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Reference frames are essential when analyzing relative motion. In this exercise, we have two key reference frames: the earth and the trolley. Understanding how to use these frames helps us calculate the velocities and displacement correctly.

The earth is considered an inertial reference frame, meaning it's stationary for our calculation purposes. The trolley, however, is moving with a constant velocity relative to the earth. Here’s how we break it down:

• Earth: This frame allows us to measure the overall motion of both the trolley and the man.
• Trolley: This frame helps us understand the motion of the man as he runs along the trolley.

By switching between these frames, we can properly account for the velocity and displacement of the moving objects within these frames. It's like changing perspective to gain a clearer view of movement and enhance our understanding of motion dynamics.

Velocity is the rate of change of an object’s position. When dealing with multiple reference frames, analyzing velocity helps us understand how fast and in what direction an object moves relative to each frame.

In our case, the man's velocity needs to be evaluated concerning both the trolley and the earth. Here's how:

• When the man runs in the same direction as the trolley, his velocity relative to the earth becomes the sum of his velocity relative to the trolley and the trolley's velocity: \( v + 1.5v = 2.5v \).
• Conversely, when returning back against the trolley's motion, his velocity relative to the earth is the difference: \( v - 1.5v = -0.5v \).

This breakdown allows us to clearly see how his motion changes based on direction and relative points of view, which is crucial for understanding the movement within distinct reference frames.

Displacement refers to the change in position of an object and involves careful tracking of initial and final positions relative to the desired frame.

For this problem, to find the man's maximum displacement relative to the earth, we only consider the segment of his journey where he reaches the end of the trolley before he turns back.

• While running with the trolley, we calculate the displacement using the expression \( (2.5v) \times \left( \frac{L}{1.5v} \right) = \frac{5L}{3} \).
• Returning, although he does cover some distance, doesn't contribute to maximum displacement because displacement is measured before the direction reversal.

Therefore, his maximum displacement occurs when he reaches the outer end of the trolley at \( \frac{5L}{3} \), just before beginning his path back. This straightforward method clears up how we handle forward and backward motion while quantifying total movement.