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The escalator that leads down into a subway station has a length of \(30.0 \mathrm{~m}\) and a speed of \(1.8 \mathrm{~m} / \mathrm{s}\) relative to the ground. A student is coming out of the station by running in the wrong direction on this escalator. The local record time for this trick is 11 s. Relative to the escalator, what speed must the student exceed in order to beat the record?

Short Answer

Expert verified
The student must run at a speed of approximately 4.53 m/s relative to the escalator.

Step by step solution

01

Understanding the Problem

We need to find the speed at which the student should run relative to the escalator to beat the record time of 11 seconds when going against the escalator's movement, which has a speed of \(1.8 \text{ m/s}\).
02

Set Up the Equation for Time

The student must cover the entire length of the escalator, which is \(30.0 \text{ m}\), in less than 11 seconds. Let \(v_s\) be the student's speed relative to the escalator and \(v_e = 1.8 \text{ m/s}\) be the escalator's speed.
03

Determine Total Speed Needed

The student needs a speed such that his combined effort (his speed \(v_s\) against the escalator) equates to moving down the entire \(30.0 \text{ m}\) length in less than 11 seconds: \(\frac{30.0}{11}\).
04

Solve for Student's Speed

The effective speed relative to the ground (running against the direction of the escalator) is \(v_s - v_e\). The condition becomes: \(v_s - 1.8 = \frac{30.0}{11}\). Solve for \(v_s\):\[v_s = \frac{30.0}{11} + 1.8\].
05

Calculate \(v_s\)

Calculate the speed:\[v_s = \frac{30.0}{11} + 1.8 \approx 2.727 + 1.8 = 4.527 \text{ m/s}\].
06

Conclusion Interpretation

We find that the student needs a speed of approximately 4.53 m/s relative to the escalator in order to beat the record.

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Key Concepts

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

Physics Problem Solving
Approaching physics problems requires a rich blend of analytical thinking and practical application. In the given problem, the challenge is to determine the speed needed to move against an escalator's motion. Problem-solving in physics often begins with understanding the scenario and identifying known variables, such as the escalator's length and speed.
Steps to follow in physics problem solving include:
  • **Analyze** the problem statement carefully.
  • **Identify** known and unknown variables. Here, the knowns are the escalator length and speed, the unknown is the student's speed relative to the escalator.
  • **Formulate** the equations needed to solve the problem, by expressing relationships between the variables, as done with the time and speed equation.
  • **Solve** the equations systematically to find your answer.
  • **Verify** if the solution makes sense in the real-world context.
Physics problem solving can be simplified by breaking down the problem into small, manageable steps. Once you become familiar with crafting equations and analyzing results, you can apply these techniques to various scenarios, making problem solving a much easier task.
Kinematics
Kinematics deals with the description of motion without considering the forces that cause it. In this exercise, it is a crucial concept because it involves calculating how fast the student should move to outperform a previous speed record. Kinematics primarily focuses on:
  • **Distance and displacement**: Here, the 30 m length of the escalator is the distance the student needs to cover.
  • **Speed and velocity**: Determining the required speed relative to the escalator is the goal, highlighting both concepts.
  • **Time**: Given as a record to be beaten, in this case, 11 seconds.
To solve problems like these, you often use the kinematic equation: $$v = \frac{d}{t},$$ where:
  • **v** is the velocity or speed you are solving for,
  • **d** refers to the distance to be covered,
  • **t** is the time within which the task should be accomplished.
By substituting the known values into the equation, you calculate the speed needed to reach the goal, providing critical insight into motion dynamics in physical systems.
Speed and Velocity
Speed and velocity are fundamental concepts in physics, yet crucially different. Speed refers to how fast something is moving, while velocity is speed with direction. In the context of the given exercise, properly distinguishing between these terms helps solve the problem.Relative speed is essential, as the student runs against the direction of the escalator. This changes the velocity experienced by the student relative to the ground. The solution involves the formula:\[v_s = \frac{d}{t} + v_e,\]where:
  • **v_s** is the student's speed relative to the escalator.
  • **d/t** is the effective speed needed to cover the distance in the specified time.
  • **v_e** is the escalator's speed.
Understanding these calculations helps clarify the dynamics at play, showing how movement in one direction can alter speed needed in the opposite, affording a robust grasp of the concepts.

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Most popular questions from this chapter

The ball is \(26.9 \mathrm{~m}\) from the goalpost. The ball is kicked with an initial velocity of \(19.8 \mathrm{~m} / \mathrm{s}\) at an angle \(\theta\) above the ground. Between what two angles, \(\theta_{1}\) and \(\theta_{2}\), will the ball clear the \(2.74\) -m-high crossbar? (Hint: The following trigonometric identities may be useful: \(\sec \theta=1 /(\cos \theta)\) and \(\sec ^{2} \theta=1+\tan ^{2} \theta\)

(a) When a projectile is launched horizontally from a rooftop at a speed \(v_{0 x}\), does its horizontal velocity component ever change in the absence of air resistance? (b) Can you calculate the horizontal distance \(D\) traveled after launch simply as \(D=v_{0 x} t\), where \(t\) is the fall time of the projectile? (c) In calculating the fall time, is the vertical part of the motion just like that of a ball dropped from rest? Explain each of your answers. A criminal is escaping across a rooftop and runs off the roof horizontally at a speed of \(5.3 \mathrm{~m} / \mathrm{s}\), hoping to land on the roof of an adjacent building. The horizontal distance between the two buildings is \(D,\) and the roof of the adjacent building is \(2.0 \mathrm{~m}\) below the jumping- off point. Find the maximum value for \(D\).

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