Year 11 Physics Module 1 ⏱ ~30 min 5 MC · 3 Short Answer Lesson 6 of 8

Equations of Motion, SUVAT in One Dimension

A car on the Hume Highway hits the brakes at 28 m/s. The crash investigator photographs a 64 m skid mark, and from that one length, plus the road's grip, calculates exactly how fast the car was going before it braked. No video, no witnesses. Just five equations that lock together time, velocity, acceleration and distance for any uniformly accelerating object.

Today's hook: A skid mark is just a length on the road. How can it tell an investigator the car's speed before it ever started braking?
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Warm up first

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Before you read, predict

A car starts from rest and speeds up steadily, reaching 12 m/s after 4 s. You know the starting speed, the finishing speed and the time, but not the distance. Without any formula given to you, how could you work out how far the car travelled? What "average" might help?

For an object moving with uniform (constant) acceleration from rest, the average velocity over the trip equals:

Learning Intentions
goals

Know

  • The five SUVAT variables: $s$, $u$, $v$, $a$, $t$, and what each one means
  • The five equations of uniformly accelerated motion
  • The condition that makes these equations valid (constant acceleration, rectilinear)
  • The accepted value $g = 9.8$ m/s² for free fall near Earth's surface

Understand

  • How $v = u + at$ comes directly from the definition of acceleration
  • How $s = ut + \tfrac{1}{2}at^2$ comes from the area under a velocity–time graph
  • Why each equation simply leaves out one of the five variables
  • Why a consistent sign convention is essential for braking and free-fall problems

Can Do

  • List the known and unknown variables, then choose the right equation
  • Solve braking, launch and free-fall problems with full working
  • Apply a sign convention correctly to a multi-step problem
  • Check an answer is physically reasonable (sign, size, units)
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Scan these before reading

Scan these before reading
vocab
SUVATa memory aid for the five motion variables: $s$ (displacement), $u$ (initial velocity), $v$ (final velocity), $a$ (acceleration), $t$ (time)
Initial velocity ($u$)the velocity at the start of the interval being analysed (for "from rest", $u = 0$)
Final velocity ($v$)the velocity at the end of the interval being analysed
Uniform accelerationacceleration that is constant in size and direction, the only case where these five equations apply
Rectilinear motionmotion along a single straight line, so direction can be handled with + and − signs
Free fallmotion under gravity alone, with acceleration $g = 9.8$ m/s² directed downward
Decelerationan acceleration that opposes the motion, slowing an object down, it carries the opposite sign to the velocity
Cross-lesson links: The equations of motion are built on everything before them: the sign convention from L01, displacement and velocity from L02–L03, and the definition of acceleration from L04. In particular, $s = ut + \tfrac{1}{2}at^2$ is just the area under the velocity–time graph you met in L05, written as algebra. In L08 you will combine these equations with graphs to tackle full multi-stage problems.
Misconceptions to fix
The SUVAT equations work for any motion you like.They only hold when acceleration is constant. If acceleration changes during the interval (e.g. a car shifting gears), you must split the motion into stages where $a$ is constant in each.
When a car brakes, you should plug in a negative distance to show it is slowing down.The car still moves forward, so its displacement stays positive. It is the acceleration that becomes negative, because it points backwards (opposite to the velocity).

True or false: the five equations of motion can be applied directly to a journey in which the acceleration keeps changing.

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SUVAT, Five Variables, One Toolkit

Core Content

01
SUVAT, Five Variables, One Toolkit
+5 XP

Every straight-line problem with steady acceleration is described by just five quantities. Once you have any three of them, the other two are fixed, there is no freedom left. That is what makes these problems solvable: physics gives you a small set of equations that lock the five quantities together.

The five variables are remembered by the word SUVAT:

  • $s$, displacement (how far, and which way, in metres)
  • $u$, initial velocity (speed and direction at the start, in m/s)
  • $v$, final velocity (speed and direction at the end, in m/s)
  • $a$, acceleration (rate of change of velocity, in m/s²)
  • $t$, time (the duration of the interval, in seconds)

Notice that $s$, $u$, $v$ and $a$ are all vectors in one dimension they each carry a + or − sign from your chosen convention. Only $t$ is a scalar. This is exactly why the sign convention from L01 matters here: get a sign wrong and the whole answer flips.

The key habit: at the start of every problem, write the five letters $s\ u\ v\ a\ t$ in a column and fill in the values you know. Mark the one you want to find. The gap that is left tells you which equation to reach for.

The five SUVAT variables are $s$ (displacement), $u$ (initial velocity), $v$ (final velocity), $a$ (acceleration) and $t$ (time). All except $t$ are vectors carrying a sign. Knowing any three lets you find the other two, but only when acceleration is constant.

Pause, copy the five SUVAT variables, their units, and the "list-knowns-first" habit into your book before moving on.

Match each SUVAT symbol to the quantity it represents.

  • s
  • u
  • v
  • a
  • acceleration (m/s²)
  • final velocity (m/s)
  • initial velocity (m/s)
  • displacement (m)
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Where the First Two Equations Come From

02
Where the First Two Equations Come From
+5 XP

We just saw the five SUVAT variables and that any three fix the rest. That raises a question: where do the equations that link them actually come from, are they just memorised, or can we build them? This card answers it → we derive the first two straight from the definition of acceleration and the area under a v–t graph.

You do not have to take these equations on faith. The first one is just the definition of acceleration rearranged, and the second is the area under a velocity–time graph, the same area you met in Lesson 5. Building them this way makes them far easier to remember and trust.

Equation 1: $v = u + at$

Acceleration is the change in velocity per unit time: $a = \dfrac{v - u}{t}$. Multiply both sides by $t$ and rearrange:

From the definition of acceleration $a = \dfrac{v-u}{t} \;\;\Rightarrow\;\; v = u + at$

In words: the final velocity is the starting velocity plus however much the acceleration has added (or removed) over the time. No displacement appears, this equation is the one to use when you do not know and do not need $s$.

Equation 2: $s = ut + \tfrac{1}{2}at^2$

On a velocity–time graph for constant acceleration, the line is straight, sloping from $u$ up to $v$. The displacement is the area under that line. Split the area into a rectangle (height $u$, width $t$) plus a triangle (base $t$, height $v-u = at$):

Area under the v–t graph = rectangle + triangle $s = \underbrace{ut}_{\text{rectangle}} + \underbrace{\tfrac{1}{2}t\,(at)}_{\text{triangle}} = ut + \tfrac{1}{2}at^2$
Why this matters: every one of the five equations can be derived from these ideas. If you ever forget one in an exam, you can rebuild $v = u + at$ from $a = \Delta v / t$ and recover the rest by substitution.

$v = u + at$ comes straight from $a = (v-u)/t$. $s = ut + \tfrac{1}{2}at^2$ is the area under the v–t graph (rectangle $ut$ + triangle $\tfrac{1}{2}at^2$). Both assume constant acceleration.

Pause, copy the two derivations (definition of $a$ → equation 1; v–t graph area → equation 2) into your book before moving on.

Drag the right terms into the gaps.

The equation comes from rearranging the definition of acceleration, while $s = ut + \tfrac{1}{2}at^2$ is found from the under a velocity–time graph.

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The Full Set of Five, and the Missing Variable

03
The Full Set of Five, and the Missing Variable
+5 XP

We just saw the first two equations built from a definition and a graph. That raises a question: what about problems that don't give you time, or don't give you the final velocity, is there an equation for those too? This card answers it → yes; there are five equations in total, and each one is the one that omits a different variable.

There are exactly five equations of uniformly accelerated motion. The clever part is that each equation is built to leave out one of the five variables. So the variable a problem does not mention is your signpost to the right equation.

The five equations of motion (constant $a$, rectilinear) $v = u + at$  ·  $s = ut + \tfrac{1}{2}at^2$  ·  $v^2 = u^2 + 2as$  ·  $s = \tfrac{1}{2}(u+v)\,t$  ·  $s = vt - \tfrac{1}{2}at^2$

Which variable does each leave out?

No displacement / no time

$v = u + at$ has no $s$ use it when displacement is neither given nor wanted.
$v^2 = u^2 + 2as$ has no $t$ perfect for braking and skid-mark problems where time is unknown.

No final/initial velocity / no acceleration

$s = ut + \tfrac{1}{2}at^2$ has no $v$.
$s = vt - \tfrac{1}{2}at^2$ has no $u$.
$s = \tfrac{1}{2}(u+v)\,t$ has no $a$ it is just average velocity × time.

The skid-mark equation: $v^2 = u^2 + 2as$ links velocities to distance with no time at all. That is exactly why a crash investigator can use a skid length $s$ and the road's deceleration $a$ to recover the car's speed $u$ before braking, the time the skid took is never needed.

The five equations: $v=u+at$ (no $s$); $s=ut+\tfrac{1}{2}at^2$ (no $v$); $v^2=u^2+2as$ (no $t$); $s=\tfrac{1}{2}(u+v)t$ (no $a$); $s=vt-\tfrac{1}{2}at^2$ (no $u$). The variable a problem doesn't mention points you to the right equation.

Pause, copy all five equations and the "missing variable" label beside each one into your book before moving on.

A problem gives you $u$, $v$ and $a$, and asks for $s$, but gives no time. Which equation has no time in it (the one to use)?

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Choosing the Equation and Handling Signs

04
Choosing the Equation and Handling Signs
+5 XP

We just saw that each equation omits one variable. That raises a question: in a real problem, how do you turn a wall of words into the right equation, and what happens when something slows down or falls? This card answers it → with a fixed routine for selecting the equation, plus the sign rules that keep braking and free-fall correct.

Solving a SUVAT problem is a routine, not a guess. List what you know, mark what you want, and the missing variable picks the equation. The one thing that trips students up is signs, so decide your positive direction first, and never change it mid-problem.

The four-step routine

  1. State a sign convention. Choose one direction as positive (e.g. "forward = +", or "down = +" for a falling object).
  2. List SUVAT. Write the five letters and fill in the three you know, each with its sign. Mark the unknown you want.
  3. Pick the equation that contains your three knowns and your wanted unknown, i.e. the one missing the variable you don't have.
  4. Substitute and solve, then check the sign and size of the answer make physical sense.
Exam trap, signs: for a car braking while moving forward, take forward as +. Then $u$ and $v$ are positive, but $a$ is negative (it points backward). For an object thrown up with "up = +", $u$ is positive but $a = -9.8$ m/s². Mixing the signs of $a$ and $u$ is the single most common SUVAT error.
Free fall: near Earth's surface, gravity gives every object the same acceleration $g = 9.8$ m/s² downward, regardless of mass. Drop a problem into SUVAT with $a = g$ (sign set by your convention) and the same five equations handle it.

Routine: (1) state a sign convention, (2) list SUVAT with signs, (3) pick the equation missing the variable you don't have, (4) substitute, solve, sanity-check. For braking, $a$ is negative (opposes motion); for free fall, $|a| = g = 9.8$ m/s².

Pause, copy the four-step routine and the sign rules for braking and free fall into your book before moving on.

True or false: taking "forward = positive", a car braking from 20 m/s to 5 m/s while still moving forward has a negative acceleration.

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Worked Problems, Braking and Free Fall

05
Worked Problems, Braking and Free Fall
+5 XP

We just saw the routine for choosing an equation and setting signs. That raises a question: how does this play out on real exam-style problems, a braking car and a falling object? This card answers it → with two fully worked solutions you can copy as templates.

Time to put the routine to work. Watch how listing SUVAT first makes the equation choice almost automatic, and how the sign convention is stated before any number is substituted.

Worked example 1, the braking car (no time given)
  1. Problem. A car travelling at $u = 28$ m/s brakes and stops in a skid of $s = 64$ m. Find its acceleration.
  2. Sign convention. Forward = positive (+).
  3. List SUVAT. $u = +28$ m/s, $v = 0$ (it stops), $s = +64$ m, $a = ?$. Time $t$ is not given.
  4. Pick the equation. No $t$ → use $v^2 = u^2 + 2as$.
  5. Solve. $0 = 28^2 + 2a(64) \;\Rightarrow\; 0 = 784 + 128a \;\Rightarrow\; a = -\dfrac{784}{128} = -6.1$ m/s².
  6. Interpret. The negative sign means the acceleration points backward, the car is decelerating, exactly as expected. Magnitude $\approx 6.1$ m/s².
Worked example 2, free fall (using g)
  1. Problem. A ball is dropped from rest from a balcony and falls for $t = 2.0$ s. How far does it fall, and how fast is it going at that moment?
  2. Sign convention. Down = positive (+), so $a = g = +9.8$ m/s².
  3. List SUVAT. $u = 0$ (dropped from rest), $a = +9.8$ m/s², $t = 2.0$ s, $s = ?$, $v = ?$.
  4. Find $s$. No $v$ → use $s = ut + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(9.8)(2.0)^2 = 19.6$ m.
  5. Find $v$. No $s$ needed → use $v = u + at = 0 + (9.8)(2.0) = 19.6$ m/s downward.
Exam trap: don't forget the $\tfrac{1}{2}$ in $s = ut + \tfrac{1}{2}at^2$. Leaving it out turns a 19.6 m fall into 39.2 m, a classic dropped-mark error.

Template: list SUVAT, spot the missing variable, choose the equation. Braking (no $t$): $v^2=u^2+2as$ gives a negative $a$. Free fall: set $a=g=9.8$ m/s² in your sign convention, then $s=ut+\tfrac{1}{2}at^2$ and $v=u+at$.

Pause, copy both worked examples (the braking car and the free-fall ball) into your book as solution templates before moving on.

Two of these statements are true. One is a lie. Find the lie.

Choosing a SUVAT equation and deriving displacement from a velocity–time graph On the left, the five equations of motion each labelled with the variable they omit, acting as a selector. On the right, a velocity–time graph for constant acceleration, with the area under the line split into a rectangle of area u times t and a triangle of area one half a t squared, together giving displacement s. WHICH EQUATION? find the missing variable v = u + at no s s = ut + ½at² no v v² = u² + 2as no t s = ½(u + v)t no a s = vt − ½at² no u Skid mark: time unknown → use v² = u² + 2as s = AREA UNDER v–t time t velocity u v t u·t ½at²
Left: each equation omits one variable, the variable a problem leaves out is your signpost. Right: $s = ut + \tfrac{1}{2}at^2$ is just the area under the v–t graph, a rectangle ($ut$) plus a triangle ($\tfrac{1}{2}at^2$).
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Activity 1, Pick the Equation

Activities

Activity 1, Pick the Equation
UnderstandBand 3

For each scenario, list the known SUVAT variables and the one you are asked to find, then state which of the five equations you would use (and which variable it leaves out). You do not have to solve them.

  1. A car accelerates from 8 m/s to 20 m/s in 6 s. Find the displacement.
  2. A stone is dropped from rest and falls 45 m. Find its speed on landing.
  3. A train slows uniformly from 30 m/s, covering 150 m before stopping. Find its acceleration.
  4. A cyclist accelerates from rest at 1.2 m/s² for 5 s. Find the distance covered.

A car starts from rest and accelerates uniformly at 3.0 m/s² for 4.0 s. Using $s = ut + \tfrac{1}{2}at^2$, how far does it travel?

Activity 2, Solve It Fully
ApplyBand 4

Solve each problem in full, showing your sign convention, your SUVAT list and the equation chosen. Use $g = 9.8$ m/s² where needed.

  1. A motorbike accelerates uniformly from rest to 18 m/s over a distance of 81 m. Find its acceleration.
  2. A ball is thrown straight up at 14.7 m/s (take up = +). How high does it rise before stopping momentarily?
  3. A driver brakes from 25 m/s to a stop with an acceleration of −5.0 m/s². How long does the stop take, and how far does the car travel while braking?
✓

Show what you have learned

Quick recall, Equations of Motion
+5 XP

A fresh five-question set drawn from this lesson's bank, feedback shown immediately.

Pick your answer, then rate your confidence.

Multiple Choice, 5 Questions
checkpoint

1. Which equation of motion does not contain time, $t$?

  1. $v = u + at$
  2. $s = ut + \tfrac{1}{2}at^2$
  3. $v^2 = u^2 + 2as$
  4. $s = \tfrac{1}{2}(u+v)\,t$

2. A car accelerates uniformly from rest at 2.0 m/s² for 5.0 s. How far does it travel?

  1. 10 m
  2. 25 m
  3. 50 m
  4. 20 m

3. A ball is dropped from rest. Taking down as positive and $g = 9.8$ m/s², what is its velocity after 3.0 s?

  1. 3.3 m/s
  2. 9.8 m/s
  3. 19.6 m/s
  4. 29.4 m/s

4. A car travelling at 24 m/s brakes uniformly and stops in 48 m. Its acceleration is:

  1. −6.0 m/s²
  2. +6.0 m/s²
  3. −12 m/s²
  4. −0.5 m/s²

5. Why must the acceleration be constant for the five equations of motion to apply?

  1. Because otherwise the object would not move in a straight line
  2. Because the equations are derived assuming a straight (constant-gradient) velocity–time line
  3. Because constant acceleration means the velocity never changes
  4. Because time must always be unknown
Short Answer, 10 marks
+5 XP

UnderstandBand 3(3 marks) 1. State the five equations of uniformly accelerated motion, and for each one name the SUVAT variable it does not contain.

ApplyBand 4(3 marks) 2. A stone is dropped from rest from the top of a cliff and takes 3.0 s to reach the bottom. Taking down as positive and $g = 9.8$ m/s², calculate the height of the cliff and the stone's speed just before impact.

AnalyseBand 5(4 marks) 3. A crash investigator finds a 64 m straight skid mark left by a car that braked to a complete stop. Tests show the road provided a deceleration of 6.0 m/s². Calculate the car's speed just before it began to brake, showing which equation you chose and why. Then explain why the time the skid took was never needed.

Show all answers

Multiple choice

Q1, C. $v^2 = u^2 + 2as$ contains $s$, $u$, $v$ and $a$ but no time. That is exactly why it is the equation of choice for braking and skid-mark problems, where time is unknown. The other three equations all include $t$.

Q2, B. From rest, $u = 0$, so $s = ut + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(2.0)(5.0)^2 = \tfrac{1}{2}(2.0)(25) = 25$ m. The 50 m distractor comes from forgetting the $\tfrac{1}{2}$.

Q3, D. Down = +, so $v = u + at = 0 + (9.8)(3.0) = 29.4$ m/s. The 19.6 m/s distractor uses $t = 2.0$ s; 9.8 m/s uses $t = 1.0$ s.

Q4, A. No time given → use $v^2 = u^2 + 2as$. With $v = 0$, $u = 24$, $s = 48$: $0 = 24^2 + 2a(48) \Rightarrow 0 = 576 + 96a \Rightarrow a = -6.0$ m/s². The negative sign shows the acceleration opposes the motion (deceleration); the magnitude is 6.0 m/s².

Q5, B. The equations are derived from a velocity–time graph that is a straight line, and the line is straight only when acceleration (the gradient) is constant. If $a$ changes, the line bends, the simple area/algebra no longer holds, and the motion must be split into constant-$a$ stages.

Short Answer, Model Answers

Q1 (3 marks): $v = u + at$ (no $s$); $s = ut + \tfrac{1}{2}at^2$ (no $v$); $v^2 = u^2 + 2as$ (no $t$); $s = \tfrac{1}{2}(u+v)t$ (no $a$); $s = vt - \tfrac{1}{2}at^2$ (no $u$). (1 mark for all five equations correct, 1 mark for correctly identifying the omitted variables, 1 mark for completeness/accuracy of notation.)

Q2 (3 marks): Down = +, $u = 0$, $a = 9.8$ m/s², $t = 3.0$ s. Height: $s = ut + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(9.8)(3.0)^2 = \tfrac{1}{2}(9.8)(9.0) = 44.1$ m. Speed at impact: $v = u + at = 0 + (9.8)(3.0) = 29.4$ m/s. (1 mark correct equation choice with $u=0$; 1 mark height 44.1 m; 1 mark speed 29.4 m/s.)

Q3 (4 marks): Forward = +. Knowns: $v = 0$ (stops), $s = 64$ m, $a = -6.0$ m/s² (deceleration opposes motion). No time is given, so choose $v^2 = u^2 + 2as$, the only equation without $t$. Solving: $0 = u^2 + 2(-6.0)(64) \Rightarrow u^2 = 768 \Rightarrow u = 27.7$ m/s (≈ 28 m/s). The time was never needed because $v^2 = u^2 + 2as$ links the velocities directly to the distance, with $t$ entirely absent from the equation. (1 mark equation choice + justification; 1 mark correct substitution with signs; 1 mark $u \approx 27.7$ m/s; 1 mark explanation that the chosen equation contains no $t$.)

✓

Retrieve, reflect and finish

Check what actually stuck
Stretch, Think Ahead
stretch

These five equations only work along a single straight line. In Lesson 7 you will meet motion on a plane, where velocity has two perpendicular components. Predict: if a ball is thrown horizontally off a cliff, could you still use $s = ut + \tfrac{1}{2}at^2$ for its downward fall while it also travels sideways? What would be different about the value of $u$ in the vertical direction?

How did your thinking change?

The skid mark on the Hume Highway gave the investigator a displacement, $s = 64$ m. Knowing the car stopped ($v = 0$) and the road's deceleration ($a$), the equation with no time, $v^2 = u^2 + 2as$, delivered the speed $u$ before braking, no video required. That is the whole power of the equations of motion: list the three you know, spot the variable that is missing, and the right equation hands you the rest. The same toolkit handles free fall, launches and braking alike, and underpins the multi-stage problems you will tackle in L08.

← Previous: Graphs of Motion Next: Vectors in Two Dimensions & Relative Motion →

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