Scalar and vector quantities, contact and non-contact forces, weight and resultant forces, work done, forces and elasticity, distance, speed and acceleration, Newton's laws of motion, and stopping distances.

A focused answer to the AQA GCSE Combined Science: Trilogy Forces topic, covering scalars and vectors, contact and non-contact forces, weight, resultant forces and work, elasticity, distance, speed and acceleration, Newton's laws of motion, and stopping distances.

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

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What this topic is asking
Scalars, vectors and forces
Work, elasticity and motion
Newton's laws and stopping distances

What this topic is asking

AQA wants you to distinguish scalars and vectors, describe contact and non-contact forces, calculate weight, resultant forces and work, describe elasticity, use the equations for speed and acceleration, apply Newton's laws of motion, and explain stopping distances.

Scalars, vectors and forces

Weight is the force of gravity on an object, $W = mg$ , where $g$ is the gravitational field strength (about 9.8 N/kg near Earth's surface). Weight acts at the object's centre of mass and is measured with a calibrated spring balance (newtonmeter); it changes with location (you weigh less on the Moon), whereas mass is constant. A resultant force is the single force that has the same effect as all the forces acting on an object combined; when forces are balanced the resultant is zero.

Work, elasticity and motion

Motion equations to know:

Speed $= \dfrac{\text{distance}}{\text{time}}$ , so distance $=$ speed $\times$ time.
Acceleration $a = \dfrac{\Delta v}{t}$ (change in velocity divided by the time taken).
$v^2 - u^2 = 2as$ (links final speed $v$ , initial speed $u$ , acceleration $a$ and distance $s$ ).

On a distance-time graph the gradient is the speed; on a velocity-time graph the gradient is the acceleration and the area under the line is the distance travelled. Typical speeds worth recalling are about 1.5 m/s for walking and 3 m/s for running.

Newton's laws and stopping distances

The stopping distance of a vehicle $=$ thinking distance $+$ braking distance. The thinking distance is the distance travelled during the driver's reaction time and increases with speed and with anything that lengthens the reaction time (tiredness, alcohol, drugs, distractions). The braking distance is the distance travelled while the braking force acts; it increases with speed (with the square of the speed, because more kinetic energy must be removed) and with poor conditions such as wet or icy roads, worn tyres or worn brakes. Large decelerations require large braking forces, which can cause the brakes to overheat or the car to skid.

Combining Newton's second law with a motion equation

A 1500 kg car is pushed by a resultant force of 3000 N. Find its acceleration, then the distance it covers while speeding up from rest to 18 m/s.

step 1: Find the acceleration from F = ma

Rearrange to $a = \dfrac{F}{m} = \dfrac{3000}{1500} = 2$ m/s squared.

step 2: Choose the right motion equation

The known quantities are initial speed $u = 0$ , final speed $v = 18$ m/s and acceleration $a = 2$ m/s squared, and the unknown is distance $s$ , so use $v^2 - u^2 = 2as$ .

step 3: Substitute and rearrange

$18^2 - 0^2 = 2 \times 2 \times s$ , so $324 = 4s$ .

step 4: Solve and check units

$s = \dfrac{324}{4} = 81$ m. Remember to square the speed (324, not 18); forgetting this gives a distance about eighteen times too small.

Exam-style practice questions

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

AQA 20194 marksA car of mass 900 kg experiences a resultant forward force of 2700 N. Calculate its acceleration, and then calculate the distance it travels while accelerating from rest to 12 m/s at this acceleration.

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A Physics Paper 2 calculation using two equations. Method: from Newton's second law $F = ma$ , acceleration $a = \dfrac{F}{m} = \dfrac{2700}{900} = 3$ m/s squared. Then use $v^2 - u^2 = 2as$ with $u = 0$ , $v = 12$ : $12^2 - 0 = 2 \times 3 \times s$ , so $144 = 6s$ and $s = 24$ m. Markers award rearranging $F = ma$ , the correct acceleration, the use of the uniform-acceleration equation, and the final distance. Squaring the final velocity (144, not 12) is the step candidates most often miss.

AQA 20214 marksExplain how the thinking distance and the braking distance of a car each change as the car's speed increases, and give one factor that increases the braking distance other than speed.

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A Physics Paper 2 explanation on stopping distances. Reward: the thinking distance is the distance travelled during the driver's reaction time; for a given reaction time it increases in proportion to speed (at twice the speed the car travels twice as far before the brakes are applied). The braking distance is the distance travelled while the brakes act; it increases more steeply (with the square of the speed) because the greater kinetic energy must be transferred away by the braking force. A factor other than speed that increases braking distance: wet or icy roads, worn tyres or worn brakes (reduced friction). Markers credit the proportional increase in thinking distance, the larger increase in braking distance, and a valid road or vehicle condition.

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Sources & how we know this

AQA GCSE Combined Science: Trilogy (8464) specification — AQA (2016)