Linear motion and equations of motion - TCE Physics (Tasmania)

Syllabus dot point

“Use the equations of motion to analyse uniform and uniformly accelerated straight line motion.”

TCEPhysicsUnit 3: Gravity and Electromagnetism9 min readUpdated 2 Jun 2026

Quick answer

Straight-line motion is described by displacement, velocity and acceleration. When acceleration is constant you can use the four equations of motion ( $v = u + at$ , $s = ut + \tfrac12 at^2$ , $v^2 = u^2 + 2as$ , $s = \tfrac12(u+v)t$ ) to find any unknown, including free fall where $a = g = 9.8\ \text{m s}^{-2}$ downward.

What this dot point is asking

This dot point is the foundation of mechanics. Before you can analyse forces, momentum or projectiles you must be able to describe how a single object moves along a line and predict where it will be and how fast it will be travelling.

Vector quantities of motion

Position, displacement, velocity and acceleration are all vectors, so direction matters. In one dimension you handle direction with a sign: choose one direction as positive and the opposite as negative, then keep that convention for the whole problem.

Displacement $s$ is the change in position, a straight line from start to finish, not the distance travelled along the path.
Velocity $v$ is the rate of change of displacement, so average velocity is $\dfrac{\Delta s}{\Delta t}$ .
Acceleration $a$ is the rate of change of velocity, $\dfrac{\Delta v}{\Delta t}$ .

Speed and distance are the scalar partners of velocity and displacement. A car that drives a loop and returns has travelled a distance but has zero displacement, so its average velocity is zero even though its average speed is not.

Acceleration

Acceleration is the rate of change of velocity with respect to time, a vector measured in metres per second squared. A negative acceleration means the velocity is becoming more negative, which is not always slowing down.

The equations of motion

When acceleration is constant, five quantities are linked by four equations. The symbols are initial velocity $u$ , final velocity $v$ , acceleration $a$ , displacement $s$ and time $t$ :

v = u + at

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

v^2 = u^2 + 2as

s = \tfrac{1}{2}(u+v)t

Each equation leaves out one variable, so the trick is to list what you know, identify the one quantity you do not need, and pick the equation that omits it. This avoids unnecessary algebra. The equations only apply when acceleration is constant; for changing acceleration you must work from a graph or use calculus.

Free fall and vertical motion

Near Earth's surface, objects in free fall accelerate downward at $g = 9.8\ \text{m s}^{-2}$ regardless of mass, once air resistance is ignored. Vertical motion is just the equations of motion with $a = g$ . Choosing up as positive means $a = -9.8\ \text{m s}^{-2}$ , and at the top of a thrown object's flight the velocity is momentarily zero while the acceleration is still $g$ downward.

Worked example: a decelerating car

A car travelling at $25\ \text{m s}^{-1}$ brakes uniformly and comes to rest in $4.0\ \text{s}$ . Find the acceleration and the stopping distance.

Step 1: Identify the known and unknown quantities

Before choosing a formula, list all five kinematic quantities and mark which ones are given or implied. Here the car starts at $u = 25\ \text{m s}^{-1}$ , finishes at rest so $v = 0$ , and the time is $t = 4.0\ \text{s}$ . The unknowns are the acceleration $a$ and the displacement $s$ . Choosing forward as positive fixes our sign convention for the whole problem.

Step 2: Find the acceleration using $v = u + at$

This equation links $u$ , $v$ , $a$ , and $t$ , which is exactly the set of quantities we know or want. Rearranging for $a$ gives:

0 = 25 + a(4.0) \Rightarrow a = -6.25\ \text{m s}^{-2}.

The negative sign is expected because the acceleration opposes the forward velocity, meaning the car is slowing down.

Step 3: Find the stopping distance using the average-velocity equation

Now that we know $u$ , $v$ , and $t$ , the equation $s = \tfrac12(u+v)t$ is the simplest choice because it avoids substituting the newly calculated $a$ (which would carry any rounding error forward). Under constant acceleration, the average velocity is simply the mean of the initial and final velocities:

s = \tfrac12(25 + 0)(4.0) = 50\ \text{m}.

Final answer: The car decelerates at $6.25\ \text{m s}^{-2}$ and stops in $50\ \text{m}$ .

Reading motion graphs

Graphs carry the same information as the equations. On a displacement-time graph the gradient is velocity. On a velocity-time graph the gradient is acceleration and the area under the line is displacement. A straight sloped line on a velocity-time graph means constant acceleration; a curve means changing acceleration. Being able to switch between graphs and equations is a frequently tested skill.

Area under a velocity-time graph is displacement

The area enclosed between the velocity-time line and the time axis equals the displacement. Areas below the axis count as negative displacement, which matters when an object reverses direction.

Resolving angled velocities

Many TCE questions launch a projectile or move an object at an angle, then ask about the straight-line motion along one axis. The method is to resolve the velocity into horizontal and vertical components with $v_x = v\cos\theta$ and $v_y = v\sin\theta$ , then apply the equations of motion separately to each direction. The horizontal and vertical motions share only the time, which is the key to projectile problems built on top of this dot point.

Confusing deceleration with negative acceleration

Students often assume a negative acceleration always means slowing down. The sign of acceleration is set by your chosen positive direction. An object speeds up whenever velocity and acceleration share the same sign and slows down only when they have opposite signs. Always state your positive direction first.

In the exam, write down your sign convention, list the five symbols with their known values, then choose the single equation that contains your three knowns and one unknown. Check that your answer has a sensible sign and units before moving on.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

A cart of mass

1.00\ \text{kg}

is released up an inclined ramp with an initial speed of

1.50\ \text{m s}^{-1}

. The ramp makes an angle of

15^\circ

to the horizontal and a frictional force of

10\%

of the cart's weight acts throughout. Calculate the maximum displacement of the cart up the ramp.

Show worked answer →

Going up the ramp both gravity and friction oppose the motion, so the cart decelerates.

Deceleration up the ramp: $a = g\sin 15^\circ + 0.10g = (9.81)(0.259 + 0.10) = 3.52\ \text{m s}^{-2}$ .

Use $v^2 = u^2 + 2as$ with $v = 0$ at the highest point, $u = 1.50\ \text{m s}^{-1}$ , $a = -3.52\ \text{m s}^{-2}$ :

0 = (1.50)^2 + 2(-3.52)s, \quad s = \frac{(1.50)^2}{2(3.52)} = \frac{2.25}{7.04} = 0.320\ \text{m}.

The cart travels about $0.32\ \text{m}$ up the ramp before momentarily stopping. Markers want the combined gravity-plus-friction deceleration and the kinematic equation with $v = 0$ .

Practice3 marks

At an archery range, an arrow is fired at

70\ \text{m s}^{-1}

toward a target

50\ \text{m}

away at the same height. The arrow is pointed

10^\circ

above the horizontal. Calculate the vertical and horizontal components of the arrow's velocity.

Show worked answer →

Resolve the launch velocity into perpendicular components using the $10^\circ$ launch angle.

Horizontal component: $v_x = 70\cos 10^\circ = (70)(0.9848) = 68.9\ \text{m s}^{-1}$ .

Vertical component: $v_y = 70\sin 10^\circ = (70)(0.1736) = 12.2\ \text{m s}^{-1}$ upwards.

The horizontal component stays constant (no horizontal force, ignoring air resistance); the vertical component decreases at $g = 9.81\ \text{m s}^{-2}$ as the arrow rises. Markers want both components with the correct trig functions and units.

Practise this

Exam-style questions 2 on this page
Flashcards Physics deck

Sources & how we know this

Physics Level 4 (PHY415115) course document (2015 onwards) — TASC