A train moves from one station to another in two hours time. Its speed during the motion is shown in the graph Calculate (i) Maximum acceleration during the journey.
(ii) Distance covered during the time interval from 0.75 hour to 1 hour

Explanation

To solve the problem, we will break it down into two parts: calculating the maximum acceleration during the journey and finding the distance covered during the time interval from 0.75 hours to 1 hour.

### Part (i): Maximum Acceleration

1. **Understanding the VT Graph**: The velocity-time (VT) graph shows the speed of the train at different times. The slope of this graph represents acceleration.

2. **Identifying Regions**: The graph can be divided into three regions based on the changes in speed:
- Region 1: From 0 to 0.5 hours (speed is constant)
- Region 2: From 0.5 to 0.75 hours (speed increases linearly)
- Region 3: From 0.75 to 2 hours (speed decreases linearly)

3. **Calculating Acceleration for Each Region**:
- **Region 1 (0 to 0.5 hours)**: The speed is constant, so acceleration \( a_1 = 0 \).
- **Region 2 (0.5 to 0.75 hours)**: The speed increases from 20 km/h to 60 km/h over 0.25 hours.
- Height = Change in speed = \( 60 - 20 = 40 \) km/h
- Base = Time interval = \( 0.75 - 0.5 = 0.25 \) hours
- Acceleration \( a_2 = \frac{\text{Height}}{\text{Base}} = \frac{40}{0.25} = 160 \) km/h².
- **Region 3 (0.75 to 2 hours)**: The speed decreases from 60 km/h to 20 km/h over 1 hour.
- Height = Change in speed = \( 60 - 20 = 40 \) km/h
- Base = Time interval = \( 1 \) hour
- Acceleration \( a_3 = \frac{\text{Height}}{\text{Base}} = \frac{40}{1} = 40 \) km/h² (negative since it's deceleration).

4. **Finding Maximum Acceleration**: The maximum acceleration is the highest value among \( a_1, a_2, \) and \( a_3 \).
- \( a_1 = 0 \) km/h²
- \( a_2 = 160 \) km/h²
- \( a_3 = 40 \) km/h²
- Therefore, the maximum acceleration is \( a_2 = 160 \) km/h².

### Part (ii): Distance Covered from 0.75 to 1 Hour

1. **Identifying the Area Under the VT Graph**: The distance covered during a time interval can be found by calculating the area under the VT graph for that interval.

2. **Finding the Area of the Trapezium**: The area between 0.75 hours and 1 hour forms a trapezium with:
- One parallel side (speed at 0.75 hours) = 60 km/h
- Other parallel side (speed at 1 hour) = 20 km/h
- Height (time interval) = \( 1 - 0.75 = 0.25 \) hours

3. **Calculating the Area**:
- Area of trapezium = \( \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height} \)
- Area = \( \frac{1}{2} \times (60 + 20) \times 0.25 \)
- Area = \( \frac{1}{2} \times 80 \times 0.25 = 10 \) km.

4. **Final Result**: The distance covered during the time interval from 0.75 hours to 1 hour is 10 km.

### Summary of Answers:
(i) Maximum acceleration during the journey is **160 km/h²**.
(ii) Distance covered during the time interval from 0.75 to 1 hour is **10 km**.