SUVAT Equations of Motion

Spec mapping: OCR H556 Module 3.1 — Motion (equations of motion for constant acceleration in a straight line; their application to free fall and vehicle-safety contexts). Refer to the official OCR H556 specification document for exact wording.

The four SUVAT equations are the algebraic workhorses of A-Level kinematics. They let you solve any problem involving uniform (constant) acceleration in a straight line without drawing a graph, provided you know three of the five kinematic variables. They will reappear throughout the H556 course: in projectile motion, free fall, inclined-plane dynamics, momentum collisions (where impulse $= \Delta p$=Δp produces a SUVAT-style velocity change), the approach to terminal velocity (where they cease to apply), and even the gravitational-field "escape velocity" derivations of Module 5.5.

In this lesson you will (i) derive all four equations from first principles, (ii) learn a robust strategy for picking the right equation, (iii) work through six exam-style problems including a quadratic-trap and a meeting-point problem, and (iv) apply SUVAT to the vehicle-safety context that OCR examines via thinking and braking distances.

Key Definition: A SUVAT equation is a kinematic relation among the variables $s$s, $u$u, $v$v, $a$a, $t$t valid only when acceleration $a$a is constant in magnitude and direction throughout the interval. For non-constant $a$a, return to graphical or calculus methods.

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The Five SUVAT Variables

Every uniform-acceleration problem involves at most five quantities:

Symbol	Quantity	SI unit
$s$s	displacement	m
$u$u	initial velocity	m s$^{-1}$−1
$v$v	final velocity	m s$^{-1}$−1
$a$a	acceleration	m s$^{-2}$−2
$t$t	time interval	s

The mnemonic SUVAT is "$s$s, $u$u, $v$v, $a$a, $t$t". A typical exam problem hands you three of these and asks for one of the remaining two. There are exactly four independent SUVAT equations, each of which uses four of the five variables (omitting one). Your job is to identify the unmentioned variable and pick the equation that also omits it.

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Derivation From First Principles

Assume uniform acceleration $a$a, starting at time $t = 0$t=0 with velocity $u$u and ending at time $t$t with velocity $v$v after displacement $s$s.

Equation 1 — $v = u + at$v=u+at

By definition of acceleration (constant $a$a):

$a = \frac{v - u}{t} \quad \Longrightarrow \quad \boxed{v = u + at}$a=tv−u​⟹v=u+at​

This equation omits $s$s.

Equation 2 — $s = \tfrac{1}{2}(u + v)\,t$s=21​(u+v)t

For uniform acceleration, the velocity grows linearly with time, so the average velocity over the interval is the arithmetic mean of the endpoints:

$\bar{v} = \frac{u + v}{2}$vˉ=2u+v​

Displacement is then average velocity multiplied by time:

$\boxed{s = \tfrac{1}{2}(u + v)\,t}$s=21​(u+v)t​

This equation omits $a$a. Geometrically, this is the area of the trapezium under the $v$v–$t$t graph: $\frac{1}{2}(a + b)h$21​(a+b)h with $a = u$a=u, $b = v$b=v, $h = t$h=t.

Equation 3 — $s = ut + \tfrac{1}{2} a t^2$s=ut+21​at2

Substitute Equation 1 ($v = u + at$v=u+at) into Equation 2:

$s = \tfrac{1}{2}(u + u + at)\,t = \tfrac{1}{2}(2u + at)\,t = ut + \tfrac{1}{2} a t^2$s=21​(u+u+at)t=21​(2u+at)t=ut+21​at2

$\boxed{s = ut + \tfrac{1}{2} a t^2}$s=ut+21​at2​

This equation omits $v$v. Geometrically, it splits the trapezium under the $v$v–$t$t graph into a rectangle of area $ut$ut and a triangle of area $\frac{1}{2} a t \cdot t$21​at⋅t.

Equation 4 — $v^2 = u^2 + 2as$v2=u2+2as

Square Equation 1:

$v^2 = (u + at)^2 = u^2 + 2uat + a^2 t^2 = u^2 + 2a\left(ut + \tfrac{1}{2} a t^2\right) = u^2 + 2as$v2=(u+at)2=u2+2uat+a2t2=u2+2a(ut+21​at2)=u2+2as

$\boxed{v^2 = u^2 + 2as}$v2=u2+2as​

This equation omits $t$t — making it the go-to whenever a problem neither gives nor asks for time. It is also the SUVAT form most closely allied to energy conservation: multiply both sides by $\frac{1}{2}m$21​m and recognise $\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mas = Fs$21​mv2−21​mu2=mas=Fs, i.e. the work-energy theorem.

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Decision Table — Which Equation When

Equation	Omits	Use when …
$v = u + at$v=u+at	$s$s	Displacement is neither given nor asked
$s = \tfrac{1}{2}(u + v)\,t$s=21​(u+v)t	$a$a	Acceleration unknown or irrelevant
$s = ut + \tfrac{1}{2} a t^2$s=ut+21​at2	$v$v	Final velocity unknown or irrelevant
$v^2 = u^2 + 2as$v2=u2+2as	$t$t	Time unknown or irrelevant

flowchart TD
  A[Read question] --> B[List SUVAT variables given]
  B --> C[Identify the unknown asked for]
  C --> D{Which of s, u, v, a, t is NEVER mentioned?}
  D -->|s missing| E[Use v = u + at]
  D -->|a missing| F[Use s = half u+v t]
  D -->|v missing| G[Use s = ut + half a t squared]
  D -->|t missing| H[Use v squared = u squared + 2as]
  E --> I[Substitute and solve]
  F --> I
  G --> I
  H --> I

Exam Tip: Write down the SUVAT list as a column at the start of every problem with values beside each known variable and a "?" beside the asked-for unknown. Circle the variable that is neither given nor asked — that variable's absence picks the right equation.

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Worked Example 1 — Stopping Distance

A car travelling at 28 m s$^{-1}$−1 brakes and decelerates uniformly at 6.5 m s$^{-2}$−2. Calculate the distance to stop.

• $u = 28$u=28 m s$^{-1}$−1, $v = 0$v=0, $a = -6.5$a=−6.5 m s$^{-2}$−2, $s = ?$s=?, $t$t is neither given nor asked.
• Use $v^2 = u^2 + 2as$v2=u2+2as:

$0^2 = 28^2 + 2(-6.5)s \Rightarrow 0 = 784 - 13s \Rightarrow s = \frac{784}{13} = 60.3 \text{ m}$02=282+2(−6.5)s⇒0=784−13s⇒s=13784​=60.3 m

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Worked Example 2 — Two-Stage Motion

A motorcycle starts from rest and accelerates uniformly at $3.2$3.2 m s$^{-2}$−2 for $6.0$6.0 s, then maintains the velocity reached for a further $10$10 s. Find the total distance.

Stage 1 ($u = 0$u=0, $a = 3.2$a=3.2 m s$^{-2}$−2, $t = 6.0$t=6.0 s):

• $v = u + at = 0 + 3.2 \times 6.0 = 19.2$v=u+at=0+3.2×6.0=19.2 m s$^{-1}$−1.
• $s_1 = ut + \tfrac{1}{2} a t^2 = 0 + \tfrac{1}{2} \times 3.2 \times 36 = 57.6$s1​=ut+21​at2=0+21​×3.2×36=57.6 m.

Stage 2 (constant 19.2 m s$^{-1}$−1 for 10 s):

• $s_2 = 19.2 \times 10 = 192$s2​=19.2×10=192 m.

Total $= 57.6 + 192 = 249.6$=57.6+192=249.6 m $\approx 250$≈250 m (3 s.f.).

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Worked Example 3 — Catch the Train

A train leaves a station from rest with uniform acceleration $0.50$0.50 m s$^{-2}$−2. A passenger arrives $4.0$4.0 s after the train departed and immediately starts running at a steady $7.0$7.0 m s$^{-1}$−1 along the platform, in the direction of the train. After how long does the passenger catch the train (if at all)?

Let $t$t = time elapsed since the passenger starts running. Take the station as origin.

• Passenger's position: $s_p = 7.0\,t$sp​=7.0t.
• Train's position (measured from station, $\tau = t + 4$τ=t+4 s after train started): $s_T = \tfrac{1}{2} \times 0.50 \times (t + 4)^2 = 0.25\,(t + 4)^2$sT​=21​×0.50×(t+4)2=0.25(t+4)2.

Setting $s_p = s_T$sp​=sT​:

$7.0\,t = 0.25(t + 4)^2 \quad \Longrightarrow \quad 28\,t = (t + 4)^2 = t^2 + 8t + 16$7.0t=0.25(t+4)2⟹28t=(t+4)2=t2+8t+16

$t^2 - 20\,t + 16 = 0 \Rightarrow t = \frac{20 \pm \sqrt{400 - 64}}{2} = \frac{20 \pm \sqrt{336}}{2}$t2−20t+16=0⇒t=220±400−64​​=220±336​​

The discriminant is positive ($336 > 0$336>0), so the passenger does catch the train. The two roots are $t \approx 0.83$t≈0.83 s and $t \approx 19.17$t≈19.17 s. Both are mathematically valid: the smaller root is the moment the passenger first reaches the train, the larger root is when the still-accelerating train catches the (constant-velocity) passenger again. The physically meaningful answer to "how long until first catch?" is $t \approx 0.83$t≈0.83 s.

This problem rewards two skills: forming a quadratic from kinematic positions, and reading both roots in physical context.

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Worked Example 4 — Quadratic Trap: Ball Returns to Start

A ball is thrown vertically upward from ground level at $u = 18$u=18 m s$^{-1}$−1. Take upward positive, $g = 9.81$g=9.81 m s$^{-2}$−2. How long until it returns to ground level?

Set $s = 0$s=0 at start and at return:

$0 = 18\,t + \tfrac{1}{2}(-9.81)\,t^2 = t(18 - 4.905\,t)$0=18t+21​(−9.81)t2=t(18−4.905t)

Solutions: $t = 0$t=0 (moment of throwing) and $t = 18 / 4.905 = 3.67$t=18/4.905=3.67 s. Total time of flight: 3.67 s.

Note that the $t = 0$t=0 root is mathematically valid but physically uninteresting — the question asks for after throwing. Recognising and rejecting trivial roots is part of the algebraic discipline OCR expects.

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Worked Example 5 — Meeting Point

Object A starts from rest at $x = 0$x=0 and accelerates at $2.0$2.0 m s$^{-2}$−2 in the positive $x$x-direction. Object B starts at $x = 60$x=60 m with velocity $-4.0$−4.0 m s$^{-1}$−1 and zero acceleration. When and where do they meet?

• $x_A(t) = \tfrac{1}{2} \times 2.0 \times t^2 = t^2$xA​(t)=21​×2.0×t2=t2.
• $x_B(t) = 60 - 4.0\,t$xB​(t)=60−4.0t.

Setting $x_A = x_B$xA​=xB​:

$t^2 + 4t - 60 = 0 \Rightarrow t = \frac{-4 \pm \sqrt{16 + 240}}{2} = \frac{-4 \pm 16}{2}$t2+4t−60=0⇒t=2−4±16+240​​=2−4±16​

Physical root: $t = 6.0$t=6.0 s. Meeting position: $x = 36$x=36 m. The negative root ($-10$−10 s) is the (imaginary) past time at which the trajectories would have intersected had the motion been extrapolated backwards under the same equations — a useful check on root choice.

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Worked Example 6 — Cross-Check via Energy

Re-examine the stopping-distance problem (Worked Example 1) using the work-energy theorem.

• Initial kinetic energy: $E_k = \frac{1}{2}mu^2$Ek​=21​mu2. Final $E_k = 0$Ek​=0.
• Work done by the braking force: $W = F \times s$W=F×s.
• Work-energy theorem: $W = -\Delta E_k \Rightarrow F \times s = \frac{1}{2}mu^2 \Rightarrow s = \frac{u^2}{2(F/m)} = \frac{u^2}{2a}$W=−ΔEk​⇒F×s=21​mu2⇒s=2(F/m)u2​=2au2​.

Substituting: $s = 28^2 / (2 \times 6.5) = 784 / 13 = 60.3$s=282/(2×6.5)=784/13=60.3 m. Identical to the SUVAT answer.

This is no accident — SUVAT's fourth equation is the work-energy theorem in disguise. Cross-checking by an independent method is the surest way to catch arithmetic slips under exam pressure.

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Common Exam Pitfalls (in working through SUVAT)

• Sign convention slippage. If upward is positive, then $g = -9.81$g=−9.81 m s$^{-2}$−2 for a freely falling body. Half of all sign errors stem from inconsistent or unstated axes.
• Applying SUVAT to non-uniform acceleration. A skydiver with drag, a car with engine power proportional to velocity, a parachutist after deployment — none are uniform-$a$a scenarios. SUVAT is invalid. Use graphs or calculus.
• Ignoring the second root. A quadratic in $t$t produces two roots; both have physical meaning (and ignoring one can lose marks).
• Rounding too early. Keep extra significant figures throughout; round only on the final answer. Premature rounding propagates errors and may cost the final mark in OCR's "to 3 s.f." style of asking.
• Unit mixing. Convert km h$^{-1}$−1 to m s$^{-1}$−1 ($\div 3.6$÷3.6), and cm to m ($\div 100$÷100), before substitution.

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Application — Stopping Distance and Vehicle Safety

SUVAT has a directly practical use in road-safety physics. The total stopping distance of a vehicle is the sum of two parts:

• Thinking distance — the distance travelled during the driver's reaction time, at constant velocity: $s_{\text{think}} = v \times t_{\text{react}}$sthink​=v×treact​ For an alert driver, $t_{\text{react}} \approx 0.7$treact​≈0.7 s.
• Braking distance — the distance required to decelerate from $v$v to 0, found from $v^2 = u^2 + 2as$v2=u2+2as with $u = v$u=v, $v_{\text{final}} = 0$vfinal​=0: $s_{\text{brake}} = \frac{v^2}{2 a_{\text{brake}}}$sbrake​=2abrake​v2​

Total:

$s_{\text{stop}} = v \, t_{\text{react}} + \frac{v^2}{2 a_{\text{brake}}}$sstop​=vtreact​+2abrake​v2​

Worked Example 7 — Highway Code at 30 mph

A car travels at $30$30 mph $\approx 13.4$≈13.4 m s$^{-1}$−1. With $t_{\text{react}} = 0.7$treact​=0.7 s and $a_{\text{brake}} = 6.75$abrake​=6.75 m s$^{-2}$−2:

• $s_{\text{think}} = 13.4 \times 0.7 = 9.4$sthink​=13.4×0.7=9.4 m.
• $s_{\text{brake}} = 13.4^2 / (2 \times 6.75) = 13.3$sbrake​=13.42/(2×6.75)=13.3 m.
• Total $\approx 22.7$≈22.7 m (Highway Code quotes 23 m).

Doubling the speed to $60$60 mph ($26.8$26.8 m s$^{-1}$−1) quadruples braking distance (it scales as $v^2$v2) but only doubles thinking distance. Total stopping distance becomes $18.8 + 53.2 \approx 72$18.8+53.2≈72 m. The non-linearity in $v$v is precisely why motorway speeds make a moment's distraction so dangerous.

Factors that increase thinking distance: alcohol, drugs, fatigue, mobile-phone distraction, passenger conversation.

Factors that increase braking distance: worn tyres (reduced friction $\Rightarrow$⇒ smaller deceleration), wet or icy road, worn brake pads, downhill gradient, heavier load.

OCR routinely examines stopping-distance qualitatively in the context of road-safety policy. Be ready to argue, for example, why a 20 mph zone outside a school dramatically reduces collision severity: at $v = 8.9$v=8.9 m s$^{-1}$−1, total stopping distance is $6.2 + 5.9 = 12.1$6.2+5.9=12.1 m, less than half the $30$30-mph value, and the kinetic energy involved in any collision (which scales as $v^2$v2) is just $44\%$44% of the $30$30-mph case.

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