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 covered by the lift, we need to calculate the area under the velocity/time graph. The area can be divided into rectangles and triangles.

The graph shows three distinct segments:

1. From 0 to 4 seconds (constant velocity of 4 m/s):

   • Area = base × height = 4 seconds × 4 m/s = 16 m

2. From 4 to 10 seconds (constant velocity of 0 m/s):

   • Area = base × height = 6 seconds × 0 m/s = 0 m

3. From 10 to 14 seconds (constant velocity of 4 m/s):

   • Area = base × height = 4 seconds × 4 m/s = 16 m

4. From 14 to 18 seconds (constant velocity of 0 m/s):

   • Area = base × height = 4 seconds × 0 m/s = 0 m

Adding these areas together gives:

$ext{Total distance} = 16 ext{ m} + 0 ext{ m} + 16 ext{ m} + 0 ext{ m} = 32 ext{ m}$