When the acceleration depends on position rather than time, gravity from a planet, a spring restoring force, a particle dragged by a position-dependent field, the chain rule gives a powerful re-write: $a = v\dfrac{dv}{dx} = \dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$. This single identity turns hard kinematics into separable integration in $x$.
+5 XP warm-up A particle has acceleration $a = -2x$.
A particle has acceleration $a = -2x$. Before checking try to integrate $a = dv/dt$ directly. What goes wrong? Now apply the chain rule to write $a$ in terms of $v$ and $dv/dx$. Sketch your reasoning below.
+5 XP to read When acceleration is given as a function of position, two habits unlock the problem: replace using the chain...
When acceleration is given as a function of position, two habits unlock the problem: replace $a$ using the chain rule $a = v\dfrac{dv}{dx}$ (or equivalently $\dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$), then separate variables in $x$ and $v$ so each side has only one variable.
The chain-rule trick: starting from $a = \dfrac{dv}{dt}$ and using $\dfrac{dv}{dt} = \dfrac{dv}{dx}\cdot\dfrac{dx}{dt} = v\dfrac{dv}{dx}$, we eliminate $t$ entirely. Equivalently, $v\dfrac{dv}{dx} = \dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$, so integrating with respect to $x$ gives $\tfrac{1}{2}v^2 = \int a(x)\,dx + C$.
$a = v\dfrac{dv}{dx} = \dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$
Start from $a = \dfrac{dv}{dt}$.
Start from $a = \dfrac{dv}{dt}$. By the chain rule, since $v$ depends on $x$ which depends on $t$,
Now observe that $\dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right) = v\dfrac{dv}{dx}$ by the chain rule again. So $a = \dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$. Integrating both sides with respect to $x$:
Worked through the hook: $a = -4x$ with $v = 6$ when $x = 0$. Apply $v\dfrac{dv}{dx} = -4x$, so $\dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right) = -4x$. Integrate: $\tfrac{1}{2}v^2 = -2x^2 + C$. Use $v = 6$ at $x = 0$: $\tfrac{1}{2}(36) = 0 + C \Rightarrow C = 18$. So $v^2 = 36 - 4x^2$. Maximum displacement is where $v = 0$: $x^2 = 9$, $|x| = 3$. The particle oscillates between $x = -3$ and $x = 3$, simple harmonic motion.
$a = v\dfrac{dv}{dx} = \dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$ · Integrate w.r.t. $x$: $\tfrac{1}{2}v^2 = \int a(x)\,dx + C$ · Fix $C$ using the value of $v$ at a known $x$ · For $a = -kx$: $v^2 = k(A^2 - x^2)$ where $A$ is the amplitude
Pause, copy $a = v\,dv/dx = d(\tfrac{1}{2}v^2)/dx$, the integration-w.r.t.-$x$ result $\tfrac{1}{2}v^2 = \int a(x)\,dx+C$, and the SHM application $v^2 = k(A^2-x^2)$ into your book.
Quick check: A particle has acceleration $a = -9x$, and $v = 12$ when $x = 0$. Which equation correctly gives $v^2$ in terms of $x$?
We just saw the identity $a = v\,dv/dx = d(\tfrac{1}{2}v^2)/dx$, which converts $a = a(x)$ into a separable ODE in $v^2$ and $x$.
We just saw the identity $a = v\,dv/dx = d(\tfrac{1}{2}v^2)/dx$, which converts $a = a(x)$ into a separable ODE in $v^2$ and $x$. That raises a question: given that $a$ can depend on $t$, $v$, or $x$, which DE form should you use for each case? This card answers it → $a = a(t)$: integrate w.r.t. $t$; $a = a(v)$: separate variables; $a = a(x)$: use $v\,dv/dx$.
The three faces of acceleration each suit a different problem:
Choosing the wrong form is a common time-sink. The decisive rule: look at which variable $a$ depends on first, then pick the form whose integration variable is that variable.
$a = a(t)$: integrate w.r.t. $t$ · $a = a(v)$: separate variables, $\int dv/a(v) = \int dt$ · $a = a(x)$: use $v\,dv/dx = a(x)$, integrate w.r.t. $x$ · The form you pick is dictated by what $a$ depends on
Pause, copy the three-case classification ($a = a(t)$: integrate; $a = a(v)$: separate; $a = a(x)$: use $v\,dv/dx$) and the rule that the form of $a$ dictates the DE choice into your book.
Did you get this? True or false: the identity $v\dfrac{dv}{dx}$ is most useful when the acceleration is given as a function of position $x$.
Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
A particle moves with $a = -16x$ m s$^{-2}$. At $x = 0$ it has speed $8$ m s$^{-1}$. Find $v^2$ as a function of $x$, and find the amplitude (max displacement).
A particle of mass 1 kg is attracted toward the origin with $a = -\dfrac{8}{x^2}$ m s$^{-2}$ for $x > 0$. It starts from rest at $x = 4$ m. Find $v^2$ as a function of $x$, and the speed when it reaches $x = 1$.
A particle moves in a straight line with acceleration $a = -\dfrac{1}{(x+1)^2}$ m s$^{-2}$ for $x \geq 0$. Initially the particle is at $x = 0$ with velocity $v = 1$ m s$^{-1}$ (away from the origin). Show that $v^2 = \dfrac{2}{x+1} - 1$, and find the maximum value of $x$ reached.
Fill the gap: The chain rule gives $\dfrac{dv}{dt} = v\dfrac{d\square}{d\square}$, and this equals $\dfrac{d}{dx}\!\left(\tfrac{1}{2}\square^2\right)$. Fill: $v\dfrac{d}{dx}$ of with respect to , which equals $\dfrac{d}{dx}$ of $\tfrac{1}{2}$ times squared.
Misconceptions to fix · the 3 traps that cost marks
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: for a particle with $a = -kx$ ($k > 0$) starting from rest at $x = A$, the equation $v^2 = k(A^2 - x^2)$ holds for all subsequent positions.
Activities · practice with the ideas
Activities · practice with the ideas
A particle has $a = -25x$. At $x = 0$, $v = 10$. Find $v^2$ as a function of $x$ and the amplitude of motion.
A particle has acceleration $a = 2x + 3$. It starts from rest at $x = 0$. Find $v^2$ in terms of $x$ and the speed when $x = 3$.
A particle moves with $a = -\dfrac{2}{x^3}$ for $x > 0$. It starts from rest at $x = 1$. Find $v^2$ in terms of $x$ and decide whether the particle ever reaches the origin.
A particle has $a = -4x + 8$. At $x = 0$, $v = 0$. Find $v^2$ in terms of $x$ and the value(s) of $x$ where $v = 0$ again.
(HSC-style) A spacecraft moves along a line with $a = -\dfrac{32}{(x+4)^2}$ (taking the centre of the planet as origin). Initially $x = 0$ and $v = 4$ (moving in the positive direction). Find $v^2$ as a function of $x$ and the farthest position reached.
Odd one out: Three of these are equivalent expressions for the acceleration when $v$ and $x$ are related by motion along a straight line. Which one is NOT?
Earlier you tackled $a = -4x$ with $v = 6$ at $x = 0$, finding the right identity, the resulting DE, and the maximum displacement.
The right move is $a = v\dfrac{dv}{dx} = \dfrac{d}{dx}\!\left(\tfrac{1}{2}v^2\right)$, giving $\tfrac{1}{2}v^2 = -2x^2 + C$. Initial condition fixes $C = 18$, so $v^2 = 36 - 4x^2$. Maximum displacement is where $v = 0$: $|x| = 3$. The crucial insight is that when acceleration depends on $x$, time disappears from the problem, you work entirely in terms of $x$ and $v$, and recover the amplitude immediately from $v^2 = 0$. This $v$-on-$v$ identity is one of the most useful tools in Module 16.
Multiple choice, then short answer under exam conditions.
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 moves with acceleration $a = -9x$. At $x = 0$, $v = 6$. Find $v^2$ as a function of $x$. (2 marks)
Q2. A particle has acceleration $a = -\dfrac{18}{x^2}$ ($x > 0$) and starts from rest at $x = 6$. Find $v^2$ as a function of $x$ and the speed when the particle reaches $x = 2$. (3 marks)
Q3. A particle moves with $a = 2x - 6$. At $x = 1$, $v = 0$. (a) Find $v^2$ in terms of $x$. (b) Find the other position at which the particle is momentarily at rest. (c) Find the maximum speed and the position at which it occurs. (3 marks)
Activity answers:
1. $\tfrac{1}{2}v^2 = -\tfrac{25}{2}x^2 + C$; $v(0) = 10 \Rightarrow C = 50$. So $v^2 = 100 - 25x^2$. Amplitude: $v = 0 \Rightarrow x^2 = 4$, $A = 2$ m.
2. $\tfrac{1}{2}v^2 = x^2 + 3x + C$; $v = 0$ at $x = 0 \Rightarrow C = 0$. So $v^2 = 2x^2 + 6x$. At $x = 3$: $v^2 = 18 + 18 = 36$, $|v| = 6$ m s$^{-1}$.
3. $\tfrac{1}{2}v^2 = \tfrac{1}{x^2} + C$ (since $\int -2x^{-3}\,dx = x^{-2}$); $v = 0$ at $x = 1 \Rightarrow 0 = 1 + C \Rightarrow C = -1$. So $v^2 = \tfrac{2}{x^2} - 2$. As $x \to 0^+$, $v^2 \to \infty$, the particle can reach the origin (mathematically; physically it would accelerate without bound).
4. $\tfrac{1}{2}v^2 = -2x^2 + 8x + C$; $v = 0$ at $x = 0 \Rightarrow C = 0$. So $v^2 = -4x^2 + 16x = 4x(4 - x)$. $v = 0$ again at $x = 4$.
5. $\tfrac{1}{2}v^2 = \dfrac{32}{x+4} + C$; $v = 4$ at $x = 0 \Rightarrow 8 = 8 + C \Rightarrow C = 0$. So $v^2 = \dfrac{64}{x+4}$. This is always $> 0$ for $x \geq 0$, so $v$ never reaches zero, the spacecraft has escape velocity and travels to infinity.
Q1 (2 marks): $v\dfrac{dv}{dx} = -9x \Rightarrow \tfrac{1}{2}v^2 = -\tfrac{9}{2}x^2 + C$ [1]. $v(0) = 6 \Rightarrow C = 18$, so $v^2 = 36 - 9x^2$ [1].
Q2 (3 marks): $\tfrac{1}{2}v^2 = \dfrac{18}{x} + C$ [1]. $v = 0$ at $x = 6 \Rightarrow C = -3$, so $v^2 = \dfrac{36}{x} - 6$ [1]. At $x = 2$: $v^2 = 18 - 6 = 12$, $|v| = 2\sqrt{3}$ m s$^{-1}$ [1].
Q3 (3 marks): (a) $\tfrac{1}{2}v^2 = x^2 - 6x + C$; $v = 0$ at $x = 1 \Rightarrow 0 = 1 - 6 + C \Rightarrow C = 5$. So $v^2 = 2x^2 - 12x + 10 = 2(x-1)(x-5)$ [1]. (b) $v = 0$ also at $x = 5$ [1]. (c) Max speed occurs at $x$ where $\dfrac{d(v^2)}{dx} = 4x - 12 = 0$, i.e. $x = 3$. But $v^2(3) = 2(2)(-2) = -8 < 0$, the particle does not reach $x = 3$ between the rests. The motion is confined to where $v^2 \geq 0$: from the factor form $v^2 = 2(x-1)(x-5)$, this requires $x \leq 1$ or $x \geq 5$. The given motion stays in one regime, and max speed within that regime is at the endpoint, for $x \leq 1$ the speed grows as $x$ decreases; no finite max (open boundary) [1].
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