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.
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.
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: $a = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2}$
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: $\dot{x} = v$, $\ddot{x} = a$
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
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 $\dfrac{d^2x}{dt^2} = f(t)$
If the position of a particle is given by a function $x(t)$, then velocity and acceleration follow by differentiation:
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.
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?
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:
- Write the DE: $\dfrac{d^2x}{dt^2} = a(t)$ (when $a$ depends only on $t$).
- Integrate once: $v(t) = \displaystyle\int a(t)\,dt + C_1$, with $C_1$ fixed by $v(0)$.
- 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).
$\ddot{x} = a(t)$ 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 $x(t) = t^3 + 2t$.
Worked examples · 3 in a row, reveal as you go
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 $t = 4$.
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 $x(t)$.
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.
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 $t$.
Misconceptions to fix · the 3 traps that cost marks
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.
Activities · practice with the ideas
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.
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.
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.
A particle moves with $x(t) = 2 \sin(3t) + \cos(3t)$. Find $v(t)$ and $a(t)$ and verify that $a = -9x$.
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?
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.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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)
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)
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)
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].
Five timed questions on $x(t), v(t), a(t)$ chains, sign conventions and setting up the DE of motion. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering quick kinematics questions. Lighter alternative to the boss.
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