Resisted Motion — Horizontal
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.
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.
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 $v^2$.
Key facts
- Newton's second law: $m\ddot x = $ net force; for horizontal motion under resistance only, net force $= -mkv$ or $-mkv^2$
- $\dfrac{dv}{dt}$, $v\dfrac{dv}{dx}$ and $\dfrac{d}{dt}(\tfrac12 v^2)$ are equivalent forms of acceleration
- $\dot v = -kv \Rightarrow v = u e^{-kt}$
- $\dot v = -kv^2 \Rightarrow \dfrac{1}{v} = \dfrac{1}{u} + kt$
Concepts
- Why a linear-drag particle never reaches rest in finite time
- Why quadratic drag gives unbounded distance even though $v \to 0$
- When to integrate w.r.t. $t$ versus w.r.t. $x$
Skills
- Solve $\dot v = -kv$ and $\dot v = -kv^2$ for $v(t)$ and $x(t)$
- Use $v\,dv/dx$ to get $v$ as a function of $x$
- Find the limiting distance (linear drag) and check terminal behaviour
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 $v(0) = u$:
- Velocity: $\ln v = -kt + \ln u \;\Rightarrow\; v = u e^{-kt}$.
- Position (with $x(0) = 0$): integrate $v$ to get $x = \dfrac{u}{k}(1 - e^{-kt})$.
- Velocity as a function of $x$: use $v\,dv/dx = -kv$, giving $dv/dx = -k$, so $v = u - kx$.
Worked through the hook: Boat with $u = 20$, $k = 0.1$.
- $v(t) = 20 e^{-0.1 t}$. At $t = 10$: $v = 20 e^{-1} \approx 7.36$ m/s — closer to $5$ than to $10$.
- The boat never strictly stops (exponential is positive), but the total distance $x_\infty = u/k = 20/0.1 = 200$ m is finite.
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. 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 $v(0) = u$:
- Velocity: $\dfrac{1}{v} = \dfrac{1}{u} + kt \;\Rightarrow\; v = \dfrac{u}{1 + ukt}$.
- $v$ in terms of $x$: use $v\,dv/dx = -kv^2$, so $dv/v = -k\,dx$, giving $v = u e^{-kx}$.
- Position: integrate $v(t) = u/(1+ukt)$ to get $x(t) = \dfrac{1}{k}\ln(1 + ukt)$.
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
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 $v = u\,e^{}$ and the limiting distance is $x_\infty = $.
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
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.
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)
Comprehensive answers (click to reveal)
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].
Five timed questions on linear and quadratic drag, velocity-time and velocity-distance forms. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering quick mechanics questions. Lighter alternative to the boss.
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