To solve this problem, we need to calculate the distance the train traveled at a velocity greater than 20 m/s. Here's how we can break down the steps:

1. Analyze the graph:

• The graph shows velocity (m/s) on the y-axis and time (s) on the x-axis.
• We need to find the time intervals during which the train's velocity was greater than 20 m/s.

Looking at the graph:

• The train’s velocity increases from 0 to 60 m/s over the first 60 seconds.
• After 60 seconds, the velocity decreases back to 0 m/s by 180 seconds.

2. Identify the time period where velocity > 20 m/s:

• From the graph, it appears the train reaches 20 m/s after approximately 20 seconds.
• The velocity drops back below 20 m/s after about 120 seconds (from the 60-second peak to the end of the journey).

Thus, the time intervals with velocity greater than 20 m/s are from:

• 20 seconds to 120 seconds (as the velocity exceeds 20 m/s in that range).

3. Calculate the distance traveled:

The area under the velocity-time graph represents the distance traveled. For the region where velocity is greater than 20 m/s, we need to calculate the area under the curve from 20 seconds to 120 seconds.

This region forms two distinct shapes:

• Trapezoid 1: From 20 to 60 seconds (increasing from 20 m/s to 60 m/s)
• Trapezoid 2: From 60 to 120 seconds (decreasing from 60 m/s to 20 m/s)

Trapezoid 1 (20 to 60 seconds):

• Height 1 = 20 m/s
• Height 2 = 60 m/s
• Base = 60 - 20 = 40 seconds

Using the trapezoid area formula:
$A = \frac{1}{2} \times (h_1 + h_2) \times \text{base}$
$A_1 = \frac{1}{2} \times (20 + 60) \times 40 = \frac{1}{2} \times 80 \times 40 = 1600 \, \text{m}$

Trapezoid 2 (60 to 120 seconds):

• Height 1 = 60 m/s
• Height 2 = 20 m/s
• Base = 120 - 60 = 60 seconds

$A_2 = \frac{1}{2} \times (60 + 20) \times 60 = \frac{1}{2} \times 80 \times 60 = 2400 \, \text{m}$

4. Total Distance:

Adding the areas of the two trapezoids gives the total distance: $\text{Total Distance} = 1600 + 2400 = 4000 \, \text{m}$

Thus, the train traveled 4000 meters at a velocity greater than 20 m/s.

Would you like further clarification or more details on the steps? Here are some follow-up questions you could explore:

1. What if the train's velocity peaked at 70 m/s instead? How would that change the calculation?
2. Can you find the total distance traveled by the train over the entire journey?
3. What is the average velocity of the train when it is moving faster than 20 m/s?
4. How would the graph change if the train's deceleration was faster after the peak?
5. Could this method be applied to a different type of velocity-time graph, such as a non-linear one?

Tip: When dealing with velocity-time graphs, always remember that the area under the curve represents distance traveled.