4.10b - Modelling with differential equations

You already know

For a particle of constant mass, Newton's second law is resultant force equals mass times acceleration. Choose a positive direction before assigning force and velocity signs. Integrating acceleration gives velocity; integrating velocity gives displacement.

Kinematics and Newton's second law

For constant mass $m>0$, in a fixed direction,

and

The force and velocity are signed components along that direction.
Not in the formula booklet: learn it.

Worked example 1: a force proportional to time

Question. A particle of mass $2\,\mathrm{kg}$ moves along a straight line. Its resultant force at time $t$ seconds is $6t\,\mathrm{N}$ in the positive direction. Initially $v=1\,\mathrm{m\,s^{-1}}$ and $x=0$. Find $v$ and $x$ in terms of $t$.

Plan. Use Newton's second law, integrate for velocity and displacement, and use each initial condition.

Why this step? The given force is the resultant, so no extra force is added to it; divide by the mass to obtain acceleration.

Answer. $v=1+\dfrac32t^2\,\mathrm{m\,s^{-1}}$ and $x=t+\dfrac12t^3\,\mathrm{m}$, for $t\ge0$.

Check. Differentiating gives $\dot{x}=1+3t^2/2=v$ and $2\dot{v}=6t$, and both initial conditions hold.

Worked example 2: a signed changing resultant

Question. The same particle now has resultant force $(6t-4)\,\mathrm{N}$, measured positively in the same direction, with the same initial conditions. Find $v$ and $x$.

Plan. Retain the negative force term when forming acceleration, then integrate twice.

Why this step? A negative resultant means negative acceleration, not automatically negative velocity; the initial motion still matters.

Answer. $v=1-2t+\dfrac32t^2\,\mathrm{m\,s^{-1}}$ and $x=t-t^2+\dfrac12t^3\,\mathrm{m}$, for $t\ge0$.

Check. At $t=0$, acceleration is $-2\,\mathrm{m\,s^{-2}}$ while velocity is $+1\,\mathrm{m\,s^{-1}}$, consistent with initially slowing down.

Watch out

Wrong: $2\dot{v}=|6t-4|$, replacing the signed resultant by its size. Right: $2\dot{v}=6t-4$ throughout the model's stated interval.

Terminal velocity

A terminal velocity is a limiting steady velocity approached as the resultant force tends to zero. In the model $m\dot{v}=F-kv$, it is $F/k$. For vertical falling motion with downward positive, the corresponding model is $m\dot{v}=mg-kv$, giving terminal velocity $mg/k$.

Worked example 3: approaching terminal velocity

Question. A $3\,\mathrm{kg}$ particle starts from rest on a horizontal line. A constant drive of $12\,\mathrm{N}$ acts positively and the resistance has signed component $-3v\,\mathrm{N}$, where $v$ is in $\mathrm{m\,s^{-1}}$. Find $v(t)$ and its limiting value.

Plan. Resolve forces, divide by the mass and solve the linear equation with its initial condition.

Why this step? Resistance is part of the resultant, so the equation is $3\dot{v}=12-3v$, rather than $3\dot{v}=12$.

Answer. $v=4(1-e^{-t})\,\mathrm{m\,s^{-1}}$, tending to $4\,\mathrm{m\,s^{-1}}$ as $t\to\infty$.

Check. $3\dot{v}=12e^{-t}=12-3[4(1-e^{-t})]$. At the limiting velocity, resistance is $3(4)=12\,\mathrm{N}$, balancing the drive.

Worked example 4: changing the drive

Question. The same particle, starting from rest, has drive $18\,\mathrm{N}$ and the same resistance law. Find its velocity and the time to reach half its terminal velocity.

Plan. Solve the new linear equation, then set the velocity equal to half its limiting value.

Why this step? Half-terminal velocity is half the new limit $6$, rather than half the previous limit $4$.

Answer. $v=6(1-e^{-t})\,\mathrm{m\,s^{-1}}$; the required time is $\ln2\,\mathrm{s}$.

Check. At $t=\ln2$, $e^{-t}=1/2$, so $v=3\,\mathrm{m\,s^{-1}}$.