3.5 - Variable forces depending on displacement | Cambridge A-Level Further Mathematics

When a force depends on position, the useful acceleration form is often $v\,\mathrm{d}v/\mathrm{d}x$ . You will derive that form, use it to connect velocity with displacement, and follow the physically correct branch up to rest and through a turning point.

Acceleration as v dv by dx

If the force is known at each position, can you find the speed there without first finding the time? The chain rule provides exactly that route.

Imagine a $2\,\mathrm{kg}$ particle whose signed force is $4x\,\mathrm{N}$ along a horizontal axis. Newton’s law gives $\mathrm{d}v/\mathrm{d}t=2x$ . Integrating this with respect to time would be awkward because $x$ also changes. Instead, connect the two rates:

\frac{\mathrm{d}v}{\mathrm{d}t} =\frac{\mathrm{d}v}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t} =v\frac{\mathrm{d}v}{\mathrm{d}x}.

Acceleration using position

On a branch of motion where velocity can be expressed as a differentiable function of position,

a=\frac{\mathrm{d}v}{\mathrm{d}t} =v\frac{\mathrm{d}v}{\mathrm{d}x} =\frac12\frac{\mathrm{d}(v^2)}{\mathrm{d}x}.

For a force depending only on position, Newton’s law becomes

mv\frac{\mathrm{d}v}{\mathrm{d}x}=F(x).

Not in the formula booklet: learn it.

Use $\mathrm{d}v/\mathrm{d}t$ when time is the convenient independent variable; use $v\,\mathrm{d}v/\mathrm{d}x$ when position is convenient. Both describe the same signed acceleration. The extra factor $v$ is essential.

The force need not depend only on position for this form to help. Suppose a particle falls from rest under gravity, with resistance of magnitude $mcv^2$ opposite to its motion, where $c>0$ . Take downwards as positive, let $x$ be the distance fallen, and use the positive-velocity falling branch. Weight acts downwards and resistance upwards, so

mv\frac{\mathrm{d}v}{\mathrm{d}x}=mg-mcv^2.

Separating gives an integral with a numerator proportional to the derivative of its denominator:

\begin{aligned} \int\frac{v}{g-cv^2}\,\mathrm{d}v&=\int1\,\mathrm{d}x \\ -\frac{1}{2c}\ln(g-cv^2)&=x+C \end{aligned}

Here $g-cv^2>0$ on the accelerating branch from rest. Using $x=0$ , $v=0$ , then rearranging the logarithm, gives

\ln\left(\frac{g}{g-cv^2}\right)=2cx, \qquad v^2=\frac{g}{c}\bigl(1-e^{-2cx}\bigr).

For $x>0$ , choose the positive square root. The formula extends to the initial rest point by continuity. Its speed stays below $\sqrt{g/c}$ and approaches that force-balance speed as the distance fallen increases indefinitely. This is a logarithmic substitution from ordinary separation: the same acceleration form can answer a distance question even when the force was specified using velocity.

Worked example 1: force away from the origin

Question. A particle of mass $2\,\mathrm{kg}$ moves along a horizontal axis. The signed force is $F=4x\,\mathrm{N}$ , where $x$ is its coordinate in metres. When $x=1$ , its velocity is $3\,\mathrm{m\,s^{-1}}$ in the positive direction. Find its speed when $x=2$ .

Plan. Use $a=v\,\mathrm{d}v/\mathrm{d}x$ , integrate with respect to $x$ , and use the initial position and velocity together.

\begin{aligned} 2v\frac{\mathrm{d}v}{\mathrm{d}x}&=4x \\ \int v\,\mathrm{d}v&=\int2x\,\mathrm{d}x \\ \frac12v^2&=x^2+C \end{aligned}

At $x=1$ , $v=3$ , so $9/2=1+C$ , giving $C=7/2$ . Hence

v^2=2x^2+7.

At $x=2$ , $v^2=15$ . The velocity stays positive on this rightward phase.

Answer. The speed is $\sqrt{15}\,\mathrm{m\,s^{-1}}$ .

Check. The work done from $x=1$ to $x=2$ is $\int_1^2 4x\,\mathrm{d}x=6\,\mathrm{J}$ . The increase in kinetic energy is $\tfrac12(2)(15-9)=6\,\mathrm{J}$ , agreeing with the work–energy principle.

Now reverse only the force direction. It becomes a restoring force towards the origin, so the same rightward displacement reduces the speed.

Worked example 2: force towards the origin

Question. Keep the mass, initial position and initial velocity from Worked example 1, but change the signed force to $F=-4x\,\mathrm{N}$ . Find the speed when $x=2$ .

Plan. Keep the positive axis fixed and carry the negative force sign through the integration.

\begin{aligned} 2v\frac{\mathrm{d}v}{\mathrm{d}x}&=-4x \\ \frac12v^2&=-x^2+C \end{aligned}

The initial data give $9/2=-1+C$ , so $C=11/2$ . Therefore

v^2=11-2x^2.

At $x=2$ , $v^2=3$ . This is positive, so the particle reaches that position before its first rest.

Answer. The speed is $\sqrt3\,\mathrm{m\,s^{-1}}$ .

Check. The work done is now $-6\,\mathrm{J}$ . The kinetic energy falls from $9\,\mathrm{J}$ to $3\,\mathrm{J}$ , a loss of $6\,\mathrm{J}$ . This confirms the force sign and the smaller speed.

The work check also explains the form $\tfrac12m(v^2-u^2)=\int F(x)\,\mathrm{d}x$ : integrating Newton’s law over position measures the change in kinetic energy. You can use that as an alternative route when appropriate.

Watch out

The line $m\mathrm{d}v/\mathrm{d}x=F(x)$ omits a factor of velocity. It also has the wrong dimensions: $\mathrm{d}v/\mathrm{d}x$ has units $\mathrm{s^{-1}}$ , whereas acceleration has units $\mathrm{m\,s^{-2}}$ .

Apply the restoring-force model with new initial data.

Solving for speed and position

The relation for $v^2$ determines speed, but a sign choice is still needed before you can use $\mathrm{d}x/\mathrm{d}t=v$ to find position.

For example, $v^2=4(1+x)^2$ gives speed $2|1+x|$ . On a region where $x>-1$ , this is $2(1+x)$ . The velocity can be either $+2(1+x)$ or $-2(1+x)$ : initial direction decides which applies. Squaring has removed that information.

A branch of the motion

A branch is a portion of the motion on which the chosen velocity sign and force law apply. On a positive-velocity branch, $v=+\sqrt{v^2}$ ; on a negative-velocity branch, $v=-\sqrt{v^2}$ . Speed is always the nonnegative value $\sqrt{v^2}$ .

Once the branch gives $v=f(x)$ , separate $\mathrm{d}x/\mathrm{d}t=f(x)$ to find position. This needs only another first-order separation, rather than a second-order equation.

Worked example 3: the positive velocity branch

Question. A particle of mass $1\,\mathrm{kg}$ has signed horizontal force $4(1+x)\,\mathrm{N}$ in the region $x>-1$ . At $t=0$ , $x=0$ and $v=2\,\mathrm{m\,s^{-1}}$ . Find its velocity in terms of position, and its position in terms of time.

Plan. First find $v^2$ , select the positive branch, then separate the position equation.

\begin{aligned} v\frac{\mathrm{d}v}{\mathrm{d}x}&=4(1+x) \\ \frac12v^2&=4x+2x^2+C \end{aligned}

The initial data give $C=2$ , so

v^2=4x^2+8x+4=4(1+x)^2.

Initially $v>0$ and $x>-1$ , so choose $v=2(1+x)$ . Hence

\begin{aligned} \frac{\mathrm{d}x}{1+x}&=2\,\mathrm{d}t \\ \ln(1+x)&=2t+B \end{aligned}

Since $x(0)=0$ , $B=0$ .

Answer. $v=2(1+x)$ and $x=e^{2t}-1$ , for $t\ge0$ . In particular, $x\ge0$ , so the specified force region remains valid.

Check. Directly differentiating the position twice gives $v=2e^{2t}$ and $a=4e^{2t}$ . The force divided by mass is $4(1+x)=4e^{2t}$ ; both initial data also hold.

Change only the initial direction of motion. The force is still rightwards in the region $x>-1$ , but now it slows a leftward-moving particle.

Worked example 4: the negative velocity branch

Question. For the same particle and force law, take $x=0$ and $v=-2\,\mathrm{m\,s^{-1}}$ when $t=0$ . Find $x$ and $v$ in terms of $t$ , and the distance travelled up to time $t$ .

Plan. The initial speed is unchanged, so the relation for $v^2$ is unchanged; select the negative branch before integrating for position.

Using $x=0$ , $v^2=4$ again gives

v^2=4(1+x)^2.

On the leftward branch in $x>-1$ ,

\begin{aligned} \frac{\mathrm{d}x}{\mathrm{d}t}&=-2(1+x) \\ \ln(1+x)&=-2t+B \end{aligned}

The initial position gives $B=0$ , so

x=e^{-2t}-1, \qquad v=-2e^{-2t}.

For every finite $t\ge0$ , $-1<x\le0$ and $v<0$ . The particle therefore travels left throughout this interval. Its displacement is $x$ , whereas its distance travelled is $-x$ .

Answer. $x=e^{-2t}-1$ , $v=-2e^{-2t}$ , and distance travelled $1-e^{-2t}\,\mathrm{m}$ .

Check. Integrating speed directly gives $\int_0^t2e^{-2s}\,\mathrm{d}s=1-e^{-2t}$ , agreeing with the distance. Also the positive acceleration $4e^{-2t}$ opposes the negative velocity, so the speed decreases.

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3.5 - Variable forces depending on displacement

Acceleration as v dv by dx

Solving for speed and position

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