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

To find the distance travelled by the scooter in the time interval from $t=12$ to $t=36$, we will calculate the area under the velocity curve using trapezoidal approximation. The area will be calculated as the sum of four trapezoids:

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

   • Height 1 = $5.2$
   • Height 2 = $6.2$
   • Base length = $20 - 12 = 8$
   • Area = rac{1}{2} (5.2 + 6.2) \times 8 = 44.8

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

   • Height 1 = $6.2$
   • Height 2 = $8.0$
   • Base length = $30 - 20 = 10$
   • Area = rac{1}{2} (6.2 + 8.0) \times 10 = 71.0

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

   • Height 1 = $8.0$
   • Height 2 = $6.0$
   • Base length = $36 - 30 = 6$
   • Area = rac{1}{2} (8.0 + 6.0) \times 6 = 42.0

4. Trapezium 4: For the interval $[36, 50]$ is not included in this interval.

Thus, the total area can be summed up as: $ext{Total Area} = 44.8 + 71.0 + 42.0 = 157.8 \text{ m}$