A hotel elevator ascends 200m with maximum speed of 5m/s. its acceleration and deceleration both have a magnitude of 1.0m/s2. part a how far does the elevator move while accelerating to full speed from rest?

Answer :

In this case, the elevator acceleration is 1m/s2 and the maximum speed is 5m. You asked how far does the elevator move to accelerate until maximal speed. First, we need to calculate the time needed to accelerate until maximal speed. The calculation would be:

time = speed / acceleration= (5m/s) / (1m/s2)= 5s

Then to calculate the distance, the formula would be:
distance= (initial speed + final speed)/2 * time= (1+5)/2 * 5 = 15m

The distance covered by the elevator as it ascends from rest to its maximum speed is [tex]\boxed{12.5\text{ m}}[/tex].

Explanation:

Given:

The maximum speed of the hotel elevator is [tex]5\text{ m/s}[/tex].

The acceleration of the elevator is [tex]1\text{ m/s}^2[/tex].

Concept:

The elevator starts from rest and its speed increases as it moves up under the acceleration. The acceleration of the body is defined as the rate of change of velocity of the body.

The motion of the elevator under the acceleration is given by the third equation of motion.

[tex]\boxed{v^2_{f}={v^2_{i}}+2aS}[/tex]

Since the elevator starts from rest, the initial velocity of the elevator is zero and the final velocity of the elevator will be [tex]5\text{ m/s}[/tex].

Substitute the value of initial velocity, final velocity and acceleration in above expression.

[tex]\begin{aligned}(5)^2&=(0)^2+2(1)S\\S&=\dfrac{25}{2}\text{ m}\\&=12.5\text{ m}\end{aligned}[/tex]

Thus, The distance covered by the elevator as it ascends from rest to its maximum speed is [tex]\boxed{12.5\text{ m}}[/tex].

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Laws of motion

Keywords:

elevator, ascends, acceleration, deceleration, initial velocity, final velocity, magnitude, maximum speed, distance.

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