Linear motion under a variable force

Why a fast object slows differently

• A bicycle usually feels more air resistance when it moves faster. Its drag 阻力 may depend on speed 速率. A different force might depend on position 位置 or time; these are different models.
• A variable force 变力 can produce changing acceleration. Start with Newton's law and set up a differential equation 微分方程 for the motion; constant-acceleration formulae need not apply.

Choose a positive direction first

• Take displacement $x$ positive to the right, and write velocity 速度 $v=dx/dt$. It can be negative. Speed is $|v|$, so a negative velocity does not mean a negative speed.
• Add signed forces on the particle: $m\,dv/dt=\sum F$. For motion to the right, resistance is negative. After reversal, ordinary resistance reverses too; a prescribed force towards a fixed point follows its own direction rule.

Change the variable using the chain rule

• If velocity is written as a function of position along a branch of the motion, the chain rule 链式法则 gives

• This is useful for finding $v$ at a position without first finding time. At a turning point 转折点 $v(x)$ may have a vertical tangent; do not assume $a=0$ merely because $v=0$. Use the original force law there.

Choose the variables for the requested answer

• For a time-dependent force, $m\,dv/dt=F(t)$ often gives $v(t)$ directly. For a position-dependent force, $mv\,dv/dx=F(x)$ often gives $v(x)$ directly.
• If force depends on velocity, either form may help: use $dv/dt$ for a time question and $v\,dv/dx$ for a distance question. These are choices of variables for the same Newton equation, not different laws.

Use each stated condition

• A separable differential equation 可分离微分方程 can be rearranged so each side contains one variable and its differential. Separate 分离 the variables, integrate 积分, include a constant, then use an initial condition 初始条件 to determine it.
• A condition at a specified position is often called a boundary condition 边界条件. Here use only separable 可分离 equations and Pure Mathematics 3 calculus. Check any value divided out: division by a factor that can be zero may remove a constant solution.

A force depending on time can reverse the velocity

• A $2\ \text{kg}$ particle has resultant force $6t\ \text{N}$ to the right. At $t=0$, $v=-1\ \dfrac{\text{m}}{\text{s}}$ and $x=0$. Newton's law gives $2\,dv/dt=6t$, so $v=\tfrac32t^2+C$; the initial velocity gives $C=-1$.
• Now integrate $dx/dt=-1+\tfrac32t^2$. The initial position gives $x=-t+\tfrac12t^3$. There are two integrations and two initial conditions. The particle initially moves left even though acceleration is to the right for $t>0$.

Displacement and distance differ across a turn

• In that example, $v=0$ first at $t_* =\sqrt{2/3}\ \text{s}$. Its position is $x_*=-2t_*/3\ \text{m}$. At $t=2\ \text{s}$, displacement is $x=2\ \text{m}$.
• Total distance is the leftward distance $-x_*$ plus the rightward distance $2-x_*$. Thus distance is $2+4\sqrt{2/3}/3\ \text{m}$, greater than the displacement. Integrating signed velocity gives displacement; split at each reversal to find distance.

A position-dependent force links work to speed

• With $F=mv\,dv/dx$, separation gives $mv\,dv=F(x)\,dx$. Integrating between two known states gives

• The signed area under a force–position graph is work 功. It equals the change in kinetic energy for the resultant force. This relation determines squared speed; use the direction of motion to choose the sign of velocity.

Work through a polynomial force

• A particle of mass $8\ \text{kg}$ starts at $x=0$ with $v=1\ \dfrac{\text{m}}{\text{s}}$. For its outward motion the force is $(x^3+4x)\ \text{N}$. Newton's law gives $8v\,dv/dx=x^3+4x$.
• Separate and integrate: $4v^2=x^4/4+2x^2+C$. Initially $C=4$, so $v^2=x^4/16+x^2/2+1=(x^2/4+1)^2$. The outward velocity is $v=x^2/4+1$; the negative square root would describe a different direction.

A second separation can give position in time

• For that outward branch, $dx/dt=1+x^2/4$. Separate as $dt=4\,dx/(x^2+4)$ and integrate: $t=2\tan^{-1}(x/2)+C$. The initial position gives $C=0$.
• Hence $x=2\tan(t/2)$ on the branch $0\leq t<\pi$. The mathematical model predicts unbounded position as $t\to\pi$ from below. Do not continue the periodic tangent formula across that boundary or assume a real system can sustain the force model indefinitely.

A turning root limits a position formula

• A $1\ \text{kg}$ particle starts at $x=0$ with velocity $2\ \dfrac{\text{m}}{\text{s}}$, under force $-x\ \text{N}$. Newton's law gives $v\,dv/dx=-x$, hence $v^2=4-x^2$ from the initial state.
• It first stops at $x=2\ \text{m}$, where acceleration is $-2\ \dfrac{\text{m}}{\text{s}^2}$, so it turns back. Use $v=+\sqrt{4-x^2}$ before that turn and $v=-\sqrt{4-x^2}$ immediately afterwards. The region $|x|>2\ \text{m}$ would require negative squared speed and is inaccessible at this energy.

Linear resistance gives an exponential speed

• A $2\ \text{kg}$ particle moves horizontally with initial velocity $3\ \dfrac{\text{m}}{\text{s}}$, opposed by resistance of magnitude $4v\ \text{N}$ while $v>0$. Newton's law gives $2\,dv/dt=-4v$, so $dv/v=-2\,dt$.
• Integration gives $\ln v=-2t+C$. The initial value gives $v=3e^{-2t}$. It remains positive for every finite $t\geq0$ and tends to zero as time grows. An exponential is not zero merely because a calculator rounds it to zero.

A limiting distance is not a finite stopping point

• Integrating $v=3e^{-2t}$ with $x(0)=0$ gives $x=\tfrac32(1-e^{-2t})$. Thus the particle travels towards a limiting displacement of $1.5\ \text{m}$.
• It never reaches that limit in finite time under this ideal model. Equivalently $v=3-2x$ on the moving branch, but putting $v=0$ in that relation gives a limit, not a finite arrival time. Check the time law before describing a stop.

Quadratic resistance has a different distance limit

• Suppose horizontal motion satisfies $dv/dt=-v^2/2$ with $v(0)=4$. Separation gives $-1/v=-t/2+C$, so $v=4/(1+2t)$ for $t\geq0$.
• With $x(0)=0$, integration gives $x=2\ln(1+2t)$. Speed tends to zero, but displacement has no finite upper limit. A decreasing speed does not by itself imply a finite total distance.

A cubic resistance gives another power law

• Suppose $dv/dt=-3v^3$ with $v(0)=2$. Separate as $v^{-3}\,dv=-3\,dt$. Integration and the initial value give $1/v^2=6t+1/4$.
• Choose the positive branch: $v=2/\sqrt{1+24t}$. With $x(0)=0$, integration gives $x=(\sqrt{1+24t}-1)/6$. Both this model and the quadratic model slow forever while covering an unbounded distance; their rates of slowing differ.

Falling with linear drag approaches a balance

• Take downward as positive. A mass $m$ falling under gravity and linear resistance $kv$ obeys $m\,dv/dt=mg-kv$. Its terminal velocity 终端速度 is $V=mg/k$, where the resultant is zero.
• For $m=2\ \text{kg}$, $k=4\ \dfrac{\text{N}\,\text{s}}{\text{m}}$ and $g=10\ \dfrac{\text{m}}{\text{s}^2}$, $V=5\ \dfrac{\text{m}}{\text{s}}$. Released from rest, separation gives $v=5(1-e^{-2t})$ and downward distance $y=5t-\tfrac52(1-e^{-2t})$.

Check a constant solution before dividing

• In $dv/dt=(k/m)(V-v)$, dividing by $V-v$ excludes $v=V$. Test it in the original equation: it is a valid constant-velocity solution if that is the initial velocity.
• An initial downward velocity below $V$ increases towards $V$; one above $V$ decreases towards it. At terminal velocity the particle is moving with zero acceleration. It is not at rest, and the model need not reach the limit in finite time.

Separate the falling quadratic-drag equation

• Take downward speed $v\geq0$ with resistance per unit mass $gv^2/V^2$. Then $dv/dt=g(1-v^2/V^2)$. From rest, $0\leq v<V$, so separation gives $dv/(V^2-v^2)=(g/V^2)\,dt$.

  $$\frac{1}{V^2-v^2}=\frac{1}{2V}\left(\frac{1}{V+v}+\frac{1}{V-v}\right).$$

• Partial fractions 部分分式 give $\ln[(V+v)/(V-v)]=2gt/V$ after using $v(0)=0$. Rearranging gives $v=V(e^{2gt/V}-1)/(e^{2gt/V}+1)$. The formula approaches $V$ from below; keep logarithm arguments and the initial branch consistent.

Falling speed can be found from distance instead

• For the same quadratic-drag fall from rest, choose $v\,dv/dy=g(1-v^2/V^2)$ with downward distance $y$. Separation and the initial state give

• With $g=10\ \dfrac{\text{m}}{\text{s}^2}$ and $V=10\ \dfrac{\text{m}}{\text{s}}$, this is $v^2=100(1-e^{-y/5})$. Gravity's lost potential energy does not all become kinetic energy: resistance does negative work.

Upward motion puts gravity and drag together

• Launch a particle vertically upwards with speed $u$. Take upwards as positive on the ascent, so $v>0$. Gravity and resistance both act downwards. For quadratic resistance per unit mass $kv^2$, the equation is $dv/dt=-g-kv^2$.
• Define $V^2=g/k$ to simplify the algebra. Here $V$ is the terminal speed for a later downward fall, not a possible steady upward speed. Using $g-kv^2$ during ascent would reverse the resistance incorrectly.

Find maximum height from the position equation

• On ascent use $v\,dv/dx=-g-kv^2$. Separate and use the launch state in the integration:

• Rearranging gives $v^2=(u^2+V^2)e^{-2kx}-V^2$ on ascent until its first zero. For $g=10$, $k=1/10$ and $u=10$ in consistent SI units, $V=10$ and $v^2=200e^{-x/5}-100$. Maximum height is $H=5\ln2\ \text{m}$. It is less than the no-drag height $u^2/(2g)=5\ \text{m}$.

Find ascent time by an inverse tangent

• Separate $dv/dt=-k(V^2+v^2)$. Integration gives $\tan^{-1}(v/V)=\tan^{-1}(u/V)-gt/V$. Use radians for the inverse tangent in this calculus calculation.
• In the example $u=V=10$ and $g=10$, $v=10\tan(\pi/4-t)$. The first zero gives ascent time $t=\pi/4\ \text{s}$. Stop using this ascent equation at the top; resistance reverses during the descent.

Join the descent to the top state

• At the top, velocity is continuous and equals zero. Restart a downward coordinate $y$ from the top, with downward speed $w\geq0$. The new equation is $dw/dt=g-kw^2$, because resistance now acts upwards.
• The falling-distance result gives $w^2=V^2(1-e^{-2ky})$. The ascent and descent each have a consistent sign convention; record the new initial state rather than carrying a negative ascent speed into an unchanged force equation.

Compare the return speed with the launch speed

• In the example $u=V=10$, $k=1/10$ and $H=5\ln2$, the particle returns to launch height after falling distance $H$. Its downward squared speed is $w^2=100(1-e^{-H/5})=50$.
• The return speed is $5\sqrt2\ \dfrac{\text{m}}{\text{s}}$, less than the launch speed $10\ \dfrac{\text{m}}{\text{s}}$. Gravity has the same potential energy at both endpoints, but resistance has removed mechanical energy during both parts.

A force towards a point is not always drag

• Suppose acceleration is $-(v^2+9)/3$ while a particle is to the right of $O$. It is prescribed to point towards $O$, even after the particle reverses and moves left. It then speeds up leftwards rather than resisting that motion.
• With $x(0)=0$ and $v(0)=3$, separation gives $\tan^{-1}(v/3)=\pi/4-t$, so $v=3\tan(\pi/4-t)$. The particle first stops at $t=\pi/4$. A negative velocity after this time correctly describes the return towards $O$.

Integrate the return branch with its domain

• Integrating that velocity and using $x(0)=0$ gives $x=3\ln[\cos(t-\pi/4)]+\tfrac32\ln2$ before the next return to $O$. At the turn the maximum displacement is $\tfrac32\ln2$.
• The next zero of $x$ is $t=\pi/2$, found from the first return branch, not a later periodic tangent value. At $O$ the direction rule changes; a formula derived for $x>0$ is not automatically valid beyond that event.

Infer a force from a velocity relation

• If a $2\ \text{kg}$ particle has velocity $v=1+x^2$ on its outward branch, then $a=v\,dv/dx=(1+x^2)(2x)$. Its resultant force is $F=ma=4x(1+x^2)$ in consistent SI units.
• To infer unknown parameters, substitute measured states into the integrated model and solve for the constants. Then check another state when available. A force coefficient has units chosen so the whole expression is a force, not necessarily newtons alone.

Check a proposed solution in the original equation

• Differentiate the proposed $v(t)$ or $v(x)$, substitute into Newton's law, and check the initial value. For $v=4/(1+2t)$, differentiation gives $dv/dt=-8/(1+2t)^2=-v^2/2$, as required.
• Check signs, units and domains as well as algebra. A positive drag magnitude must oppose velocity; a squared-speed expression must stay nonnegative; logarithms need valid arguments; a new branch needs its own physical assumptions.

Decide which event ends the current model

• Mark any turning point, crossing of a force-direction boundary, change of resistance law, or endpoint of the allowed motion. Decide whether a zero-speed state occurs at a finite time or only as a limit.
• A terminal solution can disappear when dividing by a zero factor, and a square root can hide a return branch. Restore such possibilities by checking the original equation and the order of events.

Keep the force law tied to the motion