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The main elevator \(A\) of the CN Tower in Toronto rises about \(350 \mathrm{m}\) and for most of its run has a constant speed of \(22 \mathrm{km} / \mathrm{h}\). Assume that both the acceleration and deceleration have a constant magnitude of \(\frac{1}{4} g\) and determine the time duration \(t\) of the elevator run.

Short Answer

Expert verified
The elevator run time is approximately 57.33 seconds.

Step by step solution

01

Convert Speeds to Consistent Units

First, we need to convert the speed from km/h to m/s for consistency. We know that 1 km/h = \(\frac{1000}{3600}\) m/s. Thus, 22 km/h is converted as follows: \[ 22 \times \frac{1000}{3600} = 6.11 \text{ m/s} \].
02

Calculate Acceleration and Deceleration

The problem states that the acceleration and deceleration of the elevator are \(\frac{1}{4}g\), where \(g\) is approximately \(9.81 \text{ m/s}^2\). Hence, the acceleration and deceleration will be \[ a = \frac{1}{4} \times 9.81 = 2.4525 \text{ m/s}^2 \].
03

Determine Time for Acceleration Phase

Assuming the elevator accelerates until it reaches its constant speed, we use the equation \(v = at\), where \(v = 6.11 \text{ m/s}\) is the top speed. Solving for time \(t_1\), we have: \[ t_1 = \frac{v}{a} = \frac{6.11}{2.4525} \approx 2.49 \text{ seconds} \].
04

Calculate Distance During Acceleration Phase

We use the formula for distance traveled during acceleration: \(d = \frac{1}{2} a t_1^2\). Substituting the values, we find: \[ d_1 = \frac{1}{2} \times 2.4525 \times (2.49)^2 \approx 15.11 \text{ meters} \].
05

Determine Time for Deceleration Phase

The time for deceleration will be the same as for acceleration because the acceleration magnitude is the same. Therefore, \(t_3 = t_1 = 2.49\) seconds.
06

Calculate Distance During Deceleration Phase

Similar to acceleration, the distance during deceleration \(d_3\) is also same: \[ d_3 = 15.11 \text{ meters} \].
07

Calculate Distance Traveled at Constant Speed

Subtract the combined distances during acceleration and deceleration from the total distance (350 m) to find the distance traveled at constant speed: \[ d_2 = 350 - (d_1 + d_3) = 350 - 30.22 = 319.78 \text{ meters} \].
08

Determine Time at Constant Speed

Use \(d_2 = v t_2\) to find the time \(t_2\) during the constant speed phase: \[ t_2 = \frac{d_2}{v} = \frac{319.78}{6.11} \approx 52.35 \text{ seconds} \].
09

Calculate Total Time

The total time \(t\) is the sum of the times for acceleration, constant speed, and deceleration: \[ t = t_1 + t_2 + t_3 = 2.49 + 52.35 + 2.49 \approx 57.33 \text{ seconds} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematics
Kinematics is a branch of physics that describes the motion of objects without considering the forces causing the motion. In this exercise, we analyze the movement of an elevator following specific parameters like distance and time. Kinematics focuses on quantities such as velocity, displacement, acceleration, and time.
  • Velocity is the speed of an object in a specific direction.
  • Displacement refers to the change in position of the object.
  • Acceleration is the rate at which velocity changes over time.
For the elevator, we calculated its point-to-point travel characteristics, isolating stages like acceleration, constant speed, and deceleration. Understanding how these fundamental principles interact helps clarify the elevator's journey from start to finish.
Uniform Acceleration
Uniform acceleration occurs when an object's velocity changes at a constant rate. In this problem, the elevator experiences uniform acceleration and deceleration, both equal to \( \frac{1}{4}g \). This constant acceleration is crucial because it allows for the straightforward calculation of time and distance during these phases of motion.
When dealing with uniformly accelerated motion, you use the equations of motion, such as:
\[ v = at, \] \[ d = \frac{1}{2}at^2. \] Applying these to the elevator's motion helps determine how long each phase lasts and the distance covered. This simplification provides a precise way to predict and analyze object movement, illustrating the elegance of physics in practical scenarios.
Unit Conversion
In physics, having consistent units throughout calculations is essential to ensure accuracy. In this problem, the elevator's speed is initially given in kilometers per hour (km/h). To make it compatible with meters and seconds used in other parts of the problem, the speed needs converting to meters per second (m/s).
  • To convert km/h to m/s, use the conversion factor: \[ 1 \text{ km/h} = \frac{1000}{3600} \text{ m/s}. \]
  • Thus, converting 22 km/h to m/s gives: \[ 22 \times \frac{1000}{3600} \approx 6.11 \text{ m/s}. \]
These conversions are simple but vital, preventing potential errors and ensuring all calculations are notably precise.
Calculating Time
Calculating the time of an event is a fundamental part of analyzing motion in physics. In cases of constant acceleration, time for each phase of motion can be derived from the equations of kinematics.
Here, time calculations involved:
  • Time for acceleration \( (t_1) \) using \( v = at \).
  • Time at constant speed \( (t_2) \) using the distance and constant speed formula \( d = vt \).
  • Time for deceleration mirrors the acceleration time due to symmetric acceleration/deceleration phases.
Adding these times gives the total duration of the elevator's run. These calculations demonstrate breaking down complex motions into simpler, manageable parts for analysis.
Elevator Motion
The motion of an elevator is fascinating as it incorporates multiple physics principles within a constrained space. Here, our focus was the main elevator of the CN Tower, which entails a full run of motion from rest, accelerating to constant velocity, and slowing down to a stop.
Elevator motion encapsulates:
  • Initial acceleration to reach cruising speed.
  • Constant speed for the majority of the travel distance.
  • Deceleration phase for a smooth stop.
Understanding elevator motion through this example encourages deeper insights into how physics applies to everyday mechanical systems, making complex concepts approachable and actionable.

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