Kinematics

Speed = how fast (scalar). Velocity = how fast **and which direction** (vector).

**Average** = displacement ÷ total time (over the whole trip). **Instantaneous** = the velocity at one moment — the speedometer reading right now (exams call it the 'rate of change of position').

Distance = total path travelled (scalar). Displacement = straight line start→finish, with direction (vector).

A **vector** — it has size and direction.

$v = \dfrac{\Delta s}{\Delta t}$ — displacement ÷ time.

**m s⁻¹** (metres per second).

Distance = 7 m; displacement = 5 m (straight line).

The **rate of change of velocity** — how much the velocity changes each second. Unit: m s⁻².

**m s⁻²** (metres per second, every second).

The **acceleration**.

The **change in velocity**. From rest, that area is the velocity reached.

Constant velocity → **zero** acceleration.

The object is **slowing down** — a negative acceleration (deceleration).

**Yes** — velocity includes direction, so changing direction changes the velocity.

$a = \dfrac{v - u}{t}$ — change in velocity ÷ time.

The **displacement** — how far the object travels.

The **acceleration**. (Area = displacement, slope = acceleration — don't swap them.)

$s = \dfrac{u + v}{2}\,t$ — average velocity × time (the trapezium area).

The **average velocity** — halfway between the start velocity u and the final velocity v.

**½ × base × height** = ½ × time × final velocity.

**speed × time** — a constant velocity gives a rectangular area.

**Split it** into a rectangle + a triangle, work out each, then **add** them.

**Negative** displacement — the object is moving backwards. Subtract it from the forward area for the net displacement.

Rectangle area = 10 × 3.0 = **30 m**.

Triangle area = ½ × 4.0 × 12 = **24 m**.

Height (m s⁻¹) × width (s) = m s⁻¹ × s = **m** — exactly a displacement.

The five constant-acceleration quantities: **s** displacement, **u** initial velocity, **v** final velocity, **a** acceleration, **t** time.

Only when the **acceleration is constant** (a straight velocity–time line).

v = u + at · s = ut + ½at² · v² = u² + 2as · s = ½(u + v)t — all four are **given** in the data booklet.

Write your **three knowns** + the unknown, then pick the equation that contains those four letters and **leaves out the fifth**.

**v² = u² + 2as** — use it when the time is unknown (e.g. stopping distance).

**s = ut + ½at²** — use it to find displacement from time.

**s = ½(u + v)t** — displacement from the average of the two speeds.

The **final velocity v = 0**.

**a = −5 m s⁻²** (negative, because the object is slowing down).

The **initial velocity u = 0** (it kills the ut term in s = ut + ½at²).

Use v² = u² + 2as: 0 = 400 − 10s → s = 40 m.

The equations come from a **straight** v–t line; a changing acceleration curves the line, so they no longer hold.

Motion where **gravity is the only force** acting — air resistance is ignored.

**g = 9.81 m s⁻²**, directed **downward** (given on the data booklet).

**No** — with no air resistance every object accelerates at the same g = 9.81 m s⁻².

It is constant-acceleration motion with **a = g**. Take up as positive, so a = −9.81 m s⁻².

**Velocity = 0** for an instant; **acceleration = 9.81 m s⁻² downward** (still g).

Time up to the top = time back down. Total flight time = **2 × time to the top**.

**Zero** — it ends where it started; it lands at the **same speed**, moving downward.

Same speed **u**, but downward: velocity = **−u** (up positive).

$v = gt$ — e.g. after 2.0 s, v = 9.81 × 2.0 ≈ 20 m s⁻¹.

Going up the velocity is positive; at the top it is zero; coming down it is negative — same slope (g) throughout.

An object moving through the air with **only gravity** acting on it (e.g. a thrown ball). Air resistance is ignored at SL.

Split it into **two independent parts**: horizontal (constant velocity) and vertical (free fall, a = g). They share the same time.

It stays **constant** — there is no sideways force.

It **increases** downward at g = 9.8 m s⁻² (free fall).

The **time** — it is the **same** for both columns.

From the **vertical** drop only: use s = u_y t + ½gt² (with u_y = 0 for a horizontal launch).

**Range = horizontal velocity × time of flight** (R = u_x·t), using the time from the vertical part.

**Together** — same height and same vertical start, so identical fall time. The throw only adds sideways distance.

**No** — horizontal speed adds range but does not change the vertical fall time.

A **parabola** — constant horizontal steps combined with growing vertical drops.

Both gain the same **vertical** speed, but the horizontal launch also keeps its **horizontal** velocity, so the combined speed is bigger.

A friction-like force from the air or liquid an object moves through. It always acts **against the motion** and **grows with speed**.

The **constant** velocity a falling object reaches when the **drag equals its weight**, so the resultant force (and acceleration) is zero.

It **increases** — drag grows with speed.

**Drag = weight** → resultant force = 0 → acceleration = 0.

**Zero** — weight and drag are equal and opposite, so they cancel.

**No** — weight and drag both act; they are **balanced**, so they cancel.

It **starts steep**, then **bends over and goes flat** — the flat value is the terminal velocity.

Just after release it is **near g** (drag tiny); at terminal velocity it has fallen to **zero**.

$F_g = mg$ — mass × gravitational field strength.

**Lower** — going up, drag adds to gravity, so the ball decelerates faster and rises less far.

**No** — the weight stays mg the whole way down; it is the **drag** that grows to match it.