Problem 69 - College Physics

Question

A shopper is in a hurry to catch a bargain in a department store. She walks up the escalator, rather than letting it carry her, at a speed of \(1.0 \mathrm{~m} / \mathrm{s}\) relative to the escalator. If the escalator is \(10 \mathrm{~m}\) long and moves at a speed of \(0.50 \mathrm{~m} / \mathrm{s}\), how long does it take for the shopper to get to the next floor?

Step-by-Step Solution

Answer

The shopper takes approximately 6.67 seconds to reach the next floor.

Step 1: Understand the Problem

We need to determine how much time it takes for the shopper to reach the next floor while moving on an escalator. The shopper's speed relative to the escalator and the escalator's own speed need to be considered to find the total speed relative to the stationary ground.

Step 2: Calculate Total Speed

The shopper's speed relative to the escalator is given as \(1.0 \mathrm{~m/s}\). The escalator itself is moving at \(0.50 \mathrm{~m/s}\). To find the combined speed relative to the ground, we add these two speeds: \(1.0 + 0.50 = 1.5 \mathrm{~m/s}\).

Step 3: Use Distance-Speed-Time Relationship

The formula relating distance \(d\), speed \(v\), and time \(t\) is \(d = vt\). Here, the escalator distance \(d\) is \(10 \mathrm{~m}\) and the total speed \(v\) is \(1.5 \mathrm{~m/s}\). We need to find the time \(t\). Rearrange the formula to get \(t = \frac{d}{v}\).

Step 4: Calculate the Time

Substitute the known values into the rearranged formula to find the time: \(t = \frac{10}{1.5}\). Performing the division gives \(t \approx 6.67 \mathrm{~s}\).

Key Concepts

Relative Motion

Understanding relative motion is all about analyzing how an object's position changes in relation to another object. In our exercise, the shopper moving on the escalator is an example of relative motion. She moves at a certain speed but so does the escalator she's on. If we were to stand still on the ground, we would observe the combined speed she travels: her own speed plus the escalator's.

Relative motion involves multiple entities, and it's often used to describe how one moving object is perceived by an observer in another moving object. This results in what we call 'relative speed.'

• In our case, the shopper's speed relative to the escalator is 1.0 m/s.
• The escalator itself moves at 0.50 m/s relative to the ground.
• Their combined 'relative speed' to the ground is 1.5 m/s.

Understanding this concept is crucial for solving such physics problems, as failing to properly account for one component's influence on another could lead to incorrect conclusions.

Speed and Velocity

Though often used interchangeably in everyday language, speed and velocity have distinct meanings in physics. Speed is a scalar quantity representing how fast an object is moving, while velocity is a vector that not only indicates speed but also the direction of the movement.

In our example involving the shopper and the escalator, we focused primarily on speed since the direction was straightforward and didn't change. The shopper's speed is given as 1.0 m/s relative to the escalator, indicating her pace irrespective of direction. The escalator adds another 0.50 m/s, culminating in a total speed of 1.5 m/s relative to a stationary observer.

This simple breakdown helps us grasp how different components of motion contribute to the overall speed. Whenever you solve physics problems involving moving platforms, understanding the distinction between speed and velocity ensures more accurate results.

Distance-Time Relationship

The distance-time relationship is a fundamental concept in physics kinematics and is governed by the formula: \(d = vt\), where:

• \(d\) is the distance traveled,
• \(v\) is the speed or velocity, and
• \(t\) is the time taken.

This formula ties together how far something travels (distance), how fast it travels (speed), and the duration of the travel (time).

For the shopper on the escalator, the distance to be covered is 10 meters. Using her combined speed of 1.5 m/s, we were able to determine the time by rearranging the formula to \(t = \frac{d}{v}\). Substituting the known values, the time calculated is approximately 6.67 seconds.

Understanding this relationship is key to solving such problems quickly and effectively. Knowing how to juggle the formula depending on which variable you need to solve for can serve you well in various physics scenarios.

Kinematics

Kinematics is the branch of physics that deals with motion, without worrying about what causes the motion (that's another branch called dynamics). Kinematics helps us describe the position, velocity, and acceleration of points or objects in motion.

In the exercise provided, kinematics allows us to describe how the shopper moves in relation to the escalator and the ground. We calculated how long it takes for her to reach the next floor by using kinematic principles, combined speed, and the distance-time formula.

• Kinematics helps break down complex motions into understandable elements.
• Identifying initial conditions like speeds, distances, and time frames are crucial.
• The calculations are then performed to predict or analyze motion outcomes.

Embracing these foundations of kinematics can enable students to approach problems systematically, identifying which equations or methods apply to different conditions in motion-based challenges.