Question

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, maintains that velocity for, then decelerates foruntil 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.

Short answer

Acceleration is obtained by the slope of the velocity vs. time graph.

Step-by-step solution

Reading graph

From the above graph and the data given, it is clear that the velocity increases from time t = 0 s to t = 3 s.

Figure: 2.58

From time t = 3 s to time t = 18 s, the velocity of the body is not at all changing.

Hence the velocity is constant, so the body is undergoing uniform motion.

From t = 18 s to t = 23 s, the velocity of the body decreases.

Position versus Time graph

Hence, in the initial case, the velocity increases rapidly up to $\mathbf{3}\mathbf{seconds}$, then the velocity is constant.

So the position versus time graph will be as below:

Figure: Position versus time

Hence the graph of position versus time will be similar to this.

Acceleration versus Time graph

Here the person is traveling with constant velocity from 3 to 18 s. hence there will not be a change in the velocity.

If there is no change in velocity, then acceleration will be zero.

The person's velocity is increasing from 0 to $\mathbf{3}\mathbf{}\mathbf{seconds}$. Hence there will be a constant acceleration.

In the other case, 18 to 23 s the person speed is decreasing hence there will be retardation.

Figure: Acceleration versus time

Acceleration is obtained by the slope of the velocity vs. time graph.