Problem 29 - Fundamentals of Physics

Question

A certain elevator cab has a total run of \(190 \mathrm{~m}\) and a maximum speed of \(305 \mathrm{~m} / \mathrm{min},\) and it accelerates from rest and then back to rest at \(1.22 \mathrm{~m} / \mathrm{s}^{2} .\) (a) How far docs the cab move whilc accelerating to full speed from rest? (b) How long does it take to make the nonstop \(190 \mathrm{~m}\) run, starting and ending at rest?

Step-by-Step Solution

Answer

(a) Acceleration distance is approximately 32.5 m. (b) Total time is about 62.7 seconds.

Step 1: Understand the Problem

The elevator needs to go through three phases: acceleration, constant speed, and deceleration. We need to calculate the distance covered during acceleration from rest to full speed (max speed) and also find the total time taken to cover the 190 m distance starting and ending at rest.

Step 2: Calculate Distance During Acceleration

Use the equation of motion \( v^2 = u^2 + 2as \), where \( v = 305 \text{ m/min} = \frac{305}{60} \text{ m/s} \), \( u = 0 \), and \( a = 1.22 \text{ m/s}^2 \). Solving for \( s \):\[\left(\frac{305}{60}\right)^2 = 0 + 2 \cdot 1.22 \cdot s\]\[s = \frac{\left(\frac{305}{60}\right)^2}{2 \cdot 1.22}\]Calculate \( s \) to find the distance during acceleration.

Step 3: Calculate Total Time During Acceleration

Use the formula \( v = u + at \) to find the time \( t \) it takes to accelerate to full speed from rest:\[\frac{305}{60} = 0 + 1.22t\]\[t = \frac{\frac{305}{60}}{1.22}\]Calculate \( t \) for the duration of acceleration.

Step 4: Find Constant Speed Time

After acceleration, use the constant speed until deceleration begins. First, find the distance covered at constant speed: total distance minus twice the acceleration distance (distance during acceleration and deceleration are the same). Then, use the formula \( t = \frac{d}{v} \) where \( d \) is the remaining distance and \( v = \frac{305}{60} \text{ m/s} \).

Step 5: Calculate the Full Journey Time

The total time is the sum of the acceleration time, constant speed time, and deceleration time (which equals the acceleration time). Add these times together to find the total time for the nonstop 190 m run.

Key Concepts

Equations of Motion

Equations of motion are vital tools in kinematics that help us analyze the various phases an object undergoes as it moves. In the case of the elevator cab, these equations allow us to quantify how far the cab travels while accelerating and also to determine the time taken for its complete journey. One key equation is the one relating final velocity, initial velocity, acceleration, and distance:

• \( v^2 = u^2 + 2as \)

where:

• \( v \) is the final velocity
• \( u \) is the initial velocity
• \( a \) is the acceleration
• \( s \) is the distance traveled

This equation helped us to find the distance during the acceleration phase. Likewise, to find out how long an object takes to reach a certain speed, another equation comes into play:

• \( v = u + at \)

This equation assists in computing how long the elevator takes to accelerate to its maximum speed. These foundational equations of motion are indispensable for solving kinematic problems involving constant acceleration.

Acceleration

Acceleration refers to the rate at which an object's velocity changes with time. In our elevator example, the cab starts from rest and speeds up to a certain velocity over time. The elevator accelerates at a rate of \(1.22 \, \mathrm{m/s^{2}}\), allowing us to determine various aspects of its motion. For the elevator scenario:

• The cab accelerates from 0 m/s, allowing us to calculate how far it travels while reaching its full speed.
• It requires the use of the equation \( v = u + at \) to find out the time it takes to achieve its maximum speed of \(305 \, \mathrm{m/min}\).

The knowledge of acceleration helps in ascertaining both the physical distance traveled during this phase and the time duration involved. Acceleration ensures the elevator smoothly progresses from a standstill to its cruising speed, making it an essential component of the motion equation.

Constant Speed

After the elevator cab accelerates to its maximum speed, it moves at constant speed until it's time to start decelerating. During this phase, the velocity of the elevator doesn't change, which is what we refer to as constant speed. Constant speed indicates:

• There is no additional acceleration happening to the elevator cab.
• The acceleration distance has already been covered, so only the remaining journey needs to be calculated at this constant speed.

To find the time for which the elevator moves at this constant speed, we first subtract the distances covered during acceleration and deceleration from the total distance. The remaining distance is then covered at this steady pace, which allows us to calculate the corresponding time using:

• \( t = \frac{d}{v} \)

This phase is crucial as it forms the backbone of the elevator's efficient and smooth journey over its full 190 m route.

Deceleration

Deceleration is essentially acceleration in reverse. It involves slowing down an object from its current speed to a state of rest. In the context of our elevator, after moving at constant speed, it needs to decelerate at the same rate it accelerated, \(1.22 \, \mathrm{m/s^{2}}\), to come to a halt smoothly. Deceleration details include:

• The distance covered while decelerating is equal to the distance covered during acceleration, due to the symmetrical nature of the motion.
• It uses the same mechanics as acceleration, but in reverse.

For the elevator to stop completely at the end of its journey, we need to compute the time it spends decelerating using the same formula as for acceleration. This helps ensure precise control over stopping distance and helps maintain a comfortable experience for passengers. Deceleration is pivotal to completing the elevator's journey safely and efficiently.