A motorised scooter is travelling along a straight path with velocity $v$ m/s over time $t$ seconds as shown by the following graph - AQA - A-Level Maths Mechanics - Question 14 - 2021 - Paper 2

The coordinates at $t=12$ are $(12, 5.8)$ and at $t=20$ are $(20, 5.2)$.

Area of trapezium: $A_1 = \frac{1}{2} \times (b_1 + b_2) \times h = \frac{1}{2} \times (5.8 + 5.2) \times (20 - 12) = \frac{1}{2} \times 11 \times 8 = 44$

The coordinates at $t=20$ are $(20, 5.2)$ and at $t=30$ are $(30, 6.2)$.

Area of trapezium: $A_2 = \frac{1}{2} \times (5.2 + 6.2) \times (30 - 20) = \frac{1}{2} \times 11.4 \times 10 = 57$

The coordinates at $t=30$ are $(30, 6.2)$ and at $t=36$ are $(36, 6.0)$.

Area of trapezium: $A_3 = \frac{1}{2} \times (6.2 + 6.0) \times (36 - 30) = \frac{1}{2} \times 12.2 \times 6 = 36.6$

The coordinates at $t=36$ are $(36, 6.0)$ and at $t=41$ are $(41, 6.0)$.

Area of trapezium: $A_4 = \frac{1}{2} \times (6.0 + 6.0) \times (41 - 36) = \frac{1}{2} \times 12 \times 5 = 30$

Now we sum the areas of all trapeziums: $\text{Total Area} = A_1 + A_2 + A_3 + A_4 = 44 + 57 + 36.6 + 30 = 167.6$

Noosha estimated that the scooter travelled approximately 130 metres. The calculated total area (distance) of 167.6 metres indicates that Noosha's estimate is incorrect.