Problem 71 - Fundamentals of Physics

Question

A suspicious-looking man runs as fast as he can along a moving sidewalk from one end to the other, taking \(2.50 \mathrm{~s}\). Then security agents appear, and the man runs as fast as he can back along the sidewalk to his starting point, taking \(10.0 \mathrm{~s}\). What is the ratio of the man's running speed to the sidewalk's speed?

Step-by-Step Solution

Answer

The ratio of the man's running speed to the sidewalk's speed is \( \frac{5}{3} \).

Step 1: Define Variables

Let \( v_m \) be the speed of the man and \( v_s \) be the speed of the sidewalk. We know that the time taken to run with the sidewalk is \(2.50\, \text{s}\) and against the sidewalk is \(10.0\, \text{s}\). Let \( d \) be the distance of the sidewalk.

Step 2: Establish Equations

For running with the sidewalk, the relative speed is \( v_m + v_s \), and the equation is \( d = (v_m + v_s) \times 2.50 \). For running against the sidewalk, the relative speed is \( v_m - v_s \), and the equation is \( d = (v_m - v_s) \times 10.0 \).

Step 3: Set Equations Equal

Since the distance \( d \) is the same in both scenarios, we can set the equations equal: \( (v_m + v_s) \times 2.50 = (v_m - v_s) \times 10.0 \).

Step 4: Solve for Speeds

Expand the equation: \[ 2.50 v_m + 2.50 v_s = 10 v_m - 10 v_s \] Rearrange to solve for \( v_m \) in terms of \( v_s \): \[ 2.50 v_s + 10 v_s = 10 v_m - 2.50 v_m \] \[ 12.50 v_s = 7.50 v_m \] Divide both sides by \( 7.50 \): \[ v_m = \frac{12.50}{7.50} v_s = \frac{5}{3} v_s \].

Step 5: Calculate the Ratio

The ratio of the man's running speed to the sidewalk's speed is the coefficient of \( v_s \) in \( v_m \). Thus, the ratio \( \frac{v_m}{v_s} = \frac{5}{3} \).

Key Concepts

Relative speed

Relative speed is the combined speed when two objects move in the same direction or opposite directions. For instance, when running towards an object, the effective speed is increased by the speed of both the runner and the object. When moving against an object, the relative speed decreases, as you subtract the speed of the object from your own speed.

In our example, when the man runs along with the sidewalk, his speed combines with the speed of the moving sidewalk. This means he covers more ground faster and thus, his relative speed is increased to \( v_m + v_s \). Conversely, when he runs against the direction of the sidewalk, his relative speed is lower, \( v_m - v_s \), as he's effectively countering the sidewalk's motion. Understanding relative speed is crucial for solving many problems in physics when multiple objects are involved.

Distance-time relationship

The distance-time relationship is a fundamental concept in physics. It describes the relationship between how far an object travels and the time it takes to do so, given its speed. The equation is simply distance \( d \) equals speed \( v \) multiplied by time \( t \): \[ d = vt \].

In our scenario, when the man runs on the sidewalk, either with or against its direction, he covers the same distance \( d \) each way. The speed is different, however, due to the influence of the sidewalk's motion.

• While running with the sidewalk: \( d = (v_m + v_s) \times 2.50 \)
• While running against the sidewalk: \( d = (v_m - v_s) \times 10.0 \)

This allows us to establish two equations that are set equal to solve for the relationship between \( v_m \) and \( v_s \). Understanding this relationship is vital to solve such exercises accurately.

Ratio calculations

Ratios are mathematical expressions that compare two numbers, representing their relative size. In many physics problems, comparing speeds and other quantities is crucial. Ratio calculations offer a simplified way to express one quantity in relation to another, which can be particularly useful in complex problem solving.

In the context of the problem, we used the ratio to determine the relationship between the man's speed \( v_m \) and the sidewalk's speed \( v_s \). Through setting equations from relative speed scenarios equal, we derived a ratio of \( \frac{5}{3} \), indicating the man's speed is \( \frac{5}{3} \) times that of the sidewalk's speed.

• This can be easily interpreted as: for every 3 units of speed by the sidewalk, the man runs at 5 units of speed.
• Utilizing ratio calculations provides clarity and simplifies conveying complex data about speeds.

Concept of speed

Speed is a measure of how fast an object moves from one point to another. It’s calculated as the distance covered divided by the time taken. In formula terms, it’s usually represented as \( v = \frac{d}{t} \). Speed is a scalar quantity, meaning it only has magnitude and no direction.

In the problem, the man’s running speed and the sidewalk’s speed were essential to determining the distance-time relationship and relative speed.

Understanding the concept of speed helps in breaking down problems into manageable steps. When combined with time, speed determines how distances are covered. Knowing that the man’s running speed was in contrast with that of the sidewalk, we could then deduce an equation to solve for their individual speeds.

• Being able to decompose the exercise into principles of speed equips you with the skill to tackle similar real-world problems.
• Broadly, recognizing the interplay between speed, distance, and time is key in everyday situations, from simple running paths to understanding vehicle dynamics.