What is sign of the resistance?

Resistance always opposes the motion, so it has the opposite sign to the velocity in the equation of motion. For a falling body with down positive, resistance is subtracted from the weight.

A particle has dvdt = -2v and starts at v = 5 m s^-1. Find v as a function of t. [3 marks]

ScotlandMathematics of MechanicsSyllabus dot point

How do you analyse straight-line motion when the force varies with time, velocity or position, and how do differential equations give the motion and the terminal velocity?

Set up and solve differential equations for rectilinear motion under a variable force; use the forms of acceleration as a function of t, v or x; and find terminal velocity for motion against resistance.

A focused answer to the SQA Advanced Higher Mathematics of Mechanics content on rectilinear motion governed by differential equations, covering the three forms of acceleration, setting up the equation of motion for a variable force, solving by separation, motion against resistance, and finding the terminal velocity.

Generated by Claude Opus 4.814 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The three forms of acceleration
Setting up and solving the equation of motion
Resisted motion and terminal velocity
Try this

What this dot point is asking

When the force on a body is constant, the suvat equations are enough. When it varies, Newton's second law becomes a differential equation. The SQA wants you to set up that equation of motion for a variable force, choose the right form of acceleration depending on whether the force depends on time, velocity or position, solve by separating variables, and find the terminal velocity for motion against resistance.

The three forms of acceleration

The single most important decision is how to write the acceleration. All three forms describe the same thing but lead to different (and only sometimes solvable) equations.

The trick with $a = v\dfrac{dv}{dx}$ comes from the chain rule, $\dfrac{dv}{dt} = \dfrac{dv}{dx}\dfrac{dx}{dt} = v\dfrac{dv}{dx}$ , and it is what lets you find speed as a function of distance without involving time. Matching the form to the variable in the force is the whole skill.

Setting up and solving the equation of motion

Once the form is chosen, the method is mechanical: write $m a = (\text{resultant force})$ , separate the variables, integrate both sides, and apply the initial condition.

Resisted motion with a constant driving force

A particle of mass $m$ has resistance $mkv$ and is acted on by a constant driving force $mF_0$ . Find $v$ in terms of $t$ if it starts from rest.

Step 1: Write the equation of motion

Apply Newton's second law. The driving force acts forward and the resistance $mkv$ opposes the motion, so the net force is $mF_0 - mkv$ . Dividing both sides by $m$ gives a separable equation in $v$ and $t$ :

m\frac{dv}{dt} = mF_0 - mkv, \quad \text{so} \quad \frac{dv}{dt} = F_0 - kv.

Step 2: Separate the variables and integrate

Rearrange so $v$ is on the left and $t$ is on the right, then integrate both sides. The left side needs the standard $\ln$ integral:

\int \frac{1}{F_0 - kv}\,dv = \int dt, \quad \text{giving} \quad -\frac{1}{k}\ln(F_0 - kv) = t + c.

Step 3: Apply the initial condition and rearrange

At $t = 0$ the particle starts from rest, so $v = 0$ . Substituting gives $c = -\dfrac{1}{k}\ln F_0$ . Using $\ln A - \ln B = \ln(A/B)$ and then exponentiating:

v = \frac{F_0}{k}\left(1 - e^{-kt}\right).

Final answer: the speed rises from zero and approaches the terminal value $\dfrac{F_0}{k}$ as $t \to \infty$ .

Resisted motion and terminal velocity

For a body moving against a resistance that grows with speed (often modelled as $mkv$ or $mkv^2$ ), the acceleration falls as the speed rises. Eventually the resistance balances the driving force and the body settles to a steady terminal velocity.

The body never quite reaches the terminal velocity in finite time (the approach is exponential), but the speed gets arbitrarily close. Distinguishing a question that wants the full $v$ - $t$ relationship from one that wants only the terminal value is important: the latter needs only the algebraic condition $a = 0$ , with no integration.

Try this

Q1. A particle has $\dfrac{dv}{dt} = -2v$ and starts at $v = 5$ m s $^{-1}$ . Find $v$ as a function of $t$ . [3 marks]

Cue. Separate: $\dfrac{1}{v}\,dv = -2\,dt$ , so $\ln v = -2t + c$ ; with $v = 5$ at $t = 0$ , $v = 5e^{-2t}$ .

Q2. A body falls under gravity against resistance $3v^2$ per unit mass. Find its terminal velocity ( $g = 9.8$ ). [2 marks]

Cue. Terminal when $g = 3v^2$ , so $v = \sqrt{9.8/3} \approx 1.81$ m s $^{-1}$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AH style: resisted motion6 marksA particle of mass

m

moving in a straight line experiences a resistance of magnitude

mkv

, where

v

is its speed and

k

is constant. It starts with speed

u

and no other force acts. Find

v

as a function of time

t

Show worked answer →

The only force is the resistance, opposing motion: $m\dfrac{dv}{dt} = -mkv$ , so $\dfrac{dv}{dt} = -kv$ (2 marks).

Separate the variables: $\dfrac{1}{v}\,dv = -k\,dt$ (1 mark).

Integrate: $\ln v = -kt + c$ (1 mark). Using $v = u$ at $t = 0$ gives $c = \ln u$ (1 mark).

So $\ln\dfrac{v}{u} = -kt$ , hence $v = u e^{-kt}$ (1 mark). Markers reward forming the equation of motion, separating variables, integrating, and applying the initial condition to get the exponential decay of speed.

AH style: terminal velocity5 marksA body of mass

m

falls from rest under gravity against a resistance

mkv^2

. Write down the equation of motion and find the terminal velocity. Take downward as positive.

Show worked answer →

Downward forces: weight $mg$ down, resistance $mkv^2$ up (opposing the downward motion). Equation of motion: $m\dfrac{dv}{dt} = mg - mkv^2$ (2 marks).

So $\dfrac{dv}{dt} = g - kv^2$ (1 mark).

Terminal velocity is reached when the acceleration is zero, $\dfrac{dv}{dt} = 0$ (1 mark).

Then $g = kv^2$ , so the terminal velocity is $v = \sqrt{\dfrac{g}{k}}$ (1 mark). Markers reward the equation of motion with resistance opposing the motion, and setting the acceleration to zero for the terminal velocity.

Related dot points

Sources & how we know this

Advanced Higher Mathematics of Mechanics Course Specification (C802 77) — SQA (2019)