To find the displacement of the airplane at St=1s and St=2s based on the velocity-time graph, calculate the areas under the segments of the graph within those time intervals.
Step-by-step explanation:
To determine the displacement of the airplane at St=1s and St=2s based on the given velocity-time graph, we need to find the area under the velocity-time graph within those time intervals. Since the graph consists of straight line segments, we can find the area of each segment and sum them up.
For St=1s, the area of the triangle formed by the segment is 0.5 * (20+25) * 1 = 22.5 square units of velocity-time. Since displacement is the area under the velocity-time graph, the displacement at St=1s is also 22.5 units (taking direction into account).
For St=2s, the triangle and trapezoid formed by the segments give us areas of 0.5 * (25+20) * 1 + 0.5 * (20+15) * 1 = 37.5 + 17.5 = 55 square units of velocity-time. Therefore, the displacement at St=2s is 55 units (taking direction into account).