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.
+5 XP warm-up If $\dfrac{dv}{dt} = -kv$ with $v(0) = v_0$, what is $v(t)$?
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.
+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 $v(x)$.
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.
$\displaystyle \int \frac{dv}{f(v)} = \int dt \;\;\text{or}\;\; \int \frac{v\,dv}{f(v)} = \int dx$
When the question asks for velocity (or time) explicitly as a function of time, write $a$ as $\dfrac{dv}{dt}$ and separate variables:
When the question asks for velocity (or time) explicitly as a function of time, write $a$ as $\dfrac{dv}{dt}$ and separate variables:
The left integral produces some function of $v$, the right gives $t$. Rearrange for $v$ (if possible) and apply the initial condition to find $C$.
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 $v(t)$?
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 $t$.
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:
Note the $v$ in the numerator on the left, this is what makes the integral different. Common simplifications:
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
Worked examples · 3 in a row, reveal as you go
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 $1\text{ m/s}$.
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 $x$.
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.
Fill the gap: To find velocity as a function of position when $a = f(v)$, write $a$ as and separate to get $\displaystyle\int \dfrac{v\,dv}{f(v)} = \int \;?$, where the integrand on the right is .
Misconceptions to fix · the 3 traps that cost marks
Misconceptions to fix · the 3 traps that cost marks
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.
Activities · practice with the ideas
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 $a$?
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.
Multiple choice, then short answer under exam conditions.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. A particle has $\dfrac{dv}{dt} = -2v$ with $v(0) = 5$ m/s. Find $v(t)$. (2 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)
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)
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].
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