When acceleration is constant, five quantities, displacement $s$, initial velocity $u$, final velocity $v$, acceleration $a$ and time $t$, are locked together by just three equations. Know any three of the five and you can find the rest. These "equations of motion" are on your formula sheet and they are the workhorse of every kinematics problem from here to the end of the course.
A ball starts from rest and accelerates at $2\ \text{m s}^{-2}$ for $3\ \text{s}$.
Before reading: what do you think its final velocity will be? Write your guess and one sentence on how acceleration and time give a velocity.
Three equations tie together five quantities
For constant acceleration, the same five quantities recur in every problem, and three standard equations connect them, each one deliberately leaving out a different quantity.
The five quantities are displacement $s$, initial velocity $u$, final velocity $v$, acceleration $a$ and time $t$. The formula-sheet equations of motion are:
$$v = u + at$$ $$s = ut + \tfrac{1}{2}at^2$$ $$v^2 = u^2 + 2as$$
The first omits $s$, the second omits $v$, and the third omits $t$, so whichever quantity you are not given and do not need points you to the right equation.
For uniform acceleration, the equations of motion are $v = u + at$, $s = ut + \tfrac{1}{2}at^2$ and $v^2 = u^2 + 2as$, linking displacement $s$, initial velocity $u$, final velocity $v$, acceleration $a$ and time $t$. Each equation leaves out one quantity.
Pause, copy the highlighted equations into your book before moving on.
Did you get this? Which equation of motion does NOT contain time $t$?
List the knowns, spot the missing quantity, pick the equation
We just saw the three equations and what each omits. That raises a question: faced with a real problem, how do you know which one to grab? This card answers it → list your knowns, identify the unknown, and choose the equation that contains exactly those.
The reliable method is always the same: write down which of the five you know, mark the one you want, and pick the equation that links them without involving anything you cannot find.
A car travelling at $30\ \text{m s}^{-1}$ brakes uniformly and stops in $5.0\ \text{s}$. Find the braking distance.
To solve, list the knowns among $s$, $u$, $v$, $a$, $t$, mark the unknown, and pick the equation containing exactly those. Substitute with consistent signs. The average-velocity form $s = \left(\dfrac{u+v}{2}\right)t$ is handy when $a$ is unknown.
Add the highlighted method to your notes before the check below.
Quick check: A ball starts from rest ($u = 0$) and accelerates at $3.0\ \text{m s}^{-2}$ for $4.0\ \text{s}$. Using $v = u + at$, what is its final velocity?
Pattern A, List, select, substitute
A motorbike accelerates uniformly from $8.0\ \text{m s}^{-1}$ to $20\ \text{m s}^{-1}$ over a distance of $84\ \text{m}$. In your book: (1) list the knowns ($u$, $v$, $s$) and the unknown ($a$); (2) choose the equation that links them without $t$; (3) solve for the acceleration.
Fill the gaps. An object starting from rest has $u =$ . Under uniform acceleration, $v = u + $ and $v^2 = u^2 + $ .
Dot point: Solve problems involving uniformly accelerated motion in one dimension using v = u + at, s = ut + ½at², v² = u² + 2as
Why a minus sign handles slowing down and falling
We just learned to select and apply an equation. That raises a question: how do the equations cope with slowing down, or with motion that reverses? This card answers it → through a consistent sign convention for direction.
Because $u$, $v$, $a$ and $s$ are all vectors in 1D, you choose one direction as positive and give anything pointing the other way a negative sign, then the same three equations handle everything.
Choose a positive direction (often the direction of initial motion). Deceleration then appears as a negative acceleration, opposite to the velocity. For a falling object, taking down as positive makes $a = +g = +9.8\ \text{m s}^{-2}$; taking up as positive makes $a = -9.8\ \text{m s}^{-2}$. Keeping signs consistent is what lets one set of equations describe speeding up, slowing down and reversing.
The equations of motion handle deceleration and reversal through signs: choose a positive direction and give opposite-direction quantities a negative value. Deceleration is a negative $a$; a falling body has $a = \pm g = \pm 9.8\ \text{m s}^{-2}$ depending on the chosen positive direction.
Pause, copy the highlighted convention into your book before the check below.
Last check: Decide if the statement is true or false.
Deceleration requires a different set of equations from acceleration.
Match each problem with the equation that solves it without finding another unknown first:
Dot point: Solve problems involving uniformly accelerated motion in one dimension using v = u + at, s = ut + ½at², v² = u² + 2as
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 5(4 marks) 1. A stone is dropped from rest and falls for $2.0\ \text{s}$ (take $g = 9.8\ \text{m s}^{-2}$, ignore air resistance). Calculate (a) its velocity just before it lands and (b) the height it fell. Show which equations you use.
1 mark: $v = u + at = 0 + 9.8(2.0)$ · 1 mark: $v = 19.6\ \text{m s}^{-1}$ · 1 mark: $s = ut + \tfrac{1}{2}at^2 = \tfrac{1}{2}(9.8)(2.0)^2$ · 1 mark: $s = 19.6\ \text{m}$
ApplyBand 5(4 marks) 2. A car travelling at $20\ \text{m s}^{-1}$ brakes uniformly and stops in $50\ \text{m}$. Calculate its acceleration and the time taken to stop.
1 mark: selects $v^2 = u^2 + 2as$ · 1 mark: $a = -4.0\ \text{m s}^{-2}$ · 1 mark: selects $v = u + at$ · 1 mark: $t = 5.0\ \text{s}$
Dot point: Solve problems involving uniformly accelerated motion in one dimension using v = u + at, s = ut + ½at², v² = u² + 2as
UnderstandBand 5(3 marks) 3. Explain what the sign of the acceleration tells you in a one-dimensional problem, and why a negative acceleration does not always mean an object is slowing down.
1 mark: sign gives direction relative to the chosen positive direction · 1 mark: slowing requires acceleration opposite to velocity · 1 mark: a negative acceleration on a negative velocity speeds the object up
Dot point: Solve problems involving uniformly accelerated motion in one dimension using v = u + at, s = ut + ½at², v² = u² + 2as
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): Taking down as positive with $u = 0$ and $a = g = 9.8\ \text{m s}^{-2}$. (a) Final velocity from $v = u + at$: $v = 0 + 9.8 \times 2.0 = 19.6\ \text{m s}^{-1}$ downward. (b) Height fallen from $s = ut + \tfrac{1}{2}at^2$: $s = 0 + \tfrac{1}{2}(9.8)(2.0)^2 = \tfrac{1}{2}(9.8)(4.0) = 19.6\ \text{m}$. So the stone lands at $19.6\ \text{m s}^{-1}$ after falling $19.6\ \text{m}$.
Q2 (4 marks): The known quantities are $u = 20\ \text{m s}^{-1}$, $v = 0$ and $s = 50\ \text{m}$, with $t$ unknown, so the equation without $t$ is the right first choice: $$v^2 = u^2 + 2as ;\Rightarrow; 0 = 20^2 + 2a(50)$$ giving $a = -400/100 = -4.0\ \text{m s}^{-2}$. The negative sign shows the acceleration opposes the motion, which is what braking means. For the time, use $$v = u + at ;\Rightarrow; 0 = 20 + (-4.0)t$$ so $t = 5.0\ \text{s}$.
Q3 (3 marks): In one dimension the sign of the acceleration states its direction relative to the positive direction chosen at the start, and nothing else. An object slows down only when its acceleration is opposite in sign to its velocity, because that is what reduces the magnitude of the velocity. If an object is already moving in the negative direction, so its velocity is negative, then a negative acceleration is in the same direction as the motion and the object speeds up while becoming more negative. So "negative acceleration" and "deceleration" are not synonyms: the first is a statement about direction, the second about the two signs being opposite.
You can now select and apply the three equations of motion, handling deceleration and falling with signs. Next lesson puts them to work in a practical: measuring the acceleration due to gravity, $g$, by two graphical methods.
↩ Back to focus-area hubAt the start a ball from rest accelerated at $2\ \text{m s}^{-2}$ for $3\ \text{s}$. Using $v = u + at$: $v = 0 + 2 \times 3 = 6\ \text{m s}^{-1}$. The big idea is that three equations connect five quantities, know any three, find the rest. Revisit Card 2's "list–select–substitute" routine if choosing the equation felt uncertain.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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