Final answer:
The displacement of the airplane at At = 1.0 s is 20 m and at At = 2.0 s is 17.5 m. To find the displacement of the airplane from a velocity-time graph, one must calculate the area under the velocity-time graph for the given time intervals.
Step-by-step explanation:
To find the displacement of the airplane at At = 1.0 s and At = 2.0 s, we need to calculate the area under the velocity-time graph between these two time points. Since the graph is a straight line, the area can be calculated using the formula for the area of a trapezoid: A = 0.5 * (base1 + base2) * height.
- At At = 1.0 s, the displacement can be calculated as: A = 0.5 * (10 + 30) * 1 = 20 m
- At At = 2.0 s, the displacement can be calculated as: A = 0.5 * (30 + 5) * 1 = 17.5 m
Therefore, the displacement of the airplane at At = 1.0 s is 20 m and at At = 2.0 s is 17.5 m.
To find the displacement of the airplane from a velocity-time graph, one must calculate the area under the velocity-time graph for the given time intervals. This is because in physics, the displacement is represented by the area under such a graph when acceleration is constant.
Understanding Displacement from a Velocity-Time Graph
The question on how to find the displacement of the airplane at Δt = 1.0 s and at Δt = 2.0 s using a velocity-time graph can be answered by understanding the concept that the area under the graph represents displacement. To find the displacement at different intervals, one needs to calculate the area within those intervals. Since most velocity-time graphs provide straight lines, the calculations are relatively straightforward.
For instance, if the graph is a straight line, then for constant acceleration, the equation v = v₀ + at can be used to determine the velocity at any time. In the case of a linear relationship, the displacement can be found using the average velocity over the time interval in question. The displacement is basically the area under the velocity-time graph between the specified time intervals.