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When tackling physics problems involving speed calculation, it is essential to understand how to determine the speed of an object based on time and distance. Speed is a scalar quantity that expresses how fast an object is moving regardless of its direction. Typically, speed is calculated using the formula:\[ v = \frac{d}{t} \]where:

• \( v \) is the speed,
• \( d \) is the distance travelled, and
• \( t \) is the time taken to travel that distance.

In our problem, the man's speed relative to the ground can be considered in two contexts: moving with the sidewalk and moving against it. Knowing the time taken for both journeys allows us to calculate these speeds by considering the combined effects of the runner’s and the sidewalk’s speeds. By expressing both components, it becomes possible to identify each speed precisely.

Motion on a moving sidewalk involves analyzing relative motion. When a person moves on a sidewalk that itself is also moving, their speed relative to the ground is not just their running speed. Instead, it is either increased or decreased by the sidewalk's motion.In the given exercise:

• When running to the end, the man's speed is enhanced by the sidewalk's speed, \( v_m + v_s \).
• On returning, the man's speed is diminished by the sidewalk's speed, \( v_m - v_s \).

Understanding this concept helps in deciphering real-world situations, such as traveling on escalators or moving walkways. The result is that the effective speed is a combination of both movements, which can be intuitively grasped by observing the relative direction of movement.

The ratio of speeds informs us of the relationship between two distinct motions. In our scenario, we are tasked to find the ratio of the man's running speed to the sidewalk's speed. The method involves equating two expressions derived from time and motion equations:

• First, express the length of the sidewalk, \( L \), through equations representing both directions of the man's movement, utilizing times and effective speeds.
• Since \( L \) is identical in both cases, equating the two expressions allows solving for variables in terms of others, in this case mainly \( v_m \) and \( v_s \).

By simplifying the set equation further, one can extract the required ratio \( \frac{v_m}{v_s} \). This value gives insight into how much faster one movement is relative to another, which can be pivotal in understanding the dynamics of the scenario effectively. In this problem, the conclusion is that the man's speed is \( \frac{5}{3} \) times that of the sidewalk, which provides significant insight into the interaction of the two motions.