A puck slides across a rough table; a boat drifts after the engine cuts out; a parachutist races along a horizontal wire. In every case the resistance grows with speed, so the equation of motion is a separable ODE in $v$. This lesson teaches you to write $\dot v = -kv$ (linear drag) or $\dot v = -kv^2$ (quadratic drag), integrate twice, and read off the long-term behaviour.
+5 XP warm-up A particle on a horizontal surface satisfies $\dfrac{dv}{dt} = -kv$ with $v(0) = u$.
A particle on a horizontal surface satisfies $\dfrac{dv}{dt} = -kv$ with $v(0) = u$. Without integrating what does the sign tell you about the motion? Does $v$ ever reach $0$ exactly? Sketch $v(t)$ as best you can.
+5 XP to read Every horizontal resistance problem rewards two habits: match the form of resistance to the right derivative...
Every horizontal resistance problem rewards two habits: match the form of resistance to the right derivative (use $\dot v = dv/dt$ for time, $v\,dv/dx$ for distance), then separate variables and integrate. Choosing the wrong form is the single biggest cause of error.
The resistance-form-derivative reading: (1) write the resistance ($-kv$ or $-kv^2$), (2) decide whether you want $v$ as a function of $t$ or $x$, (3) pick $dv/dt$ for $v(t)$ or $v\,dv/dx$ for $v(x)$.
Time: $\dfrac{dv}{dt} = -kv$ · Distance: $v\dfrac{dv}{dx} = -kv$ · Quadratic: replace $v$ with $v^2$.
A particle of mass moves horizontally with no driving force; the only horizontal force is resistance of magnitude opposite...
A particle of mass $m$ moves horizontally with no driving force; the only horizontal force is resistance of magnitude $mkv$ opposite to motion. Newton's second law gives:
Separate variables: $\dfrac{dv}{v} = -k\,dt$. Integrate from initial conditions $v(0) = u$:
Worked through the hook: Boat with $u = 20$, $k = 0.1$.
Linear drag (horizontal): $\dot v = -kv \;\Rightarrow\; v = u e^{-kt}$ · Position: $x = \dfrac{u}{k}(1 - e^{-kt})$, limit $x_\infty = u/k$ · $v$ in terms of $x$: $v = u - kx$ (a straight line!) · Particle approaches rest but never reaches it
Pause, copy linear drag results: $v = ue^{-kt}$, $v = u-kx$ (linear in $x$), $x = (u/k)(1-e^{-kt})$, limit $u/k$, and the "never-reaches-rest" property into your book.
Quick check: A particle satisfies $\dot v = -2v$ with $v(0) = 6$. What is $v$ when $t = \tfrac12 \ln 3$?
We just saw linear horizontal drag $\dot v = -kv$: $v = ue^{-kt}$, $v = u - kx$ (linear in $x$!), position $x = (u/k)(1-e^{-kt})$ approaching $u/k$ asymptotically.
We just saw linear horizontal drag $\dot v = -kv$: $v = ue^{-kt}$, $v = u - kx$ (linear in $x$!), position $x = (u/k)(1-e^{-kt})$ approaching $u/k$ asymptotically. That raises a question: under quadratic drag $\dot v = -kv^2$, what is the $v$-vs-$x$ relationship, and does the particle travel a finite or infinite distance? This card answers it → $v = ue^{-kx}$ (exponential in $x$), $x = (1/k)\ln(1+ukt) \to \infty$, distance is unbounded.
For higher speeds (or denser fluids) the resistance grows like $v^2$. The equation of motion becomes:
Separate: $-v^{-2}\,dv = k\,dt$. Integrating with $v(0) = u$:
Terminal behaviour: as $t \to \infty$, $v \to 0$ like $1/t$ (much slower than exponential), and $x(t) \to \infty$, the particle travels arbitrarily far, but ever more slowly.
Quadratic drag (horizontal): $\dot v = -kv^2 \;\Rightarrow\; v = \dfrac{u}{1+ukt}$ · $v$ in terms of $x$: $v = u e^{-kx}$ (exponential in $x$, not in $t$!) · Position: $x = \dfrac{1}{k}\ln(1+ukt) \to \infty$, distance unbounded · Decay rate is algebraic ($\sim 1/t$), not exponential
Pause, copy quadratic drag results: $v = u/(1+ukt)$, $v = ue^{-kx}$ (exponential in $x$), $x = (1/k)\ln(1+ukt)$ (unbounded), and the algebraic-decay ($\sim 1/t$) vs exponential contrast into your book.
Did you get this? True or false: for the model $\dot v = -kv^2$ with $v(0) = u > 0$, the total distance travelled is finite.
Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
A particle moves horizontally with initial speed $u = 12$ m/s. Resistance gives $\dot v = -3v$. Find $v(t)$, $x(t)$ (taking $x(0) = 0$), and the limiting distance.
A puck slides on ice with $\dot v = -\tfrac14 v$ and $v(0) = 8$ m/s. Use $v\,dv/dx$ to find $v$ as a function of position $x$. Hence find $v$ when $x = 16$.
A car coasts horizontally; the only horizontal force is air resistance: $\dot v = -\tfrac{1}{20}v^2$. Initial speed $u = 20$ m/s. Find the time to slow to $10$ m/s, and the distance covered when $v = 10$.
Fill the gap: For linear drag $\dot v = -kv$ with $v(0) = u$, the velocity is $v = u\,e^{p}$ and the limiting distance is $x_\infty = q$. Give $p =$ and $q =$ .
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: for $\dot v = -kv$ with $v(0) = u > 0$, the particle is at rest at $t = \dfrac{1}{k}\ln u$.
Activities · practice with the ideas
Activities · practice with the ideas
A particle has $\dot v = -2v$ with $v(0) = 10$. Find $v(t)$, $x(t)$ with $x(0) = 0$, and the limiting distance.
For $\dot v = -kv$ with $v(0) = u$, use $v\,dv/dx$ to show $v = u - kx$. State the value of $x$ at which $v = 0$ "would" occur.
Quadratic drag: $\dot v = -\tfrac{1}{10}v^2$, $u = 5$. Find $v(t)$ and the time at which $v = 1$.
For $\dot v = -kv^2$ with $v(0) = u$, derive $v = u e^{-kx}$ using $v\,dv/dx$. Hence find the distance at which the speed has halved.
A boat ($u = 6$ m/s) coasts horizontally with $\dot v = -\tfrac{1}{2}v$. After how long has the boat travelled half of its limiting distance?
Odd one out: Three of these are valid expressions for the acceleration of a particle in straight-line motion. Which one is NOT?
Earlier you considered a boat with $\dot v = -0.1 v$, $u = 20$ m/s, predicting whether it ever stops, the speed at $t = 10$, and the total distance.
The boat never strictly stops: $v(t) = 20 e^{-0.1 t} > 0$ for all finite $t$. At $t = 10$, $v = 20/e \approx 7.36$, closer to $5$ than to $10$. The total distance is bounded, $x_\infty = u/k = 200$ m, even though motion technically continues forever. The big idea: linear drag tames the motion into a finite envelope through an unending exponential tail.
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 moves horizontally with $\dot v = -\tfrac14 v$ and initial velocity $u = 8$ m/s. Find $v(t)$ and hence the value of $t$ at which $v = 2$ m/s. (2 marks)
Q2. A particle of unit mass moves horizontally so that $\dot v = -kv^2$ with $v(0) = u$, $x(0) = 0$. Show that $v = \dfrac{u}{1+ukt}$ and find the position $x(t)$. (3 marks)
Q3. A particle of mass $m$ moves horizontally subject to a resistance of magnitude $m k v$, with initial speed $u$. (a) Show that the velocity satisfies $v = u - kx$. (b) Hence find the limiting displacement and explain why this displacement is never reached in finite time. (3 marks)
Activity answers:
1. $v = 10 e^{-2t}$. $x = \int_0^t 10 e^{-2s}\,ds = 5(1 - e^{-2t})$. Limiting distance $x_\infty = 5 = u/k$.
2. $v\,dv/dx = -kv \Rightarrow dv/dx = -k$ (for $v > 0$). Integrate: $v = u - kx$. $v = 0$ "occurs" at $x = u/k$ but is only approached in the limit.
3. $1/v = t/10 + 1/5 = (t+2)/10$, so $v = 10/(t+2)$. $v = 1 \Rightarrow t + 2 = 10 \Rightarrow t = 8$ s.
4. $v\,dv/dx = -kv^2 \Rightarrow dv/v = -k\,dx \Rightarrow v = u e^{-kx}$. Half-speed: $u/2 = u e^{-kx} \Rightarrow x = (\ln 2)/k$.
5. $x_\infty = u/k = 12$. $x(t) = 12(1 - e^{-t/2}) = 6 \Rightarrow e^{-t/2} = 1/2 \Rightarrow t = 2\ln 2$ s.
Q1 (2 marks): Separate $dv/v = -dt/4$ and use $v(0) = 8$: $v = 8 e^{-t/4}$ [1]. Set $v = 2$: $\tfrac{1}{4} = e^{-t/4} \Rightarrow t = 4\ln 4 = 8\ln 2$ s [1].
Q2 (3 marks): $-v^{-2}\,dv = k\,dt$; integrate with $v(0)=u$: $1/v = kt + 1/u$ [1]. Rearrange: $v = u/(1+ukt)$ [1]. Integrate to find $x$: $x = \int_0^t \dfrac{u}{1+uks}\,ds = \dfrac{1}{k}\ln(1+ukt)$ [1].
Q3 (3 marks): (a) $m\,v\,dv/dx = -mkv$, divide by $mv$: $dv/dx = -k$; integrate with $v(0)=u$: $v = u - kx$ [1]. (b) Limit: $v = 0$ at $x = u/k$ [1]. From $v(t) = u e^{-kt} > 0$ for all finite $t$, so $v > 0$ always; hence $x < u/k$ for every finite $t$ (only the limit $t \to \infty$ achieves it) [1].
Sit the module quiz, then close the lesson off.
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Tick when you've finished the practice and review.
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