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.
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.
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:
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
True or false: in an HSC kinematics answer, writing the magnitude "12" without a sign, direction or unit is fully sufficient for the marks.
Core Content
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.
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)?
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 $u$.
$x$–$t$ 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.
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.
- 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).
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).
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).
- 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.)
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:
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.
- 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.
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.
Activities
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.
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."
A fresh five-question set drawn from this module's bank — feedback shown immediately.
Pick your answer, then rate your confidence.
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
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.)
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.
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.