Consider the velocity vs. time graph of a person in an elevator shown in Figure 2.58. Suppose the elevator is initially at rest. It then accelerates for 3 seconds, maintains that velocity for 15 seconds, then decelerates for 5 seconds until it stops. The acceleration for the entire trip is not constant so we cannot use the equations of motion from Motion Equations for Constant Acceleration in One Dimension for the complete trip. (We could, however, use them in the three individual sections where acceleration is a constant.) Sketch graphs of (a) position vs. time and (b) acceleration vs. time for this trip.

Step-by-step solution 30CQ Step 1 of 4 (a) Consider the figure 2.58 provided in the textbook. From the figure the velocity of a person is increases with respect to time for 3 second, so the position graph for that will be a curve with increasing slope. After 3 second the velocity is constant for next 15 second, so the position graph will have straight line for that portion, then the velocity decrease for next 5 sec until velocity is zero, so the graph for next 5 sec will be a curve with decreasing slope. Step 2 of 4 The graph for position verses time is as shown below: