### Elevator Dynamics and Apparent Weight Analysis#### Problem Statement:An elevator starts from the third floor and moves downward. The velocity-time (v-t) graph of a person riding the elevator is given below:\[ t_1 = 0.9 \, \text{s}, \, t_2 = 1.9 \, \text{s}, \, t_3 = 4.5 \, \text{s}, \, t_4 = 7.7 \, \text{s}, \, v = -1.0 \, \text{m/s} \]The v-t graph shows the following details:- The elevator accelerates from rest and reaches a constant velocity at \( t_1 \).- It then moves at a constant downward velocity until \( t_3 \).- At \( t_3 \), it decelerates and comes to rest at \( t_4 \).#### Questions:(a) **Draw Free Body Diagrams (FBD) for the person riding the elevator for the following cases:**- Speeding- Constant Moving- Slowing(b) **What is the apparent weight (in N) of a 50.0 kg person when he/she is:**- Speeding- Constant moving- Slowing (Use \( g = 10 \, \text{N/kg} \)).#### Graph Explanation:The graph provided shows time (t) on the x-axis and velocity (v) on the y-axis. The changes in the slope of the graph indicate periods of acceleration and deceleration:1. **Interval \( t_0 \) to \( t_1 \):** Positive slope indicates acceleration.2. **Interval \( t_1 \) to \( t_3 \):** Zero slope indicates constant velocity.3. **Interval \( t_3 \) to \( t_4 \):** Negative slope indicates deceleration.#### Solution:(a) **Free Body Diagrams (FBD):**- **Speeding:** In this phase, the elevator is accelerating downward. The FBD will have: - Gravitational force (\(mg\)) acting downward. - Normal force (\(N\)) acting upward. - Since the elevator is accelerating downward, \(N < mg\).- **Constant Moving:** Here, the elevator is moving at a constant velocity. The FBD will have

We will use Newton's equation of motion to answer the questions. The detailed steps are as follows.

Answered: 3. An elevator starts from the third…

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