AQA AQA 2019-style practice: A car of mass 900\,kg accelerates from rest. A resultant force of 3600\,N acts on it. Calculate the acceleration of the car, then calculate the velocity it reaches after travelling 50\,m from rest.

First use Newton's second law F = ma rearranged to a = F/m = 3600 / 900 = 4.0\,m/s^2 (2 marks). Then use the uniform acceleration equation v^2 - u^2 = 2as, with u = 0 (starts from rest), a = 4.0\,m/s^2 and s = 50\,m: v^2 = 0 + 2 × 4.0 × 50 = 400 (2 marks), so v = sqrt(400) = 20\,m/s (1 mark). Markers reward the rearrangement of F = ma, selecting the correct uniform acceleration equation when time is not given, and remembering to take the square root. A common error is to forget that u = 0 or to omit the square root.

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How do forces change motion, and what do Newton's laws say?

Acceleration and Newton's laws: the acceleration equation, the uniform acceleration equation, velocity-time graphs, and Newton's three laws of motion.

A focused answer to AQA GCSE Physics 4.5.6, covering acceleration and its equation, the uniform acceleration equation, reading velocity-time graphs, and Newton's first, second and third laws of motion with the force, mass and acceleration relationship.

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

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

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What this dot point is asking
Acceleration
Uniform acceleration
Velocity-time graphs
Newton's laws
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What this dot point is asking

AQA wants you to use the acceleration equation and the uniform acceleration equation, interpret velocity-time graphs, and state and apply Newton's three laws of motion, including $F = ma$ .

Acceleration

Uniform acceleration

The uniform acceleration equation is useful precisely when a question gives you a distance but no time, which is common in vehicle-motion problems. It links four quantities (final velocity, initial velocity, acceleration and distance), so as long as you know three you can find the fourth. Watch the sign of $a$ : a deceleration is a negative acceleration, so a car braking has a negative value of $a$ in the equation.

Velocity-time graphs

To find the acceleration at a point on a curved velocity-time graph, you draw a tangent to the curve at that point and find its gradient. To find the distance from a graph with several stages, split the area into rectangles and triangles, work out each area, and add them. These graph skills are tested every year, so practise reading both the gradient (acceleration) and the area (distance) from the same graph.

Newton's laws

Newton's second law tells you that the same force gives a smaller acceleration to a heavier object, which is why a loaded lorry accelerates more slowly than an empty one under the same engine force. The law also defines what we mean by inertial mass: the mass in $F = ma$ measures how hard it is to change an object's motion, so a larger mass has more inertia and resists acceleration more. A subtle but examinable point is that it is the resultant (net) force that goes into the equation, not any single force; you must first combine all the forces acting before applying $F = ma$ . The third law is also often misunderstood: the equal and opposite forces of a third-law pair always act on two different objects (for example, a foot pushes back on the ground and the ground pushes forward on the foot), so they never cancel each other out on a single object.

Try this

Q1. State Newton's second law as an equation. [1 mark]

Cue. $F = ma$ .

Q2. A resultant force of $12\,N$ acts on a $3\,kg$ object. Calculate its acceleration. [2 marks]

Cue. $a = \dfrac{F}{m} = \dfrac{12}{3} = 4\,m/s^2$ .

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 20195 marksA car of mass

900\,\text{kg}

accelerates from rest. A resultant force of

3600\,\text{N}

acts on it. Calculate the acceleration of the car, then calculate the velocity it reaches after travelling

50\,\text{m}

from rest.

Show worked answer →

First use Newton's second law $F = ma$ rearranged to $a = F/m = 3600 / 900 = 4.0\,\text{m/s}^2$ (2 marks). Then use the uniform acceleration equation $v^2 - u^2 = 2as$ , with $u = 0$ (starts from rest), $a = 4.0\,\text{m/s}^2$ and $s = 50\,\text{m}$ : $v^2 = 0 + 2 \times 4.0 \times 50 = 400$ (2 marks), so $v = \sqrt{400} = 20\,\text{m/s}$ (1 mark). Markers reward the rearrangement of $F = ma$ , selecting the correct uniform acceleration equation when time is not given, and remembering to take the square root. A common error is to forget that $u = 0$ or to omit the square root.

AQA 20214 marksState Newton's first law of motion, and use it to explain why a passenger in a car continues to move forward when the car stops suddenly in a collision.

Show worked answer →

Newton's first law states that an object remains at rest, or continues to move at a constant velocity, unless a resultant force acts on it (1 mark). When the car stops suddenly, the passenger's body tends to continue moving forward at its original velocity because there is no resultant backward force acting on the passenger (1 mark) until the seatbelt or airbag provides one (1 mark). The seatbelt exerts a backward force that decelerates the passenger more gently over a longer time, reducing the force on the body (1 mark). Markers reward an accurate statement of the law and applying it to the passenger continuing at constant velocity in the absence of a resultant force.

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Sources & how we know this

AQA GCSE Physics (8463) specification — AQA (2016)