TASC TCE 2024-style practice: A cart of mass 1.00\ kg is released up an inclined ramp with an initial speed of 1.50\ m s^-1. The ramp makes an angle of 15^ 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.

Going up the ramp both gravity and friction oppose the motion, so the cart decelerates. Deceleration up the ramp: a = gsin 15^ + 0.10g = (9.81)(0.259 + 0.10) = 3.52\ m s^-2. Use v^2 = u^2 + 2as with v = 0 at the highest point, u = 1.50\ m s^-1, a = -3.52\ m s^-2: $0 = (1.50)^2 + 2(-3.52)s, s = ((1.50)^2)/(2(3.52)) = (2.25)/(7.04) = 0.320\ m.$ The cart travels about 0.32\ m up the ramp before momentarily stopping. Markers want the combined gravity-plus-friction deceleration and the kinematic equation with v = 0.

§-Syllabus dot point

TASPhysicsSyllabus dot point

How do we describe and predict motion in a straight line?

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

Displacement, velocity and acceleration as vectors, the four constant-acceleration equations of motion, and how to read motion graphs to solve straight line problems including free fall.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

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.

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.

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.

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 practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20244 marksA 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 $v = 0$ .

TCE 20233 marksAt 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.