A runner travels 100\,m in 20\,s at constant speed. Calculate the speed. [2 marks]

OCR OCR 2019-style practice: A cyclist accelerates uniformly from rest to 8\,m/s in 4\,s. Calculate the acceleration, and then calculate the distance travelled during this time.

A P2 Calculate question on the recall equation a = vt. The acceleration is a = vt = 8 - 04 = 2\,m/s^2 (2 marks for substitution and answer with units). For the distance, use the average speed: the cyclist starts at 0 and ends at 8\,m/s, so the average speed is 0 + 82 = 4\,m/s, and distance = 4 × 4 = 16\,m (2 marks). Higher candidates can confirm with the SUVAT equation v^2 = u^2 + 2as: 8^2 = 0 + 2 × 2 × s gives 64 = 4s, so s = 16\,m. Markers reward the acceleration with units and a valid method giving 16\,m.

EnglandPhysicsSyllabus dot point

How do we describe motion with speed, velocity and acceleration, and read distance-time and velocity-time graphs?

Scalar and vector quantities, distance, displacement, speed, velocity and acceleration, distance-time and velocity-time graphs, the meaning of gradient and area under a graph, and the SUVAT equation for uniform acceleration.

A focused answer to OCR Gateway GCSE Physics A topic P2 on motion, covering scalar and vector quantities, distance, displacement, speed, velocity and acceleration, distance-time and velocity-time graphs, the meaning of gradient and area, and the SUVAT equation for Higher tier.

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 topic is asking
Scalars, vectors and the quantities of motion
Distance-time graphs
Velocity-time graphs
The equation of uniform acceleration (Higher)
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What this topic is asking

OCR wants you to distinguish scalar and vector quantities, define and calculate speed, velocity and acceleration, interpret distance-time and velocity-time graphs, and (at Higher tier) use the SUVAT equation. This is topic P2.1 of the OCR Gateway Physics A (J249) specification, examined on the Paper 1 or Paper 3 side.

Scalars, vectors and the quantities of motion

The everyday quantities are:

Distance (scalar) is how far an object moves; displacement (vector) is the straight-line distance and direction from start to finish.
Speed (scalar) is how fast it moves: $\text{speed} = \dfrac{\text{distance}}{\text{time}}$ . Typical speeds to know are walking at about $1.5\,\text{m/s}$ , running at about $3\,\text{m/s}$ , cycling at about $6\,\text{m/s}$ and a car on a motorway at around $30\,\text{m/s}$ .
Velocity (vector) is the speed in a stated direction.
Acceleration (vector) is the rate of change of velocity: $a = \dfrac{\Delta v}{t}$ , the change in velocity over the time taken, measured in $\text{m/s}^2$ . Deceleration is a negative acceleration.

You must recall the speed and acceleration equations; they are not on the data sheet.

Distance-time graphs

A straight sloping line shows constant speed; a steeper straight line shows a higher constant speed. A line curving upwards (getting steeper) shows acceleration; a line curving and flattening shows deceleration.

Velocity-time graphs

To find a distance from a velocity-time graph, calculate the area under the line by splitting it into rectangles and triangles and adding them, or by counting squares. To find an acceleration, calculate the gradient (change in velocity over change in time) of a straight section.

The equation of uniform acceleration (Higher)

For motion with uniform acceleration, Higher candidates use the SUVAT equation, which is given on the data sheet:

v^2 = u^2 + 2as

where $u$ is the initial velocity, $v$ is the final velocity, $a$ is the acceleration and $s$ is the distance. This lets you find a missing quantity without knowing the time, a common Higher calculation.

Finding the distance from a velocity-time graph

A car accelerates uniformly from rest to $20\,\text{m/s}$ over $10\,\text{s}$ , then travels at a constant $20\,\text{m/s}$ for a further $15\,\text{s}$ . Calculate the total distance travelled using the area under the velocity-time graph.

step 1: Split the graph into shapes

The first part (accelerating from $0$ to $20\,\text{m/s}$ over $10\,\text{s}$ ) is a triangle. The second part (constant $20\,\text{m/s}$ for $15\,\text{s}$ ) is a rectangle.

step 2: Find the area of the triangle

Area $= \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 10 \times 20 = 100\,\text{m}$ .

step 3: Find the area of the rectangle

Area $= \text{base} \times \text{height} = 15 \times 20 = 300\,\text{m}$ .

step 4: Add the areas for the total distance

Total distance $= 100 + 300 = 400\,\text{m}$ . The area under a velocity-time graph always gives the distance, which is why splitting it into simple shapes is the standard method.

Try this

Q1. A runner travels $100\,\text{m}$ in $20\,\text{s}$ at constant speed. Calculate the speed. [2 marks]

Cue. $\text{speed} = \dfrac{\text{distance}}{\text{time}} = \dfrac{100}{20} = 5\,\text{m/s}$ .

Q2. State what the area under a velocity-time graph represents. [1 mark]

Cue. The distance travelled.

Exam-style practice questions

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

OCR 20194 marksA cyclist accelerates uniformly from rest to

8\,\text{m/s}

4\,\text{s}

. Calculate the acceleration, and then calculate the distance travelled during this time.

Show worked answer →

A P2 Calculate question on the recall equation $a = \dfrac{\Delta v}{t}$ . The acceleration is $a = \dfrac{\Delta v}{t} = \dfrac{8 - 0}{4} = 2\,\text{m/s}^2$ (2 marks for substitution and answer with units). For the distance, use the average speed: the cyclist starts at $0$ and ends at $8\,\text{m/s}$ , so the average speed is $\dfrac{0 + 8}{2} = 4\,\text{m/s}$ , and distance $= 4 \times 4 = 16\,\text{m}$ (2 marks). Higher candidates can confirm with the SUVAT equation $v^2 = u^2 + 2as$ : $8^2 = 0 + 2 \times 2 \times s$ gives $64 = 4s$ , so $s = 16\,\text{m}$ . Markers reward the acceleration with units and a valid method giving $16\,\text{m}$ .

OCR 20214 marksState what the gradient and the area under a velocity-time graph each represent, and describe how you would find the distance travelled during a journey shown on such a graph.

Show worked answer →

A P2 question on interpreting motion graphs. On a velocity-time graph the gradient represents the acceleration (a steeper line is a greater acceleration, a horizontal line is constant velocity, a negative gradient is deceleration) (1 mark), and the area under the line represents the distance travelled (1 mark). To find the distance travelled, calculate the area under the graph by splitting it into simple shapes (rectangles and triangles) and adding them, or by counting squares (2 marks for a workable method and a clear description). Markers credit gradient equals acceleration, area equals distance, and the shapes-or-squares method.

Related dot points

Sources & how we know this

OCR GCSE (9-1) Physics A (Gateway Science) J249 specification — OCR (2016)