Chapter 3.1, Problem 4PP

3

To determine

The velocity-time graph of the motion of the elevator.

Explanation of Solution

Given Info:

The upward acceleration of the elevator is, $a_u=0.5\text{m}/{\text{s}}^2$ .

The time for the upward acceleration is, $t_1=2\text{s}$ .

The constant velocity of the elevator is, $v=1\text{m}/\text{s}$

The time for the constant velocity is, $t_2=12\text{s}$

The downward acceleration of the elevator is, $a_d=0.25\text{m}/{\text{s}}^2$

The time for the upward acceleration is, $t_3=4\text{s}$ .

Formula Used:

The acceleration describes the slope of the velocity-time graph.

Calculation:

The velocity-time graph is given in the figure (1) using the given information. Here the elevator starts at rest from the ground floor of a three-story shopping mall. It is accelerating for 2 second with acceleration of $0.5\text{m}/{\text{s}}^2$ , then continues with a constant velocity of $1\text{m}/\text{s}$ for 12 seconds and then slows down with contact acceleration of $0.25\text{m}/{\text{s}}^2$ and then reached at the third floor.

Figure (1)