Acceleration — AQA GCSE Combined Science

This statement gives the relationship used to find average acceleration: acceleration equals change in velocity divided by time taken. Change in velocity is final velocity minus initial velocity, so you must record both values and subtract in the correct order. Time is the duration over which that change happens. For example, a train slowing from 30 m/s to 10 m/s in 8 s has change in velocity = 10 m/s − 30 m/s = −20 m/s, so acceleration = −20 m/s ÷ 8 s = −2.5 m/s². The negative sign shows deceleration relative to the forward direction. The unit m/s² arises because velocity in m/s is divided by time in s. This calculation gives an average acceleration; if the acceleration varies, the value applies to the whole interval. Always include direction when interpreting the sign.

This equation defines acceleration as the rate of change of velocity. The symbol Δv means the change in velocity, found by subtracting the initial velocity from the final velocity: Δv = v − u. The symbol t is the time taken for that change, measured in seconds. Acceleration a is measured in metres per second squared, m/s². For example, a cyclist speeds up from 4 m/s to 10 m/s in 3 s, so Δv = 10 − 4 = 6 m/s and a = 6 ÷ 3 = 2 m/s². A negative value shows deceleration, meaning the object slows down. Always identify u, v and t from the question before substituting, and keep units consistent.

Acceleration is the rate at which velocity changes and is measured in metres per second squared, m/s². The unit comes from dividing a velocity change in metres per second (m/s) by a time in seconds (s), giving m/s ÷ s = m/s². This means that each second, the velocity changes by the stated number of metres per second. For example, an acceleration of 3 m/s² means the velocity increases by 3 m/s every second. Because velocity is a vector, a change in direction also produces acceleration, even at constant speed. A negative value, such as −2 m/s², indicates deceleration. Always write the unit as m/s², not m/s, and include it with your numerical answer.

Acceleration measures how quickly velocity changes, so the change in velocity ∆v is the difference between the final velocity and the initial velocity. Because velocity is a vector, ∆v can be positive, negative or zero. For example, a car speeds up from 4 m/s to 16 m/s, so ∆v = 16 m/s − 4 m/s = 12 m/s. If it slows from 20 m/s to 8 m/s, ∆v = 8 m/s − 20 m/s = −12 m/s, a negative change showing deceleration. Always subtract the initial velocity from the final velocity, keep the direction consistent, and give the unit as metres per second, m/s. ∆v is then divided by the time taken to find acceleration in m/s².

In acceleration calculations, time t is the duration over which the velocity changes, measured in seconds, s. It is the interval between the initial and final velocity readings, not the total journey time unless the whole journey is the acceleration phase. For example, if a cyclist speeds up from 3 m/s to 11 m/s in 4 s, then t = 4 s and a = (11 m/s − 3 m/s) ÷ 4 s = 2 m/s². Time must be converted to seconds before substitution: 2 minutes becomes 120 s, and 0.5 minutes becomes 30 s. Using the correct time interval ensures the acceleration value is accurate and carries the unit m/s².

Acceleration is the rate of change of velocity, so it is a vector quantity with both magnitude and direction. When an object slows down, its velocity decreases in magnitude, and this is called deceleration. Deceleration is simply acceleration in the direction opposite to the object's motion. For example, a car travelling forwards at 20 m/s that reduces its velocity to 5 m/s over 3 s has an acceleration of (5 − 20) ÷ 3 = −5 m/s²; the negative sign shows deceleration. In everyday language, deceleration means slowing down, but in physics it is still an acceleration. Students should recognise deceleration from velocity–time graphs as a negative gradient and from distance–time graphs as a decreasing gradient. The skill is assessed by asking students to identify deceleration in descriptions, graphs or calculations, and to interpret the sign of acceleration correctly.

Estimating everyday accelerations means using familiar reference values to judge the size of acceleration in common situations. For example, a car accelerating from rest to 30 m/s in 10 s has an average acceleration of 3 m/s²; a bus might accelerate at about 1 m/s²; a falling object near Earth's surface accelerates at about 10 m/s² due to gravity. Students should be able to recall or estimate typical magnitudes, such as a person walking (≈0.5 m/s²), a train (≈0.5–1 m/s²), or a rocket launch (≈30 m/s²). The skill involves using the equation a = Δv ÷ t with approximate values, or comparing to known benchmarks. Assessment may ask students to choose sensible estimates, calculate approximate accelerations from given data, or explain why an estimated value is reasonable. This helps develop a sense of scale in physics.

On a velocity–time graph, velocity is plotted on the y-axis and time on the x-axis, so the gradient measures how quickly velocity changes. Acceleration is the rate of change of velocity, a = Δv ÷ Δt, so the gradient equals the acceleration. To find it, choose two points on a straight section, read the velocity change Δv and time change Δt, then divide Δv by Δt. For example, if velocity rises from 4 m/s to 16 m/s between 2 s and 8 s, Δv = 12 m/s and Δt = 6 s, giving a = 2 m/s². A steeper line means greater acceleration; a horizontal line means zero acceleration; a negative gradient means deceleration. Curved sections show changing acceleration, so a tangent is drawn at the point of interest.

This is a Higher Tier only skill. On a velocity–time graph, the area between the line and the time axis represents the distance travelled, or displacement when direction is included. This works because distance is velocity multiplied by time, and the area of a rectangle is height × width. Split the area into simple shapes: rectangles for constant velocity, triangles for uniform acceleration, and trapezia for combined changes. For example, a car accelerating from rest to 20 m/s in 10 s covers a triangle area of ½ × 10 s × 20 m/s = 100 m. Areas below the time axis represent negative displacement, so subtract them when finding overall displacement. Count squares for curved graphs, or estimate the area between the curve and the axis.

This statement combines a practical skill with an analytical one. To draw a velocity–time graph, record velocity at regular time intervals, choose scales that use most of the grid, plot each point accurately, and join points with a line of best fit. To interpret it, remember that the slope (gradient) is acceleration: gradient = change in velocity ÷ change in time, a = Δv ÷ t, in m/s². A straight line means uniform acceleration; a horizontal line means zero acceleration; a negative gradient means deceleration. For example, points (0 s, 0 m/s), (2 s, 4 m/s) and (4 s, 8 m/s) give a gradient of (8 − 0) ÷ (4 − 0) = 2 m/s². Always convert units before calculating.

On a velocity–time graph, the area between the plotted line and the time axis represents the distance travelled, or displacement when direction is included. This specific skill is assessed at Higher Tier only. Split the enclosed region into rectangles, triangles or trapezia, calculate each area, then add them. For example, a line rising from 0 m/s to 8 m/s over 4 s forms a triangle: area = ½ × 4 s × 8 m/s = 16 m. A constant 8 m/s for the next 3 s forms a rectangle: 8 m/s × 3 s = 24 m, giving 40 m in total. Areas below the time axis count as negative displacement, so subtract them when displacement is required. Always attach the unit metre (m) and check the axes' scales before calculating.

When a velocity–time graph has a curved line or an irregular shape, the enclosed area represents the distance travelled. For Higher Tier students, this area can be estimated by counting squares on the grid. First, determine what one square represents: multiply the time interval per square by the velocity interval per square (e.g., 0.5 s × 2 m/s = 1 m). Count whole squares inside the region, then combine part-squares to make whole ones. Multiply the total number of squares by the value of one square to estimate the distance. This method gives an approximate value. Counting squares is especially useful when the line is not straight and no simple area formula applies.

This equation links the change in velocity of an object to its acceleration and the distance travelled, but only when acceleration is uniform. Written as v² − u² = 2 × a × s, it avoids needing time. For example, a car accelerating uniformly from 8 m/s to 20 m/s over 60 m: v² − u² = 400 − 64 = 336 m²/s², so 2 × a × 60 = 336, giving a = 2.8 m/s². Rearranging is a key skill: make the unknown the subject before substituting. Because velocity is squared, any negative sign indicating direction is lost; therefore, direction must be handled carefully through the signs of acceleration and displacement. Always square the full velocity value, including its unit, so the unit becomes m²/s².

This statement introduces the condition for using the kinematic equation v² − u² = 2 × a × s: acceleration must be uniform, meaning it stays constant in magnitude and direction throughout the motion. Uniform acceleration produces a straight-line velocity–time graph, so the gradient is constant. For example, a ball rolling down a smooth slope may be modelled as uniform acceleration, but a car in stop-start traffic is not. When acceleration is uniform, average velocity is (u + v) / 2 and distance is that average multiplied by time. If acceleration changes, the equation cannot be applied directly; the motion must be split into stages or a graph used. Recognising this condition is part of choosing the correct equation.

This equation links the four quantities of motion when time is not known. The symbol v is final velocity in m/s, u is initial velocity in m/s, a is acceleration in m/s² and s is displacement in m. It is derived from combining a = (v − u)/t with s = ((u + v)/2)t, eliminating t. To use it, list the known values, identify the unknown, substitute into v² − u² = 2as and solve. For example, a car accelerating at 2 m/s² from rest over 25 m gives v² = 0² + 2 × 2 × 25 = 100, so v = 10 m/s. If the object decelerates, a is negative, and if it reverses direction, v may be negative. Always check units are m/s, m/s² and m before substituting.

Final velocity, v, is the velocity of an object at the end of the time interval being considered, measured in metres per second (m/s). Velocity is a vector, so v includes both speed and direction; a negative value means motion opposite to the chosen positive direction. For example, a ball thrown upwards at 15 m/s has v = 0 m/s at its highest point, then v becomes negative as it falls back. In calculations, v is the value after acceleration a has acted for time t, linked by v = u + at or v² = u² + 2as. Always identify which moment is the 'final' one, convert units if needed, and give the unit m/s with the numerical answer.

Initial velocity, u, is the velocity of an object at the moment timing starts, measured in metres per second (m/s). Velocity is a vector: it has size and direction, so u can be positive or negative depending on the chosen positive direction. For a car starting from rest, u = 0 m/s; for a ball thrown upwards at 12 m/s, taking upwards as positive gives u = +12 m/s. In the acceleration equation a = (v − u) ÷ t, u is subtracted from final velocity v, so the change in velocity is v − u. If an object slows down, v is smaller than u and the acceleration is negative (deceleration). Always convert units to m/s before substituting: 72 km/h = 72 × 1000 ÷ 3600 = 20 m/s.

Distance, s, is the length of the path travelled by an object, measured in metres (m). It is a scalar quantity, so it has magnitude only and no direction. In acceleration problems, s usually means the distance moved during the time interval over which the acceleration acts. For example, using v² − u² = 2 × a × s, if a car accelerates from 0 to 10 m/s at 2 m/s², s = 100 ÷ 4 = 25 m. Distance is not the same as displacement: displacement is the straight-line distance in a stated direction. Convert units before substituting: 250 cm = 2.5 m and 1.2 km = 1200 m. On a velocity–time graph, distance travelled equals the total area between the line and the time axis, treating all areas as positive.

Free fall means the only force acting is gravity, so air resistance is negligible. Near the Earth’s surface, all objects have the same downward acceleration of about 9.8 m/s², often rounded to 10 m/s². This acceleration is independent of mass, shape or size when drag is absent. For example, in a vacuum a coin and a feather dropped together hit the ground at the same time. You can use a = Δv / Δt to find acceleration from velocity changes, or v = u + at to predict speed after a time t. The value 9.8 m/s² is a magnitude; the direction is towards the Earth’s centre. In everyday falls, air resistance reduces the acceleration, so 9.8 m/s² is an idealised value for free fall.

When an object falls through a fluid such as air or water, gravity pulls it down, but the fluid exerts a drag force that increases with speed. Initially the weight is larger than drag, so the object accelerates. As speed rises, drag increases until it equals the weight; the resultant force is then zero. At this point the object stops accelerating and falls at a constant maximum speed called terminal velocity. For example, a skydiver accelerates until air resistance balances weight, then falls at terminal velocity until the parachute opens. The same idea explains why a ball bearing reaches a steady speed in oil. Terminal velocity depends on shape, size and the fluid’s viscosity.