Year 11 Physics Module 1 ★ Consolidation ⏱ ~35 min 5 MC · 3 Short Answer Lesson 8 of 8

Kinematics Consolidation, Problem Solving & Exam Technique

You have met every Kinematics tool: scalars and vectors (L01), distance and displacement (L02), speed and velocity (L03), acceleration (L04), motion graphs (L05), the equations of motion (L06), and relative motion on a plane (L07). The exam will not ask them one at a time, it mixes them. This lesson is the workshop where you learn to choose the right tool fast, avoid the classic traps, and write answers that earn full marks.

Today's hook: Two students get different answers to the same kinematics question, one wrote 5 m, the other −5 m. Both did "correct" maths. Only one will get the mark. What separates them?

0/5TASKS

You’re here

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.

★ Consolidation lesson

This is the final lesson of Module 1. It introduces no new physics it pulls L01–L07 together into one problem-solving toolkit. Each card below is a strategy: a decision to make under exam pressure, the method that makes it reliable, and the trap that costs marks. Have your earlier notes open beside you.

Before you read, predict

A question reads: "A ball is thrown straight up at 20 m/s. How long until it returns to the thrower's hand?" Before doing any maths, decide: which sign convention will you set, what value of $a$ will you use and in which direction, and which equation of motion needs no displacement value? Jot your plan first, the plan is where marks are won or lost.

A multi-step problem gives you initial velocity $u$, acceleration $a$ and time $t$, and asks for the final velocity $v$. Before touching numbers, the single most important first step is to:

Learning Intentions

goals

Know

The full kinematics toolkit from L01–L07 and when each tool applies
The five equations of motion and the variable each one omits
What the gradient and area of a motion graph each represent
The standard method for relative-motion (crosswind / river) problems
The five highest-frequency exam errors and how to catch them

Understand

Why naming a sign convention first prevents most lost marks
Why the equation you choose depends on which variable is missing
Why a graph and an equation can be two views of the same motion
Why relative velocity is a vector subtraction, not a number subtraction

Can Do

Plan a multi-step problem before calculating
Select the correct equation of motion in one read
Convert and interpret position–time and velocity–time graphs under time pressure
Diagnose and fix a flawed worked solution
Set out a full-mark answer with working, signs and units

You’re here

Scan these before reading

vocab

Sign conventiona stated rule (e.g. up = +) that turns every direction along a line into a + or − sign; declare it before calculating (L01)

SUVATthe five symbols $s, u, v, a, t$ of the equations of motion for uniform acceleration (L06)

Gradientthe slope of a motion graph: gradient of an $x$–$t$ graph = velocity; gradient of a $v$–$t$ graph = acceleration (L05)

Area under graphon a $v$–$t$ graph, the area between the line and the time axis = displacement (signed) (L05)

Relative velocitythe velocity of A as seen from B: $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$ (a vector subtraction) (L07)

Resultantthe single vector equal to two or more vectors combined; for perpendicular vectors use Pythagoras + trig (L07)

Magnitudethe size of a vector, always positive; a minus sign is a direction, never a smaller size (L01)

Cross-lesson links: This lesson is the keystone of Module 1. Card 1 revisits the scalar/vector split and sign conventions (L01–L03), Card 2 the motion graphs (L05) feeding back into velocity and acceleration (L03–L04), Card 3 the equations of motion (L06), Card 4 relative motion on a plane (L07), and Card 5 the misconceptions that thread through all of them. Looking forward, every tool here is reused in Module 2 (Dynamics): SUVAT feeds $F = ma$, motion graphs become force–time graphs, and vector addition becomes net-force resolution.

Misconceptions to fix

"If my arithmetic is right, my answer is right." So a missing sign or unit doesn't matter.In kinematics the direction (sign) and the unit are part of the answer. "$v = 5$" can lose marks where "$v = -5$ m/s (5 m/s downward)" earns them. Right number, wrong physics still loses.

"There is one best equation, so I should memorise which question uses which."You select the equation by the variable that is missing from the question, not by the topic. Same scenario, different unknown → different equation.

True or false: in an HSC kinematics answer, writing the magnitude "12" without a sign, direction or unit is fully sufficient for the marks.

You’re here

Strategy 1, The Vector/Scalar Decision & Sign Conventions

Core Content

Strategy 1, The Vector/Scalar Decision & Sign Conventions

+5 XP

Almost every Kinematics mistake begins in the first ten seconds, before any number is written, when a student fails to decide whether the question wants a scalar (size only) or a vector (size + direction), and fails to name a sign convention. Get these two decisions right and the maths usually follows; get them wrong and even perfect arithmetic earns nothing.

Use this two-question filter on every motion problem:

Q1, Does the answer need a direction? "How far is the car from the start?" needs a direction → displacement (vector). "How far did the car travel?" does not → distance (scalar). Same for speed (scalar) vs velocity (vector).
Q2, Which direction is positive? Write it down: "Take up = +" or "Take east = +". Every quantity then gets a sign and you can add with ordinary arithmetic instead of geometry.

Sign rule for vertical motion (the most common exam case) Take up = + → an upward throw has $u>0$ but $a = -9.8\ \text{m/s}^2$ (gravity acts downward) for the whole flight, including the top.

Exam technique: the very first line of your working should be your convention, e.g. "Let up = positive, so $a = -9.8$ m/s²." Markers reward a stated convention because it shows your signs are deliberate, not lucky.

The classic trap (today's hook): two students compute a displacement; one writes 5 m, the other −5 m. The student who declared their convention can defend "−5 m = 5 m in the negative direction" and gets the mark. The other has a bare number the marker can't trust.

Before calculating, make two decisions: (1) scalar or vector?, does the answer need a direction? (2) state a sign convention (e.g. up = +). For vertical motion, $a = -9.8$ m/s² throughout if up is positive. A declared convention turns a "lucky" sign into a defendable one.

Pause, copy the two-question filter and the "up = +, so $a=-9.8$ m/s²" sign rule into your book before moving on.

A question asks for a single quantity. Three of these wordings call for a vector answer. Which one calls for a scalar (the odd one out)?

You’re here

Strategy 2, Reading & Converting Motion Graphs Fast

+5 XP

We just saw that direction and sign decisions come first. That raises a question: what happens when the data arrives not as words but as a graph the marker expects you to read in seconds? This card answers it → a fixed routine for extracting velocity, acceleration and displacement from any motion graph.

Graph questions feel slow only because students re-derive the rules each time. Lock in what each feature means and a graph becomes faster than a worded problem. There are just two graph types and two operations, gradient and area, to remember.

Position–time ($x$–$t$) graph

Gradient = velocity. A straight line = constant velocity; a curve = changing velocity (acceleration). A horizontal line = at rest. A line crossing back toward the axis = returning toward the start.

Velocity–time ($v$–$t$) graph

Gradient = acceleration; area to the time axis = displacement. Area below the axis is negative displacement. A straight sloping line = uniform acceleration; the value where the line meets $t=0$ is $u$.

The four graph reads $x$–$t$ gradient $= v$ · $v$–$t$ gradient $= a$ · area under $v$–$t$ $= s$ · sign of area = direction of displacement

Exam technique: for "find the displacement from this $v$–$t$ graph", split the area into triangles and rectangles, find each, and add with signs (areas under the axis are negative). Always read axis units first, a graph in km/h vs s needs a conversion before the area is in metres.

Exam trap: reading the height of a $v$–$t$ graph as displacement. Height is velocity; displacement is the area. And on an $x$–$t$ graph, a high flat line does not mean "fast", a flat line means zero velocity, however high it sits.

$x$–$t$ gradient = velocity. $v$–$t$ gradient = acceleration; area under a $v$–$t$ graph = displacement (areas below the axis are negative). Read the axis units before you calculate, and split awkward areas into triangles + rectangles.

Pause, copy the four graph reads (gradient $x$–$t$, gradient $v$–$t$, area $v$–$t$, sign of area) into your book before moving on.

Complete the rule for reading a velocity–time graph.

You’re here

Strategy 3, Choosing the Right Equation of Motion

+5 XP

We just saw how to pull velocity, acceleration and displacement straight off a graph. That raises a question: when there is no graph and the motion is uniformly accelerated, which of the five equations do you reach for? This card answers it → a single selection rule based on the variable you are not given.

For uniform acceleration there are five equations of motion (SUVAT). Each one is missing exactly one of the five variables $s, u, v, a, t$. The whole skill is: list what you know and what you want, then pick the equation that does not contain the variable you neither know nor want.

The five equations of motion (and the variable each omits) $v = u + at$ (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 $u$)

Worked example, selecting in one read

Read the question. "A car brakes from 25 m/s and stops in 40 m. Find its acceleration." Take direction of motion = +.
List SUVAT. $u = 25$ m/s, $v = 0$, $s = 40$ m, $a = ?$. The variable not mentioned is $t$.
Select the equation with no $t$. That is $v^2 = u^2 + 2as$.
Solve. $0 = 25^2 + 2a(40) \Rightarrow a = -\dfrac{625}{80} = -7.8\ \text{m/s}^2$ (i.e. 7.8 m/s² opposite to motion, the negative sign is the deceleration).

Exam technique: write the SUVAT list as a little table every time ($s=\_,\ u=\_,\ v=\_,\ a=\_,\ t=\_$). The empty box that you are neither given nor asked for instantly names the equation to avoid, and so the equation to use.

Exam trap: using $a = +9.8$ for a ball thrown upward, or forgetting that "stops" means $v=0$ while "starts from rest" means $u=0$. Mixing up $u$ and $v$ is one of the most frequent SUVAT errors.

For uniform acceleration, list $s,u,v,a,t$, note the one variable you neither know nor want, then pick the equation that omits it. $v=u+at$ (no $s$), $s=ut+\tfrac12 at^2$ (no $v$), $v^2=u^2+2as$ (no $t$), $s=\tfrac12(u+v)t$ (no $a$), $s=vt-\tfrac12 at^2$ (no $u$).

Pause, copy the five equations with the omitted-variable note beside each, and the four-step selection method, into your book before moving on.

Match the SUVAT equation to the variable it does NOT contain (so it is the one to use when that variable is missing).

v = u + at
s = ut + ½at²
v² = u² + 2as
s = ½(u + v)t

contains no acceleration a
contains no time t
contains no final velocity v
contains no displacement s

You’re here

Strategy 4, Relative Motion: The Crosswind / River Method

+5 XP

We just saw the equation-selection rule for one-dimensional uniform acceleration. That raises a question: what changes when the motion leaves the straight line, a plane in a crosswind, a boat across a flowing river? This card answers it → a fixed vector-addition routine for relative motion on a plane.

Relative-motion problems trip students because they try to add speeds as numbers. The fix is to treat every velocity as a vector and combine them tip-to-tail. For a plane in wind or a boat in a current, the ground velocity is the vector sum of the craft's velocity (relative to the air/water) and the air/water's velocity (relative to the ground).

Resultant of two perpendicular velocities $\vec{v}_{\text{ground}} = \vec{v}_{\text{craft}} + \vec{v}_{\text{medium}}$ · magnitude $= \sqrt{v_1^2 + v_2^2}$ · direction $\theta = \tan^{-1}\!\left(\dfrac{v_2}{v_1}\right)$

Worked example, boat across a river

Identify the two perpendicular vectors. Boat heads straight across at $3.0$ m/s (north); river flows at $4.0$ m/s (east).
Find the resultant magnitude (Pythagoras). $v = \sqrt{3.0^2 + 4.0^2} = \sqrt{9 + 16} = 5.0$ m/s.
Find the direction (trig). $\theta = \tan^{-1}\!\left(\dfrac{4.0}{3.0}\right) = 53^\circ$ east of north.
State both parts. Resultant ground velocity = $5.0$ m/s at $53^\circ$ east of north. (Magnitude and direction, it is a vector.)

Exam technique: always sketch the two vectors tip-to-tail and label the right angle. The sketch tells you whether to use Pythagoras (perpendicular) and which side is "opposite"/"adjacent" for the angle. For "velocity of A relative to B", remember $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$, reverse $\vec{v}_B$ before adding.

Exam trap: adding the two speeds arithmetically ($3 + 4 = 7$) instead of as vectors ($\sqrt{3^2+4^2}=5$). You may only add magnitudes directly when the vectors are along the same line.

Relative/crosswind/river: ground velocity = vector sum of craft velocity + medium velocity. For perpendicular vectors, magnitude $=\sqrt{v_1^2+v_2^2}$ and direction $\theta=\tan^{-1}(v_2/v_1)$. "A relative to B" $=\vec{v}_A-\vec{v}_B$. Never add perpendicular speeds as plain numbers.

Pause, copy the resultant formula, the boat worked example, and "$\vec{v}_{AB}=\vec{v}_A-\vec{v}_B$" into your book before moving on.

A plane flies due north at 200 km/h relative to the air. A 150 km/h wind blows due east. The plane's speed relative to the ground is:

You’re here

Strategy 5, Diagnosing the Five Classic Errors

+5 XP

We just saw the vector method for relative motion. That raises a question: if the methods are clear, why do marks still slip away in exams? This card answers it → because a small set of recurring errors strikes again and again. Name them and you can catch them in your own working.

Examiners report the same handful of kinematics mistakes every year. A "spot-the-error" skill is worth as much as a calculation skill, most band-6 students earn marks by checking, not just computing. Here is the error checklist to run over every answer before you move on.

The five-point error checklist

Units. Did you mix km/h with m/s, or km with m? Convert before substituting (÷ 3.6 turns km/h into m/s).
Vector/scalar confusion. Did you give distance where displacement was wanted, or speed where velocity was wanted (missing the direction)?
Sign error. Is your acceleration's sign consistent with your stated convention? For an upward throw with up = +, did you use $a = -9.8$ throughout?
$g$ direction / "at the top". At maximum height $v = 0$ but $a = 9.8$ m/s² downward still, gravity never switches off.
$u$ vs $v$ swap. "Starts from rest" $\Rightarrow u = 0$; "comes to rest/stops" $\Rightarrow v = 0$. Don't swap them.

Exam technique: reserve the last 20 seconds of any kinematics question to run this five-point check. A wrong sign or an un-converted unit is a one-line fix that often recovers a full mark, but only if you look for it.

Run the five-point check on every answer: (1) units consistent (km/h ÷ 3.6 = m/s)? (2) vector vs scalar, direction included where needed? (3) sign of $a$ matches your convention? (4) at the top, $v=0$ but $a=9.8$ down still? (5) $u=0$ for "from rest", $v=0$ for "stops"?

Pause, copy the five-point error checklist into your book; this is your final-check routine for every exam answer.

Two of these checks are valid. One is a lie. Find the lie.

A mixed velocity–time graph. Gradient gives acceleration (positive 0–4 s, zero 4–8 s, negative 8–10 s); the shaded area between the line and the time axis gives the total displacement, split it into a triangle, rectangle and triangle.

Interactive Tool, Motion Grapher Open fullscreen ↗

What to do: adjust the motion and watch the position–time and velocity–time graphs update together. Practise reading the gradient as velocity/acceleration and the area as displacement, exactly the reads tested in the exam.

You’re here

Activity 1, Mixed Exam Problem Set

Activities

Activity 1, Mixed Exam Problem Set

ApplyBand 4–5

Work each problem the exam way: state your sign convention, list SUVAT (or identify the vectors), pick the equation/method, solve, and finish with a signed, directed, unitised answer. Then run the five-point check.

A motorcycle accelerates uniformly from rest to 24 m/s in 6.0 s. Find (a) its acceleration and (b) the distance covered.
A ball is thrown vertically upward at 19.6 m/s (take up = +, $g = 9.8$ m/s²). Find (a) the time to reach maximum height and (b) the maximum height.
Using the velocity–time graph in this lesson (rise 0–4 s to 12 m/s, flat 4–8 s, fall 8–10 s to 0), find the total displacement by computing the area.
A kayak paddles north across a river at 2.0 m/s while the current flows east at 1.5 m/s. Find the magnitude and direction of the kayak's velocity relative to the bank.

True or false: in Activity 1 problem 2, at the maximum height the ball's acceleration is momentarily zero.

Activity 2, Spot & Fix the Flawed Solution

AnalyseBand 5–6

Each student response below contains exactly one classic kinematics error. Identify the error (name which of the five it is), explain why it is wrong, and write the corrected final answer.

"A car travels 90 km/h for 10 s, so distance = 90 × 10 = 900 m."
"A ball thrown up reaches the top, where $v = 0$, so $a = 0$ there too."
"A runner does one 400 m lap back to the start, so their displacement is 400 m."
"A plane flies 300 km/h north into a 100 km/h easterly wind, so ground speed = 300 + 100 = 400 km/h."
"A car decelerates from 30 m/s to 10 m/s; using $a = v/t$ with $v = 10$ gives the acceleration."

✓

Show what you have learned

Quick recall, Kinematics Toolkit

+5 XP

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

Pick your answer, then rate your confidence.

Multiple Choice, 5 Questions

checkpoint

1. A car brakes uniformly from 20 m/s and stops after travelling 50 m. Taking the direction of motion as positive, which equation lets you find the acceleration in one step, and what is it?

$v = u + at$, giving $a = +4.0$ m/s²
$s = ut + \tfrac{1}{2}at^2$, giving $a = -8.0$ m/s²
$v^2 = u^2 + 2as$, giving $a = -4.0$ m/s²
$s = \tfrac{1}{2}(u+v)t$, giving $a = -2.0$ m/s²

2. On a velocity–time graph, a straight line slopes from 8 m/s at $t = 0$ down to 0 m/s at $t = 4$ s. Which two quantities are read correctly?

Acceleration $= +2$ m/s²; displacement $= 32$ m
Acceleration $= -2$ m/s²; displacement $= 16$ m
Acceleration $= -2$ m/s²; displacement $= 32$ m
Acceleration $= -8$ m/s²; displacement $= 16$ m

3. A swimmer heads straight across a river at 1.2 m/s while the current carries them downstream at 1.6 m/s. Their speed relative to the bank is:

0.4 m/s
1.4 m/s
2.8 m/s
2.0 m/s

4. A stone is thrown vertically upward and caught again at the same height. Taking up as positive, which statement about the motion is correct?

The acceleration is $-9.8$ m/s² for the whole flight, including at the top
The acceleration is zero at the highest point
The displacement on return equals the total distance travelled
The velocity is $-9.8$ m/s at the highest point

5. A student writes a final answer as "$\,a = 3.5\,$". A marker deducts a mark. The most likely reason is that the answer:

used the wrong equation of motion
is missing the unit (and, for a vector, the direction/sign)
did not convert km/h to m/s
confused initial and final velocity

Short Answer, 13 marks

+5 XP

ApplyBand 4(4 marks) 1. A cyclist starts from rest and accelerates uniformly at 1.5 m/s² along a straight road. (a) Calculate the cyclist's velocity after 8.0 s. (b) Calculate the distance travelled in that time. Show your sign convention and SUVAT list.

AnalyseBand 5(5 marks) 2. A light aircraft flies due north at 180 km/h relative to the air. A steady wind blows from the west (i.e. toward the east) at 48 km/h. (a) Determine the magnitude of the aircraft's velocity relative to the ground. (b) Determine its direction relative to north. (c) Explain why the ground speed is not simply $180 + 48$ km/h.

EvaluateBand 6(4 marks) 3. A student presents this solution: "A ball is dropped from rest from a 45 m roof. Using $v = u + at$ with $u = 0$, $a = 9.8$, $t = 3.0$ s, the speed on landing is $v = 29.4$ m/s." First, verify whether $t = 3.0$ s is correct for a 45 m drop. Then evaluate whether the student's final speed is valid, correcting any error in the method or result.

Show all answers

Multiple choice

Q1, C. Known: $u = 20$ m/s, $v = 0$, $s = 50$ m, $a = ?$; the missing variable is $t$, so use $v^2 = u^2 + 2as$. Then $0 = 20^2 + 2a(50) \Rightarrow a = -400/100 = -4.0$ m/s² (the negative sign shows deceleration, opposite to motion). A and D use the wrong equations/values; B has the wrong magnitude.

Q2, B. Gradient (acceleration) $= (0 - 8)/(4 - 0) = -2$ m/s². Displacement = area under the line = area of the triangle $= \tfrac{1}{2} \times 4 \times 8 = 16$ m. C uses the right gradient but the wrong area; A has the wrong sign; D misreads the gradient as the start velocity.

Q3, D. The cross-stream (1.2 m/s) and downstream (1.6 m/s) velocities are perpendicular, so combine with Pythagoras: $v = \sqrt{1.2^2 + 1.6^2} = \sqrt{1.44 + 2.56} = \sqrt{4.00} = 2.0$ m/s. Adding them as plain numbers (2.8, option C) or subtracting (0.4, option A) ignores that they are perpendicular vectors.

Q4, A. Gravity acts at $9.8$ m/s² downward for the entire flight, so with up = + the acceleration is $-9.8$ m/s² throughout, including at the highest point (where the velocity, not the acceleration, is momentarily zero). B is the classic "g = 0 at the top" error; C is wrong because the return displacement is zero while distance is non-zero; D confuses velocity with acceleration.

Q5, B. A bare number "3.5" is missing its unit (m/s²), and for a vector quantity also its direction/sign. That is the most common single-mark deduction. Nothing in the stem indicates a wrong equation, an un-converted unit, or a $u$/$v$ swap, so A, C and D are not the best answer.

Short Answer, Model Answers

Q1 (4 marks): Convention: forward = positive. SUVAT: $u = 0$, $a = 1.5$ m/s², $t = 8.0$ s.
(a) $v = u + at = 0 + (1.5)(8.0) = 12$ m/s (forward).
(b) $s = ut + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(1.5)(8.0)^2 = 0.5 \times 1.5 \times 64 = 48$ m.
(1 mark convention + SUVAT list; 1 mark correct equation/working in (a); 1 mark $v = 12$ m/s with unit; 1 mark $s = 48$ m with unit.)

Q2 (5 marks): The aircraft's velocity (north) and the wind (east) are perpendicular.
(a) Magnitude $= \sqrt{180^2 + 48^2} = \sqrt{32400 + 2304} = \sqrt{34704} \approx 186$ km/h.
(b) Direction $\theta = \tan^{-1}\!\left(\dfrac{48}{180}\right) = \tan^{-1}(0.267) \approx 15^\circ$ east of north.
(c) The ground speed is not $180 + 48 = 228$ km/h because the two velocities are perpendicular, not along the same line; only collinear vectors add as plain numbers. Perpendicular vectors must be combined by Pythagoras, giving a resultant smaller than the arithmetic sum.
(1 mark recognising perpendicular vectors; 1 mark Pythagoras set-up; 1 mark magnitude ≈ 186 km/h; 1 mark direction ≈ 15° E of N; 1 mark valid explanation in (c).)

Q3 (4 marks): Verify the time with $s = ut + \tfrac{1}{2}at^2$: $45 = 0 + \tfrac{1}{2}(9.8)t^2 = 4.9t^2 \Rightarrow t^2 = 9.18 \Rightarrow t = 3.03 \approx 3.0$ s, so the student's time is essentially correct. The method ($v = u + at$) is also valid: $v = 0 + (9.8)(3.0) = 29.4$ m/s, or more precisely $v = 9.8 \times 3.03 = 29.7$ m/s. A cleaner, time-free route is $v^2 = u^2 + 2as = 0 + 2(9.8)(45) = 882 \Rightarrow v = 29.7$ m/s. So the student's result is valid to 2 significant figures ($\approx 30$ m/s), the only refinement is that rounding $t$ to 3.0 s makes 29.4 m/s slightly low; the exact landing speed is about 29.7 m/s. (1 mark correct $t \approx 3.0$ s with working; 1 mark confirming the method is valid; 1 mark independent check via $v^2 = u^2 + 2as$; 1 mark evaluative conclusion noting the small rounding effect.)

✓

Retrieve, reflect and finish

Check what actually stuck

Stretch, One Hard Mixed Problem

stretch

Put the whole toolkit together. A drone launches from a regional clinic and flies east at 12 m/s relative to the air. A steady 5.0 m/s wind blows from the south (toward the north). (a) Find the drone's ground velocity (magnitude and direction). (b) If the clinic is 1.3 km due east of the drop point, will the drone arrive exactly on target flying this heading, or will the wind carry it off-line? Explain using your relative-motion reasoning, and suggest how the pilot should adjust the heading.

How did your thinking change?

Back to today's hook: two students computed the same displacement, one writing 5 m and the other −5 m, both with "correct" maths, yet only one earns the mark. The difference is exam technique, not arithmetic. The student who declared a sign convention first can justify "−5 m means 5 m in the negative direction"; the other has an undefended number. That is the whole lesson in miniature: in Kinematics the plan (convention, vector-vs-scalar, equation choice) and the final-check routine (units, signs, direction, $g$, $u$ vs $v$) are where marks are won, the calculator is the easy part. You now have all seven lessons' tools and the strategy to deploy them under pressure.

I have completed this lesson

← Prev: Vectors in Two Dimensions & Relative Motion Finish Module 1, back to the hub →

Module overview · Physics