A motorised scooter is travelling along a straight path with velocity $v$ m s$^{-1}$ over time $t$ seconds as shown by the following graph - AQA - A-Level Maths Pure - Question 14 - 2021 - Paper 2

We will split the area into four trapezoids:

1. For the interval $[12, 20]$ seconds:

   • Height at $t = 12$: $5$ m/s
   • Height at $t = 20$: $8$ m/s
   • Area: $ext{Trapezium 1} = \frac{1}{2} \times (5 + 8) \times (20 - 12) = \frac{1}{2} \times 13 \times 8 = 52$

2. For the interval $[20, 30]$ seconds:

   • Height at $t = 20$: $8$ m/s
   • Height at $t = 30$: $6$ m/s
   • Area: $\text{Trapezium 2} = \frac{1}{2} \times (8 + 6) \times (30 - 20) = \frac{1}{2} \times 14 \times 10 = 70$

3. For the interval $[30, 36]$ seconds:

   • Height at $t = 30$: $6$ m/s
   • Height at $t = 36$: $3.8$ m/s
   • Area: $\text{Trapezium 3} = \frac{1}{2} \times (6 + 3.8) \times (36 - 30) = \frac{1}{2} \times 9.8 \times 6 = 29.4$

4. For the interval $[36, 45]$ seconds (although not needed for this calculation, we can calculate for verification):

   • Height at $t = 36$: $3.8$ m/s
   • Height at $t = 45$: $0$ m/s
   • Area: $\text{Trapezium 4} = \frac{1}{2} \times (3.8 + 0) \times (45 - 36) = \frac{1}{2} \times 3.8 \times 9 = 17.1$

Now, summing the areas gives:

$\text{Total Area} = 52 + 70 + 29.4 = 151.4 \text{ m}$

This total indicates that the scooter travels approximately 151.4 metres between $t = 12$ and $t = 36$ seconds.

Comparing this total area with Noosha's estimate of approximately 130 metres, we conclude:

Since $151.4$ m is greater than $130$ m, Noosha's estimate is not reasonable.