Jupiter2U School
Chapter 07

Kinematics

High School
At a glance
Core ideaPosition, velocity and acceleration describe motion; the SUVAT equations link them when a is constant.
Key termAcceleration — the rate of change of velocity, not of position.
You can…Solve projectile problems by treating horizontal and vertical motion separately.
Watch outVelocity and acceleration can point opposite ways — at a throw's peak v=0 but a=g.
Theory

Describing motion without its causes

Kinematics describes how objects move, ignoring why. In one dimension, position x(t) is measured along an axis. Velocity is its rate of change and acceleration the rate of change of velocity:

v = dx/dt    a = dv/dt = d²x/dt²

Displacement, velocity and acceleration are vectors (magnitude + direction); distance and speed are scalars. Average velocity is Δx/Δt; instantaneous velocity is its limit as Δt → 0.

Speed (scalar)

  • Magnitude only — no direction
  • Always positive or zero
  • Based on total distance travelled
  • A car circling a track keeps a steady speed

Velocity (vector)

  • Magnitude and direction
  • Can be negative (opposite direction)
  • Based on displacement, not distance
  • That same car's velocity constantly changes as it turns

Uniform acceleration — the SUVAT equations

When a is constant, integrating gives four workhorse equations (u = initial velocity, v = final, s = displacement, t = time):

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

These follow from the definitions by integration; they do not apply when acceleration varies. In two dimensions, motion decomposes into independent components along ⟂ axes — the horizontal and vertical parts of a projectile evolve separately, linked only by shared time.

Explanation

Graphs, slopes and areas

Motion graphs make calculus visible. On an x–t graph, the slope is velocity. On a v–t graph, the slope is acceleration and the area under the curve is displacement. These two facts — slope and area — connect the three quantities and let you read one graph off another.

Projectile motion is the classic "independence of components" idea: gravity pulls only downward, so horizontal velocity stays constant while vertical velocity changes at −g. A bullet fired horizontally and one dropped from the same height hit the ground simultaneously, because their vertical motions are identical. That is deeply counter-intuitive and completely correct.

Watch out. Velocity and acceleration can point in opposite directions. A ball thrown up has velocity upward but acceleration g downward the entire flight — including at the top, where v = 0 but a is still 9.81 m·s⁻² down.
time (t) v area = displacement (s) slope = acceleration (a)
On a velocity–time graph, the slope of the line is acceleration and the shaded area beneath it is displacement.
Practical

Worked example — projectile range

A ball is launched at u = 20 m·s⁻¹ at θ = 30° above horizontal (g = 9.81 m·s⁻²). Find its time of flight, maximum height, and horizontal range.

  1. Split velocity: ux = u cos30° = 17.3 m·s⁻¹, uy = u sin30° = 10.0 m·s⁻¹.
  2. Time up (vy=0): tup = uy/g = 10.0/9.81 = 1.02 s. Total flight T = 2tup = 2.04 s.
  3. Max height: H = uy²/(2g) = 100/19.62 = 5.10 m.
  4. Range: R = ux · T = 17.3 × 2.04 = 35.3 m.
  5. Check with the range formula R = u²sin(2θ)/g = 400 × sin60°/9.81 = 400 × 0.866/9.81 = 35.3 m. ✓
R = u² sin(2θ) / g range on level ground

Note sin(2θ) is maximised at θ = 45°, and that 30° and 60° give the same range — a symmetry worth remembering.

Q&A
A car accelerates uniformly from rest to 30 m·s⁻¹ in 12 s. What distance does it cover?

Use s = ½(u+v)t = ½(0+30)(12) = 180 m. Equivalently a = 30/12 = 2.5 m·s⁻² then s = ½at² = ½(2.5)(144) = 180 m.

Why do a dropped and a horizontally-fired bullet land together?

Their vertical motions are governed by the same equation y = ½gt² with the same initial vertical velocity (zero). The horizontal velocity of the fired bullet affects only where it lands, not when. Time to fall depends solely on height.

An object has velocity −5 m·s⁻¹ and acceleration +2 m·s⁻². Describe its motion.

It moves in the negative direction but is decelerating (speed decreasing), because acceleration opposes velocity. It will slow, momentarily stop at t = 2.5 s, then move in the positive direction, speeding up.

From a v–t graph, how do you find distance travelled versus displacement?

Displacement is the signed area (areas below the axis are negative). Distance is the total unsigned area (treat all areas as positive). They differ whenever the velocity changes sign.

Concept mind map

How the ideas connect

Every key idea in this chapter, branching from the core concept — use it to see the whole picture at a glance.

DisplacementVelocityAccelerationSUVAT equationsMotion graphsProjectilesKinematics
Infographic

The process, step by step

Step 1List knownsWrite down u, v, a, s, t that you are given
Step 2Pick equationChoose a SUVAT equation with only one unknown
Step 3SubstitutePut in numbers with correct units and signs
Step 4SolveRearrange and compute the unknown
Step 5CheckVerify units and that the size is sensible
Solved examples

Worked problems, step by step

Follow each solution line by line, then try to reproduce it on paper before moving on.

Example 1A ball is dropped from rest. How fast is it moving after 3.0 s? (g = 9.8 m/s^2)

  1. u = 0, a = 9.8 m/s^2, t = 3.0 s
  2. v = u + a*t
  3. v = 0 + 9.8 * 3.0

Example 2A ball is thrown horizontally at 15 m/s from a 20 m high roof. How far does it land from the base? (g = 9.8)

  1. Fall time: 20 = 0.5*9.8*t^2, t^2 = 4.08, t = 2.02 s
  2. Horizontal range = 15 * 2.02
  3. = 30.3 m
Practice problem set

Now you try

Work each one out first, then tap to reveal the worked answer.

1A car speeds from 0 to 20 m/s in 5.0 s. Find its acceleration.
a = (20 - 0)/5.0 = 4.0 m/s^2.
2What does the slope of a velocity-time graph represent?
The slope is the acceleration.
3What does the area under a velocity-time graph give?
The area gives the displacement.
4A stone falls freely for 2.0 s. How far does it drop? (g = 9.8)
s = 0.5*9.8*2.0^2 = 19.6 m.
5A car at 30 m/s brakes at 6 m/s^2. How long to stop?
t = 30/6 = 5.0 s.
6At what launch angle is a projectile range greatest?
45 degrees gives the maximum range on level ground.