Forces and Motion

Distance = total path travelled (scalar, m). Displacement = straight-line distance + direction (vector, m).

Speed = distance/time (scalar, m/s). Velocity = displacement/time (vector, m/s).

Acceleration = rate of change of velocity (vector, m/s²). $a = \frac{\Delta v}{t} = \frac{v - u}{t}$

• $u$ = initial velocity (m/s)
• $v$ = final velocity (m/s)
• $t$ = time (s)

Worked example. A car goes from 0 to 20 m/s in 5 s. Acceleration? $a = (20 - 0)/5 = 4\,\text{m/s}^2$

Deceleration is just a negative a. A car slowing from 20 m/s to 0 in 4 s: a = (0 − 20)/4 = −5 m/s² (or "5 m/s² deceleration").

Typical speeds (UK):

Activity	Speed
Walking	1.5 m/s
Running	3 m/s
Cycling	6 m/s
Sprinting	10 m/s
30 mph car	13.4 m/s
60 mph motorway car	27 m/s
Plane (typical)	250 m/s
Speed of sound (air)	343 m/s

On the AQA equation sheet:

$v^2 - u^2 = 2 a s$

• $u$ = initial velocity (m/s)
• $v$ = final velocity (m/s)
• $a$ = acceleration (m/s²)
• $s$ = distance (m)

Useful when time is unknown but distance, acceleration and one velocity are known.

Worked example. A car accelerates from rest (u = 0) at 3 m/s² over 50 m. Find final speed.

$v^2 = u^2 + 2as = 0 + 2 \times 3 \times 50 = 300$ $v = \sqrt{300} \approx 17.3\,\text{m/s}$

Worked example (braking distance). A car at 20 m/s brakes with deceleration 5 m/s². Stopping distance?

• $v = 0$, $u = 20$, $a = -5$ (deceleration).
• $0 - 400 = 2 \times -5 \times s \Rightarrow s = 400/10 = 40\,\text{m}$.

Newton's first law (inertia): an object remains at rest, or in steady straight-line motion, unless acted upon by a non-zero resultant force. This is why seatbelts are needed — the car decelerates suddenly, but you keep going at the previous velocity (your inertia) until restrained.

Newton's second law: $F = m a$. The resultant force on an object equals its mass × its acceleration.

• Acceleration in the SAME direction as the resultant force.
• Larger force → larger acceleration. Larger mass → smaller acceleration.

Worked example. A 60 kg cyclist accelerates at 1.5 m/s². Net force? $F = m a = 60 \times 1.5 = 90\,\text{N}$

Worked example. A 1000 kg car has a 4000 N driving force, 1500 N drag.

• Net force = 4000 − 1500 = 2500 N.
• a = F/m = 2500 / 1000 = 2.5 m/s².

Newton's third law: for every action (force) there is an equal and opposite reaction (force). The two forces act on DIFFERENT objects.

Examples:

• Walking: your foot pushes BACK on the ground; the ground pushes FORWARD on your foot.
• Rocket: hot gas pushed downward; gas pushes rocket upward (reaction).
• Swimmer pushing water back; water pushes swimmer forward.

Inertial mass (HT): a measure of how difficult it is to change an object's velocity. From Newton's second law: $m = F/a$ The harder it is to accelerate something, the larger its inertial mass.

Total stopping distance is split into two parts:

Thinking distance = how far the car travels while the driver REACTS.

• Reaction time ~0.7 s for an alert driver.
• Distance = speed × reaction time.

Braking distance = how far the car travels while the brakes are applied, until stopped.

• Depends on initial speed, brake force, vehicle mass, road grip.

Total stopping distance = thinking distance + braking distance.

Typical UK Highway Code figures (alert driver, dry road):

Speed (mph)	Speed (m/s)	Thinking (m)	Braking (m)	Total (m)
20	9	6	6	12
30	13	9	14	23
40	18	12	24	36
50	22	15	38	53
60	27	18	55	73
70	31	21	75	96

Notice the BRAKING distance grows much faster than thinking distance — because $E_k \propto v^2$. Doubling speed gives ~4× the braking distance.

Thinking distance depends on:

• Reaction time (alertness).
• Tiredness, alcohol, drugs, distractions (phone, conversation) all INCREASE reaction time.

Braking distance depends on:

• Speed (v²).
• Mass of the vehicle.
• Brake condition (worn pads, dirty discs).
• Tyre condition (worn tyres grip less; UK minimum tread depth 1.6 mm).
• Road conditions (wet/icy/leafy roads have less grip — braking distance can DOUBLE in the wet).
• Slope (uphill: shorter; downhill: longer).

When a car brakes, its kinetic energy is transferred via friction at the brakes to the thermal store of the brake discs (and a little to the tyres/road surface).

For an emergency stop from high speed, the energy is large — discs can reach 500+ °C in F1 cars. This is why:

• Brake discs are made of cast iron or ceramic (high melting point).
• Brake fade occurs if the brakes overheat (fluid boils, loses pressure).
• Modern cars use regenerative braking (in EVs and hybrids) to recover some energy into the battery.

Worked example. A 1000 kg car stops from 20 m/s. Energy transferred to brakes (assume 100 % efficient brake-to-heat): $E_k = \tfrac{1}{2} m v^2 = \tfrac{1}{2} \times 1000 \times 20^2 = 200{,}000\,\text{J} = 200\,\text{kJ}$

This 200 kJ ends up as thermal energy in the discs.