Figure 8 is a velocity/time graph for a lift moving upwards in a tall building. - Edexcel - GCSE Physics - Question 5 - 2023 - Paper 1

To find the distance travelled by the lift, we need to calculate the area under the velocity/time graph. The graph consists of three sections:

1. From time 0 to 4 seconds: This is a trapezium with a height of 4 m/s and a width of 4 s. The area can be calculated as: $Area_1 = \frac{1}{2} \times (4 + 4) \times 4 = 32 \, m$

2. From time 4 to 12 seconds: This is a rectangle with a height of 4 m/s and a width of 8 s. The area is: $Area_2 = 4 \times 8 = 32 \, m$

3. From time 12 to 18 seconds: This is a trapezium again with a height of 4 m/s and width of 6 s: $Area_3 = \frac{1}{2} \times (4 + 4) \times 6 = 24 \, m$

Now, summing these areas gives: $Total \ Distance = Area_1 + Area_2 + Area_3 = 32 + 32 + 24 = 88 \, m$

Thus, the total distance travelled by the lift is 88 m. However, since the question requires us to round to an approximate value, any values that round to 19 m (such as 18.9 m) would also be acceptable for full marks.