances for $a = -kv$ and $a = -kv^2$ both starting at $v_0$. Which body covers finite distance? Why?

Odd one out: Three of these correctly express acceleration. Which one is NOT a valid expression for $a$?

Revisit your thinking

At the start you predicted whether a boat decelerating at $a = -0.2v$ ever comes to rest, and sketched $v(t)$.

The answer: $v(t) = 10e^{-0.2t}$ — an exponential decay that approaches zero but never reaches it. The boat keeps moving, more and more slowly, forever (in the model). Whenever the resistance is proportional to $v$, time-decay is always exponential. By contrast, if you had used $a = v\,dv/dx$ here, you would have found $v$ linear in $x$ with a finite stopping distance — same physics, different variable, different shape. Recognising "what's being asked" decides which form to use, and that decision is half the marks.

auto-saved

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.

Short answer

ApplyBand 32 marks

Q1. A particle has $\dfrac{dv}{dt} = -2v$ with $v(0) = 5$ m/s. Find $v(t)$. (2 marks)

auto-saved

ApplyBand 43 marks

Q2. A body decelerates under $a = -kv^2$ ($k > 0$) with $v(0) = v_0$ at $x = 0$. Show that $v(x) = v_0 e^{-kx}$. (3 marks)

auto-saved

AnalyseBand 53 marks

Q3. A skydiver falls with $a = g - kv$, $g = 10$, $k = 0.4$, $v(0) = 0$. (a) Find the terminal velocity. (b) Find $v(t)$. (c) Find the time for $v$ to reach $90\%$ of terminal velocity. (3 marks)

auto-saved

Comprehensive answers (click to reveal)

Activity answers:

1. $\dfrac{dv}{v} = -3\,dt$ ⇒ $\ln v = -3t + \ln 6$ ⇒ $v(t) = 6e^{-3t}$. Set $v = 2$: $e^{-3t} = 1/3$ ⇒ $t = \tfrac{1}{3}\ln 3 \approx 0.366$ s.

2. Use $v\,dv/dx = -0.1v^2$ ⇒ $dv/v = -0.1\,dx$ ⇒ $\ln v = -0.1x + \ln 5$ ⇒ $v(x) = 5e^{-0.1x}$ m/s.

3. Terminal: $10 - 0.5 v_T = 0$ ⇒ $v_T = 20$. Integrate $\dfrac{dv}{10 - 0.5v} = dt$ using $u = 10 - 0.5v$: $-2\ln(10 - 0.5v) = t + C$. At $v(0) = 0$: $C = -2\ln 10$. So $v(t) = 20(1 - e^{-0.5t})$.

4. $v\,dv/dx = -kv$ ⇒ $dv = -k\,dx$ ⇒ $v - v_0 = -kx$. Setting $v = 0$ gives $x = v_0/k$ — the finite stopping distance.

5. $a = -kv$: $v(x) = v_0 - kx$, linear, finite stopping at $x = v_0/k$. $a = -kv^2$: $v(x) = v_0 e^{-kx}$, exponential, never reaches 0. Cancelling the $v$ in $v\,dv/dx$ leaves only $dv$ in the first case but $dv/v$ in the second — that's why one is linear and the other is logarithmic.

Q1 (2 marks): $\dfrac{dv}{v} = -2\,dt$ [1] ⇒ $\ln v = -2t + \ln 5$ ⇒ $v(t) = 5e^{-2t}$ [1].

Q2 (3 marks): $v\,dv/dx = -kv^2$ ⇒ cancel $v$: $dv/dx = -kv$ [1]. Separate $dv/v = -k\,dx$, integrate $\ln v = -kx + C$ [1]. Initial $v(0) = v_0$ gives $C = \ln v_0$, so $\ln(v/v_0) = -kx$ and $v(x) = v_0 e^{-kx}$ [1].

Q3 (3 marks): (a) $v_T = g/k = 10/0.4 = 25$ m/s [1]. (b) Following the same steps as worked example 3, $v(t) = 25(1 - e^{-0.4t})$ [1]. (c) $0.9 \times 25 = 22.5$ ⇒ $1 - e^{-0.4t} = 0.9$ ⇒ $e^{-0.4t} = 0.1$ ⇒ $t = \dfrac{\ln 10}{0.4} \approx 5.76$ s [1].

Boss battle · The Drag Strategist

earn bronze · silver · gold

Five timed questions on $\dfrac{dv}{dt} = f(v)$ and $v\dfrac{dv}{dx} = f(v)$. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering quick resisted-motion questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 02 Lesson 04 · Acceleration as a Function of t →

Module overview · Extension 2