The answer
\(v = 12.5\)
O-Level E-Math 2025 Paper 1 Question 22 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 22 of the O-Level E-Math 2025 Paper 1. It tests area under speed-time graph = distance, in the Kinematics (travel graphs) area. It is worth 4 marks. It is a worded / diagram-based question, so open your Ten-Year Series (TYS) or the official paper at this question, then follow our full worked solution below.
Distance = area under the graph.
Cyclist \(A\): for \(0 \leqslant t \leqslant 60\) the speed falls linearly from 20 to \(v\) (a trapezium), then it is constant \(v\) for \(60 \leqslant t \leqslant 100\) (a rectangle): \[\text{dist}_A = \underbrace{\tfrac12(20 + v)(60)}_{\text{trapezium}} + \underbrace{v \times 40}_{\text{rectangle}} = 30(20 + v) + 40v = 600 + 70v.\]
Cyclist \(B\): constant \((v + 3)\) for 100 s: \(\text{dist}_B = (v + 3)(100) = 100v + 300\).
\(B\) travels 75 m further: \(\text{dist}_B - \text{dist}_A = 75\): \[(100v + 300) - (600 + 70v) = 75 \Rightarrow 30v - 300 = 75 \Rightarrow 30v = 375 \Rightarrow v = 12.5.\]
Answer: \(v = 12.5\)
Same structure, different numbers
Swap the constants, dress a quadratic as a length, hide a derivative inside an integral, and a student sees a brand new problem. The structure underneath is the same, and so is the method. Once a student can name the structure, a whole row of questions that look different start to open the same way.
That is where marks really leak: in choosing the method, not in the algebra that follows. We call it Lock and Key, name the lock, then the key follows.
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Genius Plus Academy · O-Level & IP Mathematics
Our O-Level E-Math tuition trains the same recognise-the-structure method these worked solutions show, taught by a team that has marked these papers for years. It runs within our weekly Secondary Math programme, Sec 1 to 4 and IP.
It is a area under speed-time graph = distance question from Kinematics (travel graphs), worth 4 marks.
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