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Module 16 · L01 of 16 ~45 min ⚡ +90 XP available

Velocity and Acceleration as Derivatives

Mechanics is calculus with a direction. When a particle moves along a straight line, its position $x(t)$, velocity $v = \dot{x}$ and acceleration $a = \ddot{x}$ are linked by differentiation, and a single sign error wrecks the whole problem. This lesson nails down the three quantities, the sign conventions, and how to set up the differential equation of motion from any starting description.

Today's hook, A particle has position $x(t) = t^3 - 6t^2 + 9t$ metres for $t \geq 0$. Before reading on, write down (a) the velocity $v(t)$, (b) the times when the particle is momentarily at rest, and (c) whether the particle is speeding up or slowing down at $t = 1.5$. Compare your answers after card 05.
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Recall, your gut answer first

+5 XP warm-up A particle moves with $x(t) = 5t - t^2$ metres for $t \geq 0$.

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Recall, your gut answer first
+5 XP warm-up

A particle moves with $x(t) = 5t - t^2$ metres for $t \geq 0$. Before checking find $v(t)$ and $a(t)$ by differentiation, then determine when the particle is momentarily at rest. What does $a$ being constant and negative tell you about the motion? Sketch your reasoning below.

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The two moves for kinematics problems

+5 XP to read Every motion problem rewards two habits: choose a positive direction and origin (this fixes the sign of $x$, $v$, $a$), then decide which derivative you have (position $\to$ differentiate; acceleration $\to$ integrate).

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The two moves for kinematics problems
+5 XP to read

Every motion problem rewards two habits: choose a positive direction and origin (this fixes the sign of $x$, $v$, $a$), then decide which derivative you have (position $\to$ differentiate; acceleration $\to$ integrate). Skipping the sign convention is the single biggest cause of error in mechanics.

The position-velocity-acceleration chain: (1) write $x(t)$, (2) differentiate once for $v = \dot{x}$, (3) differentiate again for $a = \ddot{x}$. To go the other way (from $a$ to $v$ to $x$), integrate using initial conditions to fix constants.

Position: $x(t)$  ·  Velocity: $v = \dfrac{dx}{dt}$  ·  Acceleration: $a = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2}$

Differentiate to go from displacement to velocity to acceleration.
$a = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2}$
Speed is NOT velocity
Velocity $v$ is signed (direction matters). Speed is $|v|$. A particle with $v = -3$ is moving in the negative direction at speed 3.
Speeding up vs slowing
A particle is speeding up when $v$ and $a$ have the same sign, and slowing down when they have opposite signs. Never just look at $a$ alone.
Pick the positive direction once
Mark "up" or "right" as positive on your diagram. Then forces, velocity and acceleration all use that convention. Switching halfway through is the classic NESA mark-loser.
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What you'll master
Know

Key facts

  • $v = \dfrac{dx}{dt}$ is the time derivative of position
  • $a = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2}$ is the second time derivative of position
  • Speed $= |v|$; velocity is signed
  • Dot notation: $\dot{x} = v$, $\ddot{x} = a$
Understand

Concepts

  • Why sign conventions must be fixed at the start of a problem
  • Why "speeding up" depends on the product $v \cdot a$, not on $a$ alone
  • How initial conditions determine the constants of integration
Can do

Skills

  • Differentiate $x(t)$ to obtain $v(t)$ and $a(t)$
  • Solve $v = 0$ to find times of rest and turning points
  • Set up and integrate the differential equation $\dfrac{d^2x}{dt^2} = f(t)$
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Key terms
Displacement $x(t)$Signed position of the particle measured from a fixed origin $O$, at time $t$. Units: metres. Sign indicates direction.
Velocity $v = \dot{x}$Time rate of change of displacement: $v = \dfrac{dx}{dt}$. Signed. Units: m s$^{-1}$. Positive $v$ means motion in the chosen positive direction.
Acceleration $a = \ddot{x}$Time rate of change of velocity: $a = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2}$. Signed. Units: m s$^{-2}$.
SpeedMagnitude of velocity, $|v|$. Always non-negative. Speed and velocity differ in sign convention.
Momentarily at restInstant when $v = 0$. The particle may still be accelerating, at rest is not the same as in equilibrium.
Differential equation of motionEquation of the form $\dfrac{d^2x}{dt^2} = f(t,x,v)$ giving acceleration. Solving requires initial conditions for $x(0)$ and $v(0)$.
MEX-M1NESA outcome (Applications of Calculus to Mechanics): applies calculus to model and solve problems involving the motion of a particle in a straight line.
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Position, velocity, acceleration, the derivative chain

If the position of a particle is given by a function $x(t)$, then velocity and acceleration follow by differentiation:

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Position, velocity, acceleration, the derivative chain
core concept

If the position of a particle is given by a function $x(t)$, then velocity and acceleration follow by differentiation:

$$v(t) = \frac{dx}{dt} = \dot{x}, \qquad a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2} = \ddot{x}$$

The dot notation $\dot{x}, \ddot{x}$ is standard in mechanics and is the form NESA uses in MEX-M1 questions. Going the other way, from $a$ back to $v$ back to $x$, requires integration, with constants of integration fixed by initial conditions $x(0)$ and $v(0)$.

Worked through the hook: $x(t) = t^3 - 6t^2 + 9t$.

  • Differentiate: $v(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3)$.
  • Rest when $v = 0$: $t = 1$ or $t = 3$.
  • Differentiate again: $a(t) = 6t - 12$. At $t = 1.5$, $a = -3$ and $v = 3(0.5)(-1.5) = -2.25 < 0$. Same sign $\Rightarrow$ speeding up.
Sign-convention rule. Choose a positive direction before writing any equation. Then a positive $v$ means motion in that direction; a negative $a$ means the velocity is being reduced (in the algebraic sense, not in magnitude). To decide whether the particle is speeding up, check whether $v$ and $a$ share a sign.

Three quantities: $x$ (position), $v = \dot{x}$, $a = \ddot{x}$ · Differentiate to go $x \to v \to a$; integrate to reverse · Speed $= |v|$; particle at rest when $v = 0$ (not when $a = 0$) · Speeding up $\Leftrightarrow$ $v$ and $a$ have the same sign

Pause, copy $v = \dot{x}$, $a = \ddot{x}$, the differentiate/integrate chain, the speed vs velocity distinction, and the speeding-up sign rule into your book.

Quick check: A particle has $x(t) = 2t^3 - 9t^2 + 12t$ for $t \geq 0$. At which time is the particle momentarily at rest?

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Setting up the differential equation of motion

We just saw the derivative chain: $v = \dot{x}$ and $a = \ddot{x}$; integrate once for $v$, again for $x$; each integration introduces one constant fixed by an initial condition.

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Setting up the differential equation of motion
core concept

We just saw the derivative chain: $v = \dot{x}$ and $a = \ddot{x}$; integrate once for $v$, again for $x$; each integration introduces one constant fixed by an initial condition. That raises a question: how does this chain become a differential equation, and what happens when $a$ depends on $x$ rather than $t$? This card answers it → $\ddot{x} = a(t)$ is a second-order DE needing two ICs; if $a = a(x)$, use the identity $a = v\,dv/dx$.

Many HSC questions don't give you $x(t)$, they give you a description of the acceleration, often from a physical law (Newton's second law $F = ma$ rearranged for $a$). The job is then:

  1. Write the DE: $\dfrac{d^2x}{dt^2} = a(t)$ (when $a$ depends only on $t$).
  2. Integrate once: $v(t) = \displaystyle\int a(t)\,dt + C_1$, with $C_1$ fixed by $v(0)$.
  3. Integrate again: $x(t) = \displaystyle\int v(t)\,dt + C_2$, with $C_2$ fixed by $x(0)$.

If $a$ depends on $v$ rather than $t$, use $\dfrac{dv}{dt} = a(v)$ and separate variables. If $a$ depends on $x$, use the alternative form $a = v\dfrac{dv}{dx}$ (covered in lesson 02).

$$\frac{d^2x}{dt^2} = a(t) \quad\Longrightarrow\quad v(t) = \int a(t)\,dt, \quad x(t) = \int v(t)\,dt$$
Common mistake. Forgetting the constants of integration. Each integration introduces one constant; you need both initial conditions ($x(0)$ and $v(0)$) to pin down the unique solution. Many candidates lose marks by quoting $x(t)$ without a $C_2$.

$\ddot{x} = a(t)$ is a second-order DE, needs two initial conditions · Integrate once for $v$, again for $x$; each step needs $+C$ · $v(0)$ fixes the first constant; $x(0)$ fixes the second · If $a = a(v)$: separable; if $a = a(x)$: use $a = v\,dv/dx$ (lesson 02)

Pause, copy the second-order DE structure ($\ddot{x} = a(t)$, two ICs, two $+C$ constants), the IC assignment order ($v(0)$ first, then $x(0)$), and the $a = v\,dv/dx$ identity for $a = a(x)$ into your book.

Did you get this? True or false: if a particle has $a(t) = 6t$ with $v(0) = 2$ and $x(0) = 0$, then $v(t) = 3t^2 + 2$ and $x(t) = t^3 + 2t$.

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Worked examples · 3 in a row, reveal as you go

Worked examples · 3 in a row, reveal as you go

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · DIFFERENTIATE x(t)

A particle moves so that $x(t) = t^3 - 9t^2 + 24t - 4$ metres ($t \geq 0$). Find $v(t)$ and $a(t)$, determine when the particle is at rest, and decide whether it is speeding up or slowing down at $t = 4$.

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Differentiate: $v(t) = 3t^2 - 18t + 24 = 3(t^2 - 6t + 8) = 3(t-2)(t-4)$.
Factor the velocity to expose the times when $v = 0$, those are the candidates for momentarily at rest.
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PROBLEM 2 · INTEGRATE FROM a(t)

A particle has acceleration $a(t) = 12 - 6t$ m s$^{-2}$. Initially the particle is at $x(0) = 1$ m with velocity $v(0) = -3$ m s$^{-1}$. Find $v(t)$ and $x(t)$.

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Integrate $a$ for $v$: $v(t) = \displaystyle\int (12 - 6t)\,dt = 12t - 3t^2 + C_1$.
Each integration introduces one constant. Don't lose marks by skipping $C_1$ here.
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PROBLEM 3 · SIGN OF MOTION FROM v(t)

A particle's velocity is $v(t) = t^2 - 4t + 3$ m s$^{-1}$ for $t \geq 0$. Find (a) when the particle changes direction, (b) the acceleration when this happens, (c) the time intervals during which the particle is speeding up.

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$v(t) = (t-1)(t-3)$, so $v = 0$ at $t = 1, 3$. Sign chart: $v > 0$ on $[0,1)$, $v < 0$ on $(1,3)$, $v > 0$ on $(3, \infty)$. So the particle changes direction at $t = 1$ and $t = 3$.
A change in direction requires $v$ to actually change sign (not just touch zero). The sign chart confirms both crossings are genuine.
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Fill the gap: The relationship between position, velocity and acceleration is $v = \dfrac{d\,\rule{0.6em}{0.5pt}\,}{dt}$ and $a = \dfrac{d\,\rule{0.6em}{0.5pt}\,}{dt}$. Fill: $v = \dfrac{d}{dt}$ of and $a = \dfrac{d}{dt}$ of , so $a$ is the derivative of $x$ with respect to $t$.

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Misconceptions to fix · the 3 traps that cost marks

Misconceptions to fix · the 3 traps that cost marks

Misconceptions to fix · the 3 traps that cost marks

Trap 01
Treating $a = 0$ as "at rest"
A particle is at rest when $v = 0$, not when $a = 0$. $a = 0$ means the velocity is momentarily not changing, the particle may still be moving at constant velocity. Always solve $v = 0$ for rest times.
Trap 02
Reading "speeding up" off $a$ alone
Negative $a$ does not mean slowing down. Speeding up requires $v$ and $a$ to share a sign; slowing requires opposite signs. Always evaluate both at the time of interest.
Trap 03
Dropping constants of integration
When integrating $a$ to get $v$, you need $C_1$, fixed by $v(0)$. When integrating $v$ to get $x$, you need $C_2$, fixed by $x(0)$. Skipping either step makes the final $x(t)$ wrong by a polynomial.

Did you get this? True or false: if a particle has $v = -2$ m s$^{-1}$ and $a = -3$ m s$^{-2}$ at a particular instant, then the particle is speeding up.

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Revisit your thinking

Activities · practice with the ideas

Activities · practice with the ideas

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A particle has $x(t) = t^3 - 3t^2$ for $t \geq 0$. Find $v(t)$ and $a(t)$, and determine when the particle is momentarily at rest.

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A particle has $a(t) = 4 - 2t$ m s$^{-2}$. If $v(0) = 0$ and $x(0) = 0$, find $v(t)$ and $x(t)$, and the maximum displacement reached.

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The velocity of a particle is $v(t) = 6 - 2t$ m s$^{-1}$. Determine the time intervals during which the particle is (i) moving in the positive direction, (ii) speeding up.

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A particle moves with $x(t) = 2 \sin(3t) + \cos(3t)$. Find $v(t)$ and $a(t)$ and verify that $a = -9x$.

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A particle is released from rest and falls under gravity with $a = -g$ (taking up as positive, $g = 9.8$ m s$^{-2}$). If it starts at $x(0) = 100$ m, find when it hits the ground ($x = 0$).

Odd one out: Three of these are equivalent expressions for the acceleration of a particle moving along a straight line. Which one is NOT?

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Revisit your thinking

Earlier you analysed $x(t) = t^3 - 6t^2 + 9t$, finding velocity, the times of rest, and the speeding-up status at $t = 1.5$.

Velocity is $v(t) = 3(t-1)(t-3)$, so the particle is at rest at $t = 1$ and $t = 3$. Acceleration is $a(t) = 6t - 12$; at $t = 1.5$, $v = -2.25 < 0$ and $a = -3 < 0$, so $v \cdot a > 0$, the particle is speeding up (moving in the negative direction and accelerating in the negative direction). The decisive idea is that "speeding up" is a statement about the product $v \cdot a$, not about $a$ in isolation, this single rule eliminates a huge proportion of mechanics errors.

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Show what you have learned

Multiple choice, then short answer under exam conditions.

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer
ApplyBand 32 marks

Q1. A particle has displacement $x(t) = t^3 - 4t^2 + 3t$ metres. Find the velocity $v(t)$ and determine the times at which the particle is momentarily at rest. (2 marks)

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ApplyBand 43 marks

Q2. A particle moves with acceleration $a(t) = 6t - 12$ m s$^{-2}$. Initially the particle is at rest at $x(0) = 5$. Find $x(t)$. (3 marks)

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AnalyseBand 53 marks

Q3. The velocity of a particle is $v(t) = t^2 - 5t + 6$ m s$^{-1}$. Determine (a) the times when the particle changes direction, (b) the acceleration at those times, (c) the interval(s) during which the particle is speeding up. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $v(t) = 3t^2 - 6t = 3t(t - 2)$; $a(t) = 6t - 6$. At rest when $v = 0$: $t = 0$ or $t = 2$.

2. $v(t) = 4t - t^2 = t(4 - t)$; $x(t) = 2t^2 - \tfrac{1}{3}t^3$. Max displacement when $v = 0$, $t = 4$: $x(4) = 32 - \tfrac{64}{3} = \tfrac{32}{3}$ m.

3. (i) $v > 0 \Leftrightarrow t < 3$. (ii) $a = -2$ constant; $v \cdot a > 0 \Leftrightarrow v < 0 \Leftrightarrow t > 3$. So speeding up on $t > 3$.

4. $v = 6\cos 3t - 3\sin 3t$; $a = -18\sin 3t - 9\cos 3t = -9(2\sin 3t + \cos 3t) = -9x$. Confirmed: SHM with angular frequency $\omega = 3$.

5. $v(t) = -gt$, $x(t) = 100 - \tfrac{1}{2}gt^2$. Hits ground when $x = 0$: $t = \sqrt{200/g} = \sqrt{200/9.8} \approx 4.52$ s.

Q1 (2 marks): $v(t) = 3t^2 - 8t + 3$ [1]. At rest: $t = \dfrac{8 \pm \sqrt{28}}{6} = \dfrac{4 \pm \sqrt{7}}{3}$, so $t \approx 0.45$ or $t \approx 2.22$ s (both $\geq 0$) [1].

Q2 (3 marks): $v(t) = 3t^2 - 12t + C_1$; $v(0) = 0 \Rightarrow C_1 = 0$, so $v(t) = 3t^2 - 12t$ [1]. $x(t) = t^3 - 6t^2 + C_2$ [1]; $x(0) = 5 \Rightarrow C_2 = 5$, so $x(t) = t^3 - 6t^2 + 5$ [1].

Q3 (3 marks): (a) $v = (t-2)(t-3) = 0$ at $t = 2, 3$; sign of $v$ changes at both, so direction changes there [1]. (b) $a = 2t - 5$: $a(2) = -1$, $a(3) = 1$ [1]. (c) $v > 0$ on $[0,2)\cup(3,\infty)$, $v < 0$ on $(2,3)$. $a > 0$ when $t > 2.5$. Speeding up where $v \cdot a > 0$: $(2.5, 3)$ has $v < 0, a > 0$, slowing; $(2, 2.5)$ has $v < 0, a < 0$, speeding up; $t > 3$ both positive, speeding up; $[0, 2)$ has $v > 0, a < 0$ on $(0, 2.5)$, slowing. So speeding up on $(2, 2.5) \cup (3, \infty)$ [1].