Final answer:
The elevator moves 1.35 meters while accelerating to full speed from rest utilizing the kinematic equation for acceleration s = ut + 0.5at², with an initial velocity of 0 m/s, acceleration of 1.20 m/s², and time of 1.50 s.
Step-by-step explanation:
To determine how far the elevator moves while accelerating to full speed from rest, we can apply the kinematic equations for uniformly accelerated motion.
The question falls into Physics as it relates to mechanics and specifically the topic of motion under constant acceleration. We're given that an elevator of mass 1700 kg accelerates upward from rest at a rate of 1.20 m/s² for a duration of 1.50 s.
First, we can calculate the distance using the formula:
Where:
- s is the displacement (distance moved)
- u is the initial velocity (0 m/s since the elevator starts from rest)
- a is the acceleration (1.20 m/s²)
- t is the time (1.50 s)
Plugging in the values, we get:
- s = 0 * 1.50 + 0.5 * 1.20 * (1.50)²
- s = 0 + 0.5 * 1.20 * 2.25
- s = 1.35 m
Therefore, the elevator moves 1.35 meters while accelerating to full speed from rest.