Acceleration — AQA GCSE Physics

The equation v² − u² = 2as relates final velocity v, initial velocity u, uniform acceleration a and distance s. It is provided on the Physics Equations Sheet and is particularly useful when time is not given. For example, if a cyclist speeds up uniformly from 4 m/s to 10 m/s over 21 m: a = (10² − 4²) ÷ (2 × 21) = (100 − 16) ÷ 42 = 2 m/s². Rearranging requires care: v² = u² + 2as, u² = v² − 2as and s = (v² − u²) ÷ (2a). The equation applies only for uniform acceleration in a straight line. Always use standard SI units: m/s for velocity, m/s² for acceleration and m for distance.

This equation links the change in the square of velocity to acceleration and distance for motion in a straight line with constant acceleration. The symbol v is final velocity in metres per second (m/s), u is initial velocity in m/s, a is acceleration in m/s² and s is distance in metres (m). The term v² − u² is the difference between the squares of the final and initial velocities, not the square of the difference. To use it, list the known quantities with units, rearrange for the unknown, substitute values and solve. For example, a car accelerating uniformly from rest (u = 0 m/s) over 50 m with a = 2 m/s² has v² = 0² + 2 × 2 × 50 = 200 m²/s², so v = √200 ≈ 14 m/s.

Final velocity, v, is the velocity of an object at the end of the time interval or displacement being considered. Velocity is a vector, so v has both magnitude and direction, and it is measured in metres per second (m/s). The value of v depends on the initial velocity u, the acceleration a and the time or displacement over which the motion occurs. For example, a cyclist moving at 4 m/s who accelerates at 1.5 m/s² for 6 s reaches v = u + a t = 4 + 1.5 × 6 = 13 m/s. When motion reverses, v can be negative if the chosen positive direction is opposite to the final motion. Always state the direction or sign convention when the motion is not in a single straight line.

Initial velocity, symbol 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 example, a car already rolling at 8 m/s when a stopwatch starts has u = 8 m/s; a ball thrown upwards at 12 m/s, with upwards taken as positive, has u = +12 m/s, while the same ball moving downwards has negative velocity. In acceleration calculations, u is the starting value and the final velocity v is the value after the time interval. The change in velocity is v − u, and dividing this by the time taken gives the acceleration. Always state the direction convention and convert units such as km/h to m/s before substituting.

Acceleration, symbol a, is the rate of change of velocity, measured in metres per second squared (m/s²). It is a vector: speeding up, slowing down or changing direction all involve acceleration. The defining relationship is a = (v − u) ÷ t, where v is final velocity, u is initial velocity and t is the time taken. For example, a cyclist increasing from 4 m/s to 10 m/s in 3 s has a = (10 − 4) ÷ 3 = 2 m/s². A value of 2 m/s² means the velocity changes by 2 m/s every second. Negative acceleration, sometimes called deceleration, means the velocity change opposes the chosen positive direction. On a velocity–time graph, acceleration is the gradient, so a straight rising line shows constant positive acceleration and a horizontal line shows zero acceleration.

In the AQA GCSE Physics equations for acceleration, the symbol s stands for distance travelled, measured in metres (m). Distance is the total length of the path moved, a scalar quantity, so it has magnitude but no direction. When an object accelerates, its velocity changes, and the distance moved depends on initial velocity, acceleration and time. For example, a car accelerating from rest at 2 m/s² for 3 s travels s = ½ × 2 × 3² = 9 m. Always convert centimetres or kilometres to metres before substituting, and remember that distance is not the same as displacement, which is a vector measured in a stated direction.

Near the Earth’s surface, an object falling freely under gravity accelerates downwards at about 9.8 m/s². This value, often approximated as 10 m/s² in GCSE calculations, is called the acceleration due to gravity and is represented by g. It applies when air resistance is negligible, so all objects fall with the same acceleration regardless of mass. For example, a ball dropped from rest reaches about 9.8 m/s after 1 s and about 19.6 m/s after 2 s. The acceleration is directed towards the Earth’s centre. If air resistance becomes significant, the resultant downward force decreases and acceleration is less than 9.8 m/s².

Average acceleration measures how quickly velocity changes over a time interval. Velocity is a vector: its direction matters as well as its size. To calculate average acceleration, find the change in velocity by subtracting the initial velocity from the final velocity, then divide by the time taken for that change. For example, a cyclist speeds up from 4 m/s to 10 m/s in 3 s: the change in velocity is 10 m/s − 4 m/s = 6 m/s, so the average acceleration is 6 m/s ÷ 3 s = 2 m/s². A negative result means the object is slowing down or accelerating in the opposite direction to its motion. The equation gives an average value; if acceleration varies during the interval, the calculation still describes the overall rate of change of velocity.

This equation defines average acceleration as the change in velocity divided by the time taken. In symbols, a = Δv / t, where Δv is final velocity minus initial velocity. For example, a car accelerates from 8 m/s to 20 m/s in 4 s: Δv = 20 m/s − 8 m/s = 12 m/s, so a = 12 m/s ÷ 4 s = 3 m/s². The equation applies to motion in a straight line when acceleration is average or uniform. If velocity is measured in m/s and time in s, acceleration is in m/s². A negative value shows deceleration or acceleration opposite to the chosen positive direction. Rearranging gives Δv = a × t and t = Δv ÷ a, which are useful when the question supplies different quantities.

Acceleration measures how quickly velocity changes. The equation a = Δv / t links acceleration a (m/s²), to the change in velocity Δv (m/s), and the time t taken (s). Δv is found by subtracting the initial velocity from the final velocity, so Δv = v − u. For example, a cyclist speeds up from 4 m/s to 10 m/s in 3 s: a = (10 - 4) / 3 = 2 m/s². Negative acceleration indicates a velocity change in the negative direction, meaning speeding up if velocity is negative, or slowing down if positive. Always convert units before substituting, and rearrange the equation when the question asks for Δv or t rather than a. Ensure time is in seconds and velocity in m/s to yield acceleration in m/s².

The symbol Δv stands for change in velocity, measured in metres per second (m/s). The Greek letter Δ means 'change in', so Δv = final velocity − initial velocity. Velocity is a vector: it has both size and direction, so a change in direction at constant speed is also a change in velocity. For example, a car travelling at 20 m/s that slows to 8 m/s has Δv = 8 − 20 = −12 m/s; the negative sign shows the change opposes the original motion. If a runner reverses direction at the same speed, the change in velocity is twice the speed because the sign of the velocity reverses. Always record velocities with their direction signs before subtracting.

Time, t, is the duration over which a change in velocity occurs. In acceleration calculations, t must be measured in seconds (s). If a stopwatch records 2 minutes 30 seconds, convert to 150 s before using the equation. The symbol t represents this time interval, not a clock reading. For example, a car accelerating from rest to 20 m/s in 5 s has t = 5 s. If the same change takes 0.5 minutes, t = 30 s. Always identify the start and end of the interval, then express the duration in seconds. This ensures the acceleration unit m/s² is consistent. When timing, start the stopwatch at the initial velocity and stop at the final velocity. If using light gates, the time interval is between the two gates. Record raw times and convert units carefully, remembering 1 minute = 60 s and 1 ms = 0.001 s.

This statement requires you to remember the acceleration equation and use it in calculations. The equation is a = Δv / t, where a is acceleration in metres per second squared (m/s²), Δv is the change in velocity in metres per second (m/s), and t is the time in seconds (s). To apply it, first find Δv by subtracting the initial velocity from the final velocity. Then divide by the time taken. For example, a cyclist speeds up from 4 m/s to 10 m/s in 3 s: Δv = 10 − 4 = 6 m/s, so a = 6 / 3 = 2 m/s². You may also rearrange the equation to find Δv or t. Always include units and check that the time is in seconds. This equation applies to uniform acceleration in a straight line.

Estimating means choosing sensible numbers and dividing. A family car pulling away might reach 13 m/s in about 8 s, giving roughly 1.6 m/s²; a sports car reaching 27 m/s in 5 s manages about 5 m/s². A sprinter leaving the blocks achieves something near 5 m/s² briefly. A commuter train accelerates gently, at well under 1 m/s². Anything dropped near the Earth's surface accelerates at about 10 m/s² while air resistance is small, and that value is the natural yardstick, because an acceleration much larger than it is violent and a passenger lift is designed for around 1 m/s² so that people stay comfortable. State the two values you assumed and show the division, because examiners reward a reasoned estimate rather than a remembered number.

On a velocity-time graph the vertical axis is velocity and the horizontal is time, so the gradient is a change in velocity divided by a time, which is the definition of acceleration. A horizontal line means constant velocity and zero acceleration; a straight sloping line means constant acceleration; a line sloping downwards means the object is slowing, and the gradient comes out negative. Take the gradient from a large triangle: a velocity rising from 4 m/s to 20 m/s over 8.0 s gives an acceleration of 2.0 m/s². Where the line curves the acceleration is changing, so a tangent is needed to find its value at a particular instant. Do not confuse this graph with a distance-time graph, on which the gradient is a speed.

Multiply a velocity by a time and you get a distance, which is why the area between the line and the time axis on a velocity-time graph is the distance travelled. For straight-line sections, split the area into rectangles and triangles: a rectangle is velocity multiplied by time, and a triangle is half the base multiplied by the height. An object accelerating uniformly from rest to 12 m/s in 6.0 s covers 36 m. Where the line curves, count squares instead: work out what one square represents by multiplying the value of one square on each axis, count the squares under the line, counting a part square if more than half of it lies underneath, then multiply. An area below the time axis represents motion in the opposite direction.

A velocity–time graph plots velocity on the vertical axis against time on the horizontal axis. To draw one, record velocity at regular time intervals, choose sensible scales, plot each pair of values and join the points. The slope (gradient) gives acceleration: a = Δv ÷ t. A straight horizontal line means constant velocity; a straight sloping line means uniform acceleration. For example, if velocity rises from 0 m/s to 20 m/s in 5 s, a = 4 m/s². Higher Tier students must also use a tangent to find the gradient at a point on a curved velocity–time graph, representing instantaneous acceleration. Always ensure axes are labelled with appropriate units.

On a velocity–time graph, the area enclosed between the plotted line and the time axis represents the distance travelled or displacement. To determine this area, split the space under the graph into simple geometric shapes: rectangles where velocity is constant, and triangles where velocity changes uniformly. For a rectangle, use area = base × height. For a triangle, use area = ½ × base × height. Add these separate areas together to obtain the total distance. For example, if a car travels at 15 m/s for 4 s, then accelerates uniformly to 25 m/s over 6 s, the distance is the initial rectangle (15 × 4 = 60 m) plus the acceleration phase's rectangle (15 × 6 = 90 m) and triangle (½ × 6 × 10 = 30 m), giving a total distance of 180 m.

On a velocity–time graph, the area between the line and the time axis represents distance travelled. When the line is curved or made of several straight sections, you can find this area by counting squares on the grid. First work out what one square is worth: multiply the velocity represented by one square's height by the time represented by one square's width. For example, if one square is 2 m/s tall and 0.5 s wide, each square is 1 m of distance. Count whole squares under the line, then combine part-squares to make whole ones. Multiply the total number of squares by the value of one square to give the distance. This method is appropriate when the area is not a simple triangle or rectangle, and it estimates distance from the graph.

When an object is released its speed is zero, so drag is zero. Falling through air, upthrust is negligible; the resultant force is downwards and equal to weight, so it accelerates at about 9.8 m/s². In a liquid, upthrust is not negligible, so the initial resultant force is weight minus upthrust, making initial acceleration less than g. As it speeds up, fluid resistance grows. The resultant force is now weight minus drag and upthrust (if in a liquid). This resultant force is still downwards but smaller, so the object speeds up at a steadily decreasing rate. On a velocity-time graph, this is a curve that starts steep and gradually flattens. In a denser fluid, drag builds up faster, so acceleration falls away sooner.

As the falling object speeds up, the drag from the fluid keeps increasing until it equals the weight in size. The two forces are then balanced, there is no resultant, and the object stops accelerating and carries on at a steady velocity, its terminal velocity. Note that it does not slow down; it simply stops speeding up. A heavier object of the same shape needs a larger drag to balance its greater weight, so it must reach a higher speed first and its terminal velocity is greater, which is why a large hailstone lands faster than a small one. Opening a parachute increases the drag sharply, so the resultant force acts upwards, the skydiver slows, and a new and much lower terminal velocity is reached.

Marks here come from pairing each part of the graph with a force diagram. At release there is one arrow down for the weight and nothing upwards, so the gradient is at its steepest. Part way down there is a long weight arrow and a shorter drag arrow, so the resultant is smaller and the curve is flattening. At terminal velocity the arrows are equal in length, the resultant is zero and the line is horizontal. When a parachute opens the drag arrow suddenly becomes longer than the weight, the resultant points upwards, the skydiver decelerates and the line falls steeply before levelling off at a new and lower terminal velocity. Draw the curve smoothly, with no corners except at the moment the parachute opens, and label both terminal velocities.