AQA Paper 2 2021 June — Question 14 4 marks

Exam BoardAQA
ModulePaper 2 (Paper 2)
Year2021
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeDistance from velocity-time graph
DifficultyModerate -0.8 This is a straightforward application of finding distance from a velocity-time graph by calculating the area under the curve. Students need to identify the relevant region (12 ≤ t ≤ 36) and use basic area formulas (likely trapezium rule or counting squares). The question requires minimal problem-solving—just direct application of a standard technique with simple arithmetic to verify a given statement.
Spec3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area

14 A motorised scooter is travelling along a straight path with velocity \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) over time \(t\) seconds as shown by the following graph. \includegraphics[max width=\textwidth, alt={}, center]{b7df05bf-f3fc-4705-a13c-6b562896fa9f-20_1120_1134_420_452} Noosha says that, in the period \(\mathbf { 1 2 } \leq \boldsymbol { t } \leq \mathbf { 3 6 }\), the scooter travels approximately 130 metres. Determine if Noosha is correct, showing clearly any calculations you have used.

Question 14:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Distance = area below the curve; split into four trapeziaM1 (AO3.1b) Interprets problem as area beneath curve
Trapezium \(1 = \frac{6}{2}(5.8+5.2) = 33\); Trapezium \(2 = \frac{7}{2}(5.2+6.2) = 39.9\)A1 (AO1.1b) Divides into appropriate areas; at least three polygons; at least one correct area
Trapezium \(3 = \frac{5}{2}(6.2+6) = 30.5\); Trapezium \(4 = \frac{6}{2}(6+3.8) = 30.5\); Total \(= 132.8 \approx 130\)M1 (AO1.1a) Obtains total area for interval \(12 < t < 36\); AWFW 125–135
Noosha's estimate was reasonableR1 (AO1.1b) Compares total area with Noosha's estimate to conclude result was reasonable
## Question 14:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance = area below the curve; split into four trapezia | M1 (AO3.1b) | Interprets problem as area beneath curve |
| Trapezium $1 = \frac{6}{2}(5.8+5.2) = 33$; Trapezium $2 = \frac{7}{2}(5.2+6.2) = 39.9$ | A1 (AO1.1b) | Divides into appropriate areas; at least three polygons; at least one correct area |
| Trapezium $3 = \frac{5}{2}(6.2+6) = 30.5$; Trapezium $4 = \frac{6}{2}(6+3.8) = 30.5$; Total $= 132.8 \approx 130$ | M1 (AO1.1a) | Obtains total area for interval $12 < t < 36$; AWFW 125–135 |
| Noosha's estimate was reasonable | R1 (AO1.1b) | Compares total area with Noosha's estimate to conclude result was reasonable |

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14 A motorised scooter is travelling along a straight path with velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ over time $t$ seconds as shown by the following graph.\\
\includegraphics[max width=\textwidth, alt={}, center]{b7df05bf-f3fc-4705-a13c-6b562896fa9f-20_1120_1134_420_452}

Noosha says that, in the period $\mathbf { 1 2 } \leq \boldsymbol { t } \leq \mathbf { 3 6 }$, the scooter travels approximately 130 metres.

Determine if Noosha is correct, showing clearly any calculations you have used.\\

\hfill \mbox{\textit{AQA Paper 2 2021 Q14 [4]}}