1. **State the problem:** We have a speed-time graph of a train journey with the following key points: - Speed rises linearly from 0 m/s at 0 s to 9 m/s at 40 s. - Speed remains constant at 9 m/s from 40 s to 100 s. - Speed decreases linearly from 9 m/s at 100 s to 0 m/s at 120 s. We need to find: - The total time taken for the journey in minutes. - The acceleration during the first 40 seconds. - The distance traveled between 40 s and 100 s. - The total distance between the two stations. 2. **Time taken for the journey:** The journey starts at 0 s and ends at 120 s. Convert seconds to minutes: $$\text{Time in minutes} = \frac{120}{60} = 2$$ 3. **Calculate acceleration in the first 40 seconds:** Acceleration is the rate of change of speed: $$a = \frac{\Delta v}{\Delta t} = \frac{9 - 0}{40 - 0} = \frac{9}{40} = 0.225\ \text{m/s}^2$$ 4. **Calculate distance traveled between 40 s and 100 s:** Speed is constant at 9 m/s during this interval. Time interval: $$100 - 40 = 60\ \text{seconds}$$ Distance: $$d = v \times t = 9 \times 60 = 540\ \text{meters}$$ 5. **Calculate total distance between the two stations:** Distance is the area under the speed-time graph. - From 0 to 40 s: speed increases linearly from 0 to 9 m/s, so area is a triangle: $$d_1 = \frac{1}{2} \times 40 \times 9 = 180\ \text{meters}$$ - From 40 to 100 s: constant speed segment: $$d_2 = 540\ \text{meters}$$ - From 100 to 120 s: speed decreases linearly from 9 to 0 m/s, another triangle: $$d_3 = \frac{1}{2} \times 20 \times 9 = 90\ \text{meters}$$ Total distance: $$d = d_1 + d_2 + d_3 = 180 + 540 + 90 = 810\ \text{meters}$$