Equations of motion (SUVAT) — derivations from calculus: Worked examples

The scenario matrix

Before working anything, here is the full map. We split every SUVAT problem into case classes — a case class just means "a genuinely different flavour of problem", one that needs a different trick or watch-out. A degenerate input means one of the five quantities is set to a special value (like $u = 0$ or $a = 0$ ) that makes some terms vanish. The last column names the example that covers each row.

#	Case class	What makes it different	Sign / edge trap	Example
A	Speeding up, all positive	$u > 0, a > 0$	none — the "easy" base case	Ex 1
B	Slowing to a stop	$a < 0$ , ends at $v = 0$	$a$ opposes $v$ , so $a$ is negative	Ex 2
C	Start from rest	$u = 0$ (degenerate input: a $u$ -term vanishes)	the $u$ -terms vanish	Ex 3
D	Reversal: up then down	$u > 0, a < 0$ , passes through $v = 0$	displacement ≠ distance; sign flips mid-flight	Ex 4
E	Two roots for $t$	quadratic $s (t)$ has two solutions	which root is physical?	Ex 5
F	Limiting value / $a \to 0$	acceleration shrinks toward zero (degenerate input)	formulas must reduce to constant-speed motion	Ex 6
G	Real-world word problem	must extract SUVAT letters from prose	choose positive direction first	Ex 7
H	Exam twist: two phases	accelerate, then brake	SUVAT applies per phase, not across	Ex 8

We work every row below.

Case A — the base case (all positive)

Worked example Ex 1 · Speeding up

A train leaves a station at $u = 8 m/s$ and accelerates at $a = 1.5 m/s^{2}$ for $t = 12 s$ . Find its final speed $v$ and the distance $s$ travelled.

Forecast: speed should climb well past $8$ ; distance should be more than $8 \times 12 = 96$ (because it's speeding up). Guess the numbers before reading on.

Pick the equation for $v$ . Missing variable is $s$ → use $v = u + a t$ . Why this step? We know $u, a, t$ and want $v$ ; the timeless and average equations both need something we'd have to compute first. $v = 8 + 1.5 \times 12 = 26 m/s$
Pick the equation for $s$ . Now we know $u, v, t$ → the average-velocity form is cleanest. Why this step? $s = \frac{1}{2} (u + v) t$ avoids re-squaring anything. $s = \frac{1}{2} (8 + 26) \times 12 = \frac{1}{2} \times 34 \times 12 = 204 m$

Verify: cross-check $s$ with $s = u t + \frac{1}{2} a t^{2} = 8 (12) + \frac{1}{2} (1.5) (144) = 96 + 108 = 204 m$ ✓. And $204 > 96$ as forecast. Units: $m/s \times s = m$ ✓.

Case B — braking to a stop

Worked example Ex 2 · Slowing down (sign of

$a$ ) A cyclist at $u = 12 m/s$ brakes at $3 m/s^{2}$ and stops. How long does it take, and how far do they roll?

Forecast: stopping time should be a handful of seconds; distance a couple of dozen metres.

Choose positive direction = direction of motion. Then the brake pushes backwards, so $a = - 3 m/s^{2}$ . Why this step? SUVAT doesn't know "brake" — it only knows signs. Deceleration is a negative acceleration relative to the chosen positive axis.
Find $t$ with $v = u + a t$ and $v = 0$ (stopped). Why this step? We want time and know $u, v, a$ ; this equation omits $s$ , which we don't yet need. $0 = 12 + (- 3) t \Rightarrow t = \frac{12}{3} = 4 s$
Find $s$ with $v^{2} = u^{2} + 2 a s$ , $v = 0$ . Why this step? Timeless form goes straight from speeds to distance. $0 = 1 2^{2} + 2 (- 3) s \Rightarrow s = \frac{144}{6} = 24 m$

Verify: $s = \frac{1}{2} (u + v) t = \frac{1}{2} (12 + 0) (4) = 24 m$ ✓. Positive $s$ — good, they still move forwards while stopping.

Case C — starting from rest (degenerate input $u = 0$ )

Worked example Ex 3 · Dropped from rest

A ball is dropped ( $u = 0$ ) and falls freely for $t = 2 s$ . Here $g$ is the magnitude of gravitational acceleration, $g = 9.8 m/s^{2}$ (see Free fall and g). We choose down as positive for this problem, so the acceleration is simply $a = + g = + 9.8 m/s^{2}$ . Find its speed and how far it drops.

Forecast: each second adds about $9.8$ to the speed, so ~ $20$ m/s; distance somewhere near $20$ m.

Set $u = 0$ . Every term with $u$ disappears. Why this step? "From rest" is the special case that kills the $u t$ term — this is the degenerate input the matrix warns about.
Speed: $v = u + a t = 0 + 9.8 (2) = 19.6 m/s$ (positive = downward, as chosen).
Distance: $s = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} (9.8) (2^{2}) = 19.6 m$ . Why this step? With $u = 0$ this collapses to just $\frac{1}{2} a t^{2}$ .

Verify: timeless check $v^{2} = u^{2} + 2 a s \Rightarrow 19. 6^{2} = 0 + 2 (9.8) (19.6)$ . Left $= 384.16$ , right $= 384.16$ ✓.

Case D — the reversal (displacement ≠ distance)

This is the trap that catches the most people, so it gets a figure.

Worked example Ex 4 · Thrown straight up

A stone is thrown up at $u = 20 m/s$ . Here we choose up as positive, so gravity (which pulls down) gives $a = - g = - 9.8 m/s^{2}$ . Find (a) time to the top, (b) max height, (c) its velocity at $t = 3 s$ , (d) its displacement at $t = 3 s$ .

Forecast: it rises ~2 s, peaks around 20 m, and by 3 s it's already falling back down.

Time to top: at the peak the stone is momentarily still, $v = 0$ . Use $v = u + a t$ . Why this step? The top is defined by $v = 0$ ; that's our target condition. $0 = 20 + (- 9.8) t \Rightarrow t = \frac{20}{9.8} = 2.0408 s$
Max height: $v^{2} = u^{2} + 2 a s$ with $v = 0$ . $0 = 2 0^{2} + 2 (- 9.8) s \Rightarrow s = \frac{400}{19.6} = 20.408 m$
Velocity at $t = 3$ s: $v = 20 + (- 9.8) (3) = - 9.4 m/s$ . Why this step? The negative sign tells you it's moving downward now — the formula tracks direction automatically. Look at the red arrow in the figure: it has flipped.
Displacement at $t = 3$ s: $s = u t + \frac{1}{2} a t^{2} = 20 (3) + \frac{1}{2} (- 9.8) (9) = 60 - 44.1 = 15.9 m$ . Why this step? $s$ is measured from the launch point. It's less than the peak height because the stone has started coming back — displacement is not the total path length.

Verify: at $t = 3$ s the stone is at $+ 15.9$ m (still above the hand) yet its velocity is $- 9.4$ (heading down) — consistent with "past the top, falling". Distance travelled $= 20.408$ up $+ (20.408 - 15.9) = 4.508$ down $= 24.9$ m, larger than the $15.9$ m displacement ✓ — exactly the distance-vs-displacement point.

Case E — two roots: which time is physical?

Worked example Ex 5 · When does it pass a given height?

Using the same throw ( $u = 20$ , $a = - 9.8$ , up positive), at what time(s) is the stone at height $s = 15 m$ ?

Forecast: two answers — once on the way up, once on the way down.

Set up the quadratic: $s = u t + \frac{1}{2} a t^{2}$ . $15 = 20 t + \frac{1}{2} (- 9.8) t^{2} \Rightarrow 4.9 t^{2} - 20 t + 15 = 0$ Why this step? Height as a function of time is a parabola, so asking "when is $s = 15$ ?" is asking where the parabola hits a horizontal line — that generally has two crossings.
Solve with the quadratic formula $t = \frac{20 \pm 2 0 ^{2} - 4 ( 4.9 ) ( 15 )}{2 ( 4.9 )}$ . Why this step? The equation is quadratic in $t$ (it has a $t^{2}$ term), and the quadratic formula is the general tool that returns both roots at once — exactly what we need when we suspect two crossing times. Nothing else (linear rearrangement) would find them both. $t = \frac{20 \pm 400 - 294}{9.8} = \frac{20 \pm 106}{9.8}$ $t_{1} = \frac{20 - 10.2956}{9.8} = 0.9902 s, t_{2} = \frac{20 + 10.2956}{9.8} = 3.0914 s$
Screen the roots immediately. Both $t_{1}$ and $t_{2}$ are positive and real, so both happen after launch — neither is discarded here. (If a root had come out negative, we'd throw it away at once: negative time means "before the throw", which never happened.) Why this step? Root selection is a physics decision, not algebra: keep only times that fall inside the motion we're describing.
Interpret the survivors. $t_{1} \approx 0.99$ s = passing $15$ m going up; $t_{2} \approx 3.09$ s = passing the same height coming down.

Verify: their sum should equal $2 \times$ (time to top) by symmetry of the parabola: $0.9902 + 3.0914 = 4.0816 = 2 \times 2.0408$ ✓.

Case F — limiting behaviour ( $a \to 0$ )

Worked example Ex 6 · What if there's (almost) no acceleration?

A puck slides at $u = 6 m/s$ with a tiny deceleration $a = - 0.01 m/s^{2}$ over $t = 5 s$ . Find $s$ , and show that as $a \to 0$ the result matches plain constant-speed motion.

Forecast: with such a small $a$ , distance ≈ $6 \times 5 = 30$ m, just a hair less.

Full formula: $s = u t + \frac{1}{2} a t^{2} = 6 (5) + \frac{1}{2} (- 0.01) (25) = 30 - 0.125 = 29.875 m$ .
Take the limit $a \to 0$ : the $\frac{1}{2} a t^{2}$ term vanishes, leaving $s \to u t = 30 m$ . Why this step? Constant acceleration includes zero acceleration as a boundary case. Checking that the formula degrades gracefully to $s = u t$ is a sanity check that our physics is consistent — see Velocity-time graphs, where $a = 0$ is a flat horizontal line and the area is just a rectangle $u \times t$ .

Verify: the correction term $\frac{1}{2} ∣ a ∣ t^{2} = 0.125$ m is $0.4%$ of $30$ m — small, as a "nearly-frictionless" puck should be ✓.

Case G — extracting SUVAT from a word problem

Worked example Ex 7 · The overtaking lorry

A car travelling at a steady $25 m/s$ passes a stationary police bike. The instant the car passes, the bike sets off from rest at $a = 4 m/s^{2}$ . How long until the bike catches the car, and how far have they gone?

Forecast: the bike is faster-accelerating but starts from zero, so catch-up takes several seconds.

Choose positive direction = road ahead. List each vehicle's SUVAT. Why this step? Two objects means two motions; a shared positive direction lets us compare positions directly.
- Car: $u = 25$ , $a = 0$ → position $s_{car} = 25 t$ .
- Bike: $u = 0$ , $a = 4$ → position $s_{bike} = \frac{1}{2} (4) t^{2} = 2 t^{2}$ .
Catch-up means equal positions: $s_{bike} = s_{car}$ . Why this step? "Catches up" is a position condition, not a speed one. $2 t^{2} = 25 t \Rightarrow t (2 t - 25) = 0 \Rightarrow t = 0 or t = 12.5 s$
Discard $t = 0$ (that's the instant of passing) → catch-up at $t = 12.5 s$ .
Distance: $s = 25 \times 12.5 = 312.5 m$ .

Verify: bike's distance $= 2 (12.5)^{2} = 2 \times 156.25 = 312.5 m$ — matches ✓. Bike's speed at catch-up $= a t = 4 (12.5) = 50 m/s$ , double the car's — plausible for it to have caught up.

Case H — the exam twist: two phases

Worked example Ex 8 · Accelerate then brake

A drone starts from rest, accelerates at $2 m/s^{2}$ for $6 s$ , then immediately brakes at $3 m/s^{2}$ until it stops. Find the total distance travelled.

Forecast: it reaches a decent speed, then takes a shorter distance to stop (harder braking).

Phase 1 (accelerate). $u_{1} = 0, a_{1} = 2, t_{1} = 6$ . Why split? SUVAT needs one constant $a$ at a time. Acceleration changes at the switch, so we treat each leg separately — this is the whole point of the cell. $v_{1} = 0 + 2 (6) = 12 m/s, s_{1} = \frac{1}{2} (0 + 12) (6) = 36 m$
Phase 2 (brake). The end speed of phase 1 is the start speed of phase 2: $u_{2} = 12$ , $a_{2} = - 3$ , ends at $v_{2} = 0$ . Why this step? Motion is continuous — velocity carries over even though acceleration jumps. $0 = 1 2^{2} + 2 (- 3) s_{2} \Rightarrow s_{2} = \frac{144}{6} = 24 m$
Total distance: $s = s_{1} + s_{2} = 36 + 24 = 60 m$ .

Verify: braking time $t_{2} = \frac{12}{3} = 4$ s; average speed each phase is $6$ m/s, so $s_{1} = 6 \times 6 = 36$ , $s_{2} = 6 \times 4 = 24$ — both match ✓. Total time $= 10$ s.

Recall Which cell is which?

Speeding up all-positive is cell ::: A (Ex 1). The cell where displacement differs from distance is ::: D (Ex 4, thrown up). The cell producing two valid times for the same height is ::: E (Ex 5). When $a \to 0$ , $s = u t + \frac{1}{2} a t^{2}$ reduces to ::: $s = u t$ (cell F). "Accelerate then brake" needs SUVAT applied ::: once per phase (cell H).

Connections

Equations of motion (SUVAT) — derivations from calculus — the parent; every formula used here is derived there.
Velocity-time graphs — cells A–F are just areas and slopes on a $v$ – $t$ plot.
Vectors — the signs in cells B and D come from choosing a positive axis direction.
Projectile motion — cell D is the vertical half of a projectile.
Free fall and g — cell C uses $a = g$ .
Simple Harmonic Motion — a reminder that if $a$ weren't constant, none of these cells would apply.