Application to Kinematics – OCR FSMQ Additional Maths

1 Displacement, Velocity and Acceleration

CA14: use differentiation and integration with respect to time to solve simple problems involving variable acceleration. This is calculus applied to motion — and it is the reason calculus was invented.

The chain of relationships

$s \;\xrightarrow{\text{differentiate}}\; v \;\xrightarrow{\text{differentiate}}\; a$
$a \;\xrightarrow{\text{integrate}}\; v \;\xrightarrow{\text{integrate}}\; s$

The dot notation (CA1)

$v = \dot{s} = \dfrac{\mathrm{d}s}{\mathrm{d}t}$ · $a = \dot{v} = \ddot{s} = \dfrac{\mathrm{d}^2 s}{\mathrm{d}t^2}$

Quantity	Meaning	Units
Displacement $s$	Position relative to a starting point — can be negative	m
Velocity $v$	Rate of change of displacement — has direction	m s⁻¹
Acceleration $a$	Rate of change of velocity	m s⁻²

Displacement is not distance. A ball thrown up and caught again has zero displacement but has travelled a real distance. Likewise, negative velocity means moving backwards, not slowing down.

2 Going Forwards: Differentiating

Worked Example 1 — From displacement to acceleration

A particle's displacement is $s = t^3 - 6t^2 + 9t$ metres at time $t$ seconds. Find $v$ and $a$, and the values at $t = 4$.

①$v = \dfrac{\mathrm{d}s}{\mathrm{d}t} = 3t^2 - 12t + 9$

②$a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = 6t - 12$

③At $t = 4$: $v = 48 - 48 + 9 = 9$ m s⁻¹

④$a = 24 - 12 = 12$ m s⁻²

⑤The particle is moving forwards and speeding up.

Three standard questions, and what to solve

Momentarily at rest? Set $v = 0$
Maximum or minimum velocity? Set $a = 0$
Changes direction? Find where $v$ changes sign

Worked Example 2 — When is it at rest?

For the particle in Worked Example 1, find when it is momentarily at rest, and describe its motion in the first $4$ seconds.

①At rest means $v = 0$: $3t^2 - 12t + 9 = 0$

②$3\left(t^2 - 4t + 3\right) = 3(t-1)(t-3) = 0$

③$t = 1$ s and $t = 3$ s.

④Sign of $v$: at $t=0$, $v = 9 > 0$; at $t=2$, $v = 12-24+9 = -3 < 0$; at $t=4$, $v = 9 > 0$.

⑤So it moves forwards for $0 < t < 1$, backwards for $1 < t < 3$, then forwards again after $t = 3$.

Two direction changes mean the total distance travelled is more than the change in displacement. Any "total distance" question must be split at $t = 1$ and $t = 3$.

Worked Example 3 — Minimum velocity

Find the minimum velocity of the same particle.

①Velocity is stationary when $a = 0$: $6t - 12 = 0$, so $t = 2$.

②$v = 3(4) - 24 + 9 = -3$ m s⁻¹

③$\dfrac{\mathrm{d}a}{\mathrm{d}t} = 6 > 0$, so this is a minimum.

④The minimum velocity is $-3$ m s⁻¹, at $t = 2$ s — the fastest it travels backwards.

Minimum velocity is not minimum speed. Speed is the size of velocity, so the minimum speed here is $0$, at $t = 1$ and $t = 3$. Read the question's exact word.

3 Going Backwards: Integrating

Integrating introduces a constant, and in kinematics that constant is always an initial condition: the initial velocity, or the starting position.

Worked Example 4 — From acceleration to displacement

A particle has acceleration $a = 6t - 4$ m s⁻². At $t = 0$ it has velocity $5$ m s⁻¹ and displacement $2$ m. Find $s$ in terms of $t$.

①$v = \displaystyle\int (6t-4)\,\mathrm{d}t = 3t^2 - 4t + c_1$

②At $t=0$, $v = 5$: so $c_1 = 5$ and $v = 3t^2 - 4t + 5$.

③$s = \displaystyle\int \left(3t^2-4t+5\right)\mathrm{d}t = t^3 - 2t^2 + 5t + c_2$

④At $t=0$, $s = 2$: so $c_2 = 2$.

⑤$s = t^3 - 2t^2 + 5t + 2$

Use different letters for the two constants. They are different numbers, and finding the first before integrating again is essential.

Change in displacement from velocity

$\text{change in } s = \displaystyle\int_{t_1}^{t_2} v \,\mathrm{d}t$

Worked Example 5 — Distance versus displacement

A particle has $v = t^2 - 4t$ m s⁻¹. Find (a) the change in displacement, and (b) the total distance travelled, over the first $6$ seconds.

①(a) $\displaystyle\int_0^6 \left(t^2-4t\right)\mathrm{d}t = \Bigg[\dfrac{t^3}{3} - 2t^2\Bigg]_0^6$

②$= (72 - 72) - 0 = 0$ m

③The particle ends where it started — the change in displacement is zero.

④(b) But it clearly moved. $v = t(t-4) = 0$ at $t = 0$ and $t = 4$, so it changes direction at $t = 4$. Split there.

⑤$\displaystyle\int_0^4 = \left(\dfrac{64}{3} - 32\right) - 0 = -\dfrac{32}{3}$ (moving backwards)

⑥$\displaystyle\int_4^6 = (72-72) - \left(-\dfrac{32}{3}\right) = \dfrac{32}{3}$ (moving forwards)

⑦Total distance $= \dfrac{32}{3} + \dfrac{32}{3} = \dfrac{64}{3} = 21\tfrac13$ m

This is exactly the "split at the root" technique from area questions — because distance is the area under the velocity–time graph, with signs ignored.

4 The Special Case of Constant Acceleration (CA15)

CA15: recognise the special case where the use of constant acceleration formulae is appropriate. When acceleration is a constant, calculus is unnecessary — the familiar "suvat" equations do the job faster.

The constant acceleration formulae

$v = u + at$ · $s = ut + \tfrac12 at^2$
$v^2 = u^2 + 2as$ · $s = \tfrac12(u+v)t$

How to tell which to use

$a$ is a number (or gravity is the only force) → use the suvat formulae
$a$ contains $t$ → you must use calculus

The suvat formulae are invalid when acceleration varies. Applying $s = ut + \tfrac12at^2$ to $a = 6t - 4$ is simply wrong, and no marks follow. Check whether $a$ depends on $t$ before choosing.

Worked Example 6 — Recognising the special case

A stone is dropped from rest down a well and hits the water after $2.5$ seconds. Taking $g = 9.8$ m s⁻², find the depth.

①Gravity alone, so the acceleration is the constant $9.8$ m s⁻². The suvat formulae apply.

②$u = 0$, $a = 9.8$, $t = 2.5$; we want $s$.

③$s = ut + \tfrac12 at^2 = 0 + \tfrac12(9.8)(6.25)$

④$= 30.625$, so the well is $30.6$ m deep (3 s.f.).

⑤By calculus, as a check: $v = \int 9.8\,\mathrm{d}t = 9.8t$ (since $u=0$), then $s = \int_0^{2.5} 9.8t\,\mathrm{d}t = \left[4.9t^2\right]_0^{2.5} = 30.625$ ✓

Both methods agree, but the formula took one line. That is the point of CA15 — recognising when the shortcut is legitimate.

Worked Example 7 — When the shortcut is not available

A train's acceleration is $a = 1.2 - 0.06t$ m s⁻². It starts from rest. Find its speed when the acceleration first reaches zero, and the distance travelled by then.

①$a$ depends on $t$, so the suvat formulae are not available.

②$a = 0$ when $1.2 = 0.06t$, so $t = 20$ s.

③$v = \displaystyle\int \left(1.2 - 0.06t\right)\mathrm{d}t = 1.2t - 0.03t^2 + c$, and $v = 0$ at $t = 0$ gives $c = 0$.

④At $t = 20$: $v = 24 - 12 = 12$ m s⁻¹

⑤$s = \displaystyle\int_0^{20}\left(1.2t - 0.03t^2\right)\mathrm{d}t = \Big[0.6t^2 - 0.01t^3\Big]_0^{20}$

⑥$= 240 - 80 = 160$ m

⑦Note that $t = 20$ gives the maximum velocity, since $a$ turns negative afterwards.

5 Quick Reference

Differentiate

$s \to v \to a$.

Integrate

$a \to v \to s$, with a constant each time.

Constants

Come from the initial conditions.

At rest

Set $v = 0$.

Max velocity

Set $a = 0$.

Displacement change

$\displaystyle\int_{t_1}^{t_2} v\,\mathrm{d}t$.

Total distance

Split where $v = 0$; add the sizes.

Speed

The size of velocity — never negative.

Constant $a$

Use the suvat formulae.

Variable $a$

Calculus only — suvat is invalid.

6 Practice Questions

Question 1

A particle has $s = 4t^2 + 3t$ m. Find its velocity at $t = 2$.

▶ Show solution

$v = 8t + 3$

At $t=2$: $v = 19$ m s⁻¹

Question 2

A particle has $v = 3t^2 - 2t$ m s⁻¹. Find its acceleration at $t = 3$.

▶ Show solution

$a = 6t - 2$

At $t=3$: $a = 16$ m s⁻²

Question 3

A particle has $s = t^3 - 9t$ m. Find when it is momentarily at rest.

▶ Show solution

$v = 3t^2 - 9 = 0$

$t^2 = 3$, so $t = \sqrt3 = 1.73$ s (3 s.f.), taking $t > 0$.

Question 4

A particle has $a = 4$ m s⁻² and starts from rest. Find its velocity after $6$ s.

▶ Show solution

Acceleration is constant, so use $v = u + at$:

$v = 0 + 4(6) = 24$ m s⁻¹

Question 5

A particle has $a = 12t$ m s⁻² and velocity $2$ m s⁻¹ at $t = 0$. Find $v$ in terms of $t$.

▶ Show solution

$v = \displaystyle\int 12t \,\mathrm{d}t = 6t^2 + c$

At $t=0$, $v=2$, so $c = 2$.

$v = 6t^2 + 2$

Question 6

A particle has $v = 2t + 1$ m s⁻¹. Find the distance travelled between $t = 1$ and $t = 4$.

▶ Show solution

$v > 0$ throughout, so no splitting is needed.

$\displaystyle\int_1^4 (2t+1)\,\mathrm{d}t = \Big[t^2 + t\Big]_1^4$

$= (16+4) - (1+1) = 18$ m

Question 7

A ball is thrown upwards at $14$ m s⁻¹. Taking $g = 9.8$ m s⁻², find the greatest height reached.

▶ Show solution

Constant acceleration, so use $v^2 = u^2 + 2as$ with $v = 0$ at the top, $u = 14$, $a = -9.8$:

$0 = 196 - 19.6s$

$s = \dfrac{196}{19.6} = 10$ m

Question 8

A particle has $v = t^2 - 6t + 8$ m s⁻¹. Find when it changes direction.

▶ Show solution

$t^2 - 6t + 8 = (t-2)(t-4) = 0$, so $v = 0$ at $t = 2$ and $t = 4$.

At $t=0$: $v = 8 > 0$. At $t=3$: $v = 9-18+8 = -1 < 0$. At $t=5$: $v = 25-30+8 = 3 > 0$.

The velocity changes sign at both, so the particle changes direction at $t = 2$ s and $t = 4$ s.

Question 9

A car's acceleration is $a = 3 - 0.5t$ m s⁻². It starts from rest. Find its maximum velocity.

▶ Show solution

Velocity is greatest when $a = 0$: $3 = 0.5t$, so $t = 6$ s.

$v = \displaystyle\int (3 - 0.5t)\,\mathrm{d}t = 3t - 0.25t^2 + c$, and $v(0) = 0$ gives $c = 0$.

At $t = 6$: $v = 18 - 9 = 9$ m s⁻¹

Since $a$ becomes negative after $t=6$, this is a maximum.

Question 10

A particle moves in a straight line with velocity $v = 3t^2 - 12t + 9$ m s⁻¹, starting from the origin at $t = 0$.

(a) Find the acceleration at $t = 1$. (b) Find when the particle is momentarily at rest. (c) Find its displacement at $t = 4$. (d) Find the total distance travelled in the first $4$ seconds, and explain why this exceeds the answer to (c).

▶ Show solution

(a) $a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = 6t - 12$

At $t=1$: $a = -6$ m s⁻² — the particle is decelerating.

(b) $3t^2 - 12t + 9 = 3(t-1)(t-3) = 0$

At rest at $t = 1$ s and $t = 3$ s.

(c) $s = \displaystyle\int_0^4 \left(3t^2-12t+9\right)\mathrm{d}t = \Big[t^3 - 6t^2 + 9t\Big]_0^4$

At $t=4$: $64 - 96 + 36 = 4$. At $t=0$: $0$.

Displacement $= 4$ m from the origin.

(d) The velocity changes sign at $t=1$ and $t=3$, so split the interval into three parts. Write $\mathrm{S}(t) = t^3 - 6t^2 + 9t$.

$\mathrm{S}(0) = 0$, $\mathrm{S}(1) = 1-6+9 = 4$, $\mathrm{S}(3) = 27-54+27 = 0$, $\mathrm{S}(4) = 4$

$0 \to 1$ s: displacement $+4$ m (forwards)

$1 \to 3$ s: displacement $0 - 4 = -4$ m (backwards)

$3 \to 4$ s: displacement $4 - 0 = +4$ m (forwards)

Total distance $= 4 + 4 + 4 = \mathbf{12}$ m

Why it exceeds (c): the particle went $4$ m forwards, came all the way back to the origin, then went $4$ m forwards again. Its net displacement is only $4$ m, but the odometer reading — the total path length — is $12$ m.

Displacement counts only where you ended up; distance counts every metre travelled, which is why the backwards leg must be added rather than subtracted.

Application to Kinematics (CA14–CA15) · OCR FSMQ Additional Maths · Created with MathJax