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Chapter 2: Problem 38

If the maximum acceleration that is tolerable for passengers in a subway train is \(1.34 \mathrm{~m} / \mathrm{s}^{2}\) and subway stations are located \(806 \mathrm{~m}\) apart, what is the maximum speed a subway train can attain between stations? (b) What is the travel time between stations? (c) If a subway train stops for \(20 \mathrm{~s}\) at each station, what is the maximum average speed of the train, from one start-up to the next? (d) Graph \(x, v\), and \(a\) versus \(t\) for the interval from one start-up to the next.

Short Answer

Expert verified
Max speed is 32.7 m/s; travel time is 48.8 s; avg speed is 11.72 m/s.

Step by step solution

01

Understand the Problem

We need to find the maximum speed a subway train can achieve given a maximum acceleration and a distance between stations. Additionally, we need to determine the travel time between stations, the average speed considering stop time, and plot graphs for position, velocity, and acceleration.
02

Calculate Maximum Speed

Using the equation for motion \[ v^2 = u^2 + 2as \]where \(u = 0\) (starting from rest), \(a = 1.34 \, \text{m/s}^2\), and \(s = 403\, \text{m}\) (since acceleration occurs over half the journey), we calculate:\[ v^2 = 0 + 2 \times 1.34 \times 403 \]Solving for \(v\):\[ v = \sqrt{2 \times 1.34 \times 403} \approx 32.7 \, \text{m/s} \]
03

Calculate Travel Time

The train accelerates and then decelerates over half the distance each. Using \(v = u + at\), where \(v = 32.7\, \text{m/s}\) and \(a = 1.34 \, \text{m/s}^2\), calculate time to accelerate or decelerate:\[ t = \frac{v}{a} = \frac{32.7}{1.34} \approx 24.4 \, \text{s} \]For acceleration and deceleration, the time is \(2 \times 24.4 = 48.8\, \text{s}\).
04

Consider Station Stop Time

Station stop time is an additional 20 seconds. Add this to the travel time:\[ 48.8 + 20 = 68.8 \, \text{s} \]
05

Calculate Average Speed

The distance between stations is 806 m. The total time including stop is 68.8 s.Average speed is calculated as:\[ \text{Average speed} = \frac{\text{total distance}}{\text{total time}} = \frac{806}{68.8} \approx 11.72 \, \text{m/s} \]
06

Graphing x, v, and a versus t

For graphing purposes:- \(x(t)\): Starts at zero, follows a parabola during acceleration and deceleration, peaking at mid-point.- \(v(t)\): Rises linearly to peak speed of 32.7 m/s halfway, then symmetrically drops to zero.- \(a(t)\): Constant at 1.34 m/s虏 for first half, then -1.34 m/s虏 for the second half. Include a flat portion at zero for stop time.

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Key Concepts

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

Kinematics
In the realm of physics, kinematics is the branch that delves into the motion of objects. This field is crucial for understanding the paths that objects take, how fast they travel, and how their speeds change over time.
Kinematics equations describe the relationships between distance, velocity, acceleration, and time, and do not require the analysis of forces.

Key points of kinematics include:
  • Describing motion: using position, velocity, acceleration over time.
  • Ignoring forces: focus solely on the movement itself.
  • Utilizing equations: aid in predicting future states of moving objects.
Understanding kinematics helps us solve problems like the one in the exercise, where we are concerned primarily with the subway train's motion without needing to consider the forces involved.
Acceleration
Acceleration is a fundamental concept when analyzing motion. It refers to any change in the velocity of an object over time. When a train or any vehicle increases or decreases its speed, it is undergoing acceleration.
Acceleration can be measured as either positive or negative, indicating speeding up or slowing down, respectively.

Key characteristics of acceleration include:
  • Defined as the rate of change of velocity with respect to time: expressed as meters per second squared (m/s虏).
  • Responsible for changes in speed of an object.
  • Can be constant or vary over time.
In the given problem, the subway train has a maximum tolerable acceleration of 1.34 m/s虏 that dictates how quickly it can reach its maximum speed from a stop and how smoothly it can bring the train to rest at the next station.
Velocity
Velocity extends beyond the concept of speed by adding direction to the motion.
It is not only about how fast something is moving, but also in which direction.
The specific velocity of interest in the exercise is the maximum velocity the subway train can reach between stations.

Key elements of velocity include:
  • A vector quantity, meaning it has both magnitude and direction.
  • Essential in determining how fast an object travels over a displacement.
  • Can vary over time or remain constant depending on the system.
The calculated maximum velocity for the train is crucial because it ensures passengers remain comfortable with an allowable acceleration limit while maximizing efficiency by minimizing travel time between subway stations.
Motion Equations
Motion equations are the mathematical backbone of solving kinematic problems like that of the subway train in the exercise. These equations interrelate velocity, acceleration, time, and position to describe the motion of objects.
They are powerful tools for predicting future states and uncovering unknown variables.

Important motion equations include:
  • Initial velocity (\(u\)), final velocity (\(v\)), acceleration (\(a\)), and displacement (\(s\)): \[v^2 = u^2 + 2as\]
  • Routine to find travel time (\(t\)): Using \(v = u + at\).
  • Total travel considerations: adding time for stops or constant velocity phases.
In this problem, motion equations guide us to derive the answers step-by-step, calculating key variables such as maximum speed and travel times, all while maintaining a clear view of the journey's dynamics.

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