Mathematics/Lessons/Variable Acceleration/Using Differentiation
Mechanics · Variable Acceleration · Year 1

Using Differentiation

Velocity is the rate of change of displacement, and acceleration is the rate of change of velocity — so differentiating once or twice turns s(t) into everything else you need to know about the motion.

11 min read 2 worked examples

The calculus chain

Velocity is the rate of change of displacement, and acceleration is the rate of change of velocity. Differentiating $s(t)$ once gives velocity, and differentiating again — equivalently, differentiating $v(t)$ once — gives acceleration.

Differentiating displacement

$$v = \dfrac{ds}{dt} \qquad a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}$$

Going from $s$ to $v$ to $a$ needs no extra information beyond the expression for $s(t)$ itself — unlike integrating, no constant needs to be found.

Worked example 1

Differentiating to find velocity and acceleration
A particle moves so that its displacement at time $t$ seconds is $s = t^4 - 8t^2 + 3$ metres. Find its velocity and acceleration when $t=2$.
MethodDifferentiate once for v, again for a, then substitute
1
$v = \dfrac{ds}{dt} = 4t^3 - 16t$.
Differentiate $s$ term by term.
2
$a = \dfrac{dv}{dt} = 12t^2 - 16$.
Differentiate $v$ term by term.
3
At $t=2$: $v = 4(8) - 16(2) = 32 - 32 = 0\ \text{m s}^{-1}$.
Substitute $t=2$ into $v(t)$.
4
At $t=2$: $a = 12(4) - 16 = 48-16 = 32\ \text{m s}^{-2}$.
Substitute $t=2$ into $a(t)$.
▪ solved
Reading the result

The particle is momentarily at rest at $t=2$ ($v=0$), but the acceleration there is a large $32\ \text{m s}^{-2}$ — so this is not a lasting stop, just the single instant where the direction of motion is about to change.

Speeding up or slowing down?

Comparing the signs of $v$ and $a$ at a given instant tells you not just the direction of travel, but whether the particle is speeding up or slowing down at that moment.

Same sign or opposite sign

If $v$ and $a$ have the same sign at an instant, the particle is speeding up — the acceleration is reinforcing the direction of travel. If $v$ and $a$ have opposite signs, the particle is slowing down — the acceleration is working against the direction of travel.

Worked example 2

Deciding whether the particle is speeding up
A particle's displacement is $x = t^3 - 5t^2 + 3t + 2$ metres. Find its velocity and acceleration a when $t=1$, and b when $t=2$, and in each case state whether the particle is speeding up or slowing down.
MethodDifferentiate once for v(t) and again for a(t), then compare signs at each instant
1
$v = 3t^2-10t+3$, $a = 6t-10$.
Differentiate $x$ once and twice.
a
At $t=1$: $v = 3-10+3=-4$, $a = 6-10=-4$. Both negative — same sign, so the particle is speeding up (in the negative direction).
v and a reinforce each other.
b
At $t=2$: $v = 12-20+3=-5$, $a = 12-10=2$. Signs differ — opposite sign, so the particle is slowing down.
a is now working against the direction of motion.
▪ solved
Common mistake

Assuming a positive acceleration always means "speeding up" — it only does if the velocity is also positive. A positive acceleration acting on a particle moving in the negative direction is actually slowing that particle down. Always compare the signs of $v$ and $a$ together, never look at $a$ in isolation.

Check yourself

A particle's displacement is $s = 2t^3 - 9t^2 + 5$ metres. Find its velocity and acceleration when $t=1$, and state whether it is speeding up or slowing down at that instant.
(Hint: differentiate twice, then compare the signs of v and a.)

$v = 6t^2-18t$, so $v(1) = 6-18=-12\ \text{m s}^{-1}$.

$a = 12t-18$, so $a(1) = 12-18=-6\ \text{m s}^{-2}$. Both negative — same sign — so the particle is speeding up at $t=1$.

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