The object's displacement can be calculated by finding the area under the velocity-time graph. For the time interval 0≤t≤2s, the object's displacement is 34 m. For the time interval 2≤t≤4s, the object's displacement is 46 m.
The object moves along a line at a velocity of 17 m/s for the time interval 0 ≤ t ≤ 2 s, and then at a velocity of 23 m/s for the time interval 2 ≤ t ≤ 4 s. Since the motion is along a straight line, we can determine the change in position by calculating the area under the velocity-time graph.
For the time interval 0 ≤ t ≤ 2 s, the area under the graph represents the displacement. The area is a rectangle with a base of 2 s and a height of 17 m/s, so the displacement during this interval is (2 s) * (17 m/s) = 34 m.
For the time interval 2 ≤ t ≤ 4 s, we have a different velocity. The area under the graph can be divided into two parts: a rectangle with a base of 2 s and a height of 23 m/s, and a triangle with a base of 2 s and a height of (23 m/s - 17 m/s). The displacement during this interval is (2 s) * (23 m/s) + (0.5 * 2 s) * (23 m/s - 17 m/s) = 46 m.