Here is a speed-time graph for a train journey between two stations - Edexcel - GCSE Maths - Question 22 - 2019 - Paper 2

To find the half distance traveled in the first 100 seconds, we first calculate the area under the graph up to 100 seconds. The area represents the distance traveled.

1. Calculate the area under the graph:

   • From 0 to 30 seconds, the speed is 10 m/s.

   • From 30 to 60 seconds, the speed is 15 m/s.

   • From 60 to 100 seconds, the speed decreases back to 0 m/s.

   • Area from 0 to 30s:

   $ext{Area}_1 = ext{Base} \times ext{Height} = 30 \times 10 = 300 ext{ m}$

   • Area from 30 to 60s (triangle):

   $ext{Area}_2 = \frac{1}{2} \times ext{Base} \times ext{Height} = \frac{1}{2} \times 30 \times (15 - 10) = 75 ext{ m}$

   • Area from 60 to 100s:

   $\text{Area}_3 = \frac{1}{2} \times (100-60) \times (15 - 0) = \frac{1}{2} \times 40 \times 15 = 300 ext{ m}$

   • Total Distance: $ext{Total} = ext{Area}_1 + ext{Area}_2 + ext{Area}_3 = 300 + 75 + 300 = 675 ext{ m}$

2. Calculate distance for half the distance:

   • Half of 675 m is: $\frac{675}{2} = 337.5 ext{ m}$

3. Find the time taken to travel this half distance:

   • From 0 to 30s, the first segment covers 300 m, so we still need: $337.5 - 300 = 37.5 ext{ m}$
   • After 30 seconds, the speed is 15 m/s: $\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{37.5}{15} = 2.5 ext{ seconds}$

4. Total time for half the distance: $30 + 2.5 = 32.5 ext{ seconds}$