A person walks first at a constant speed of 5.30 m/s along a straight line from

Question

A person walks first at a constant speed of 5.30 m/s along a straight line from point to point and then back along the line from to at a constant speed of 3.10 m/s.
(a) What is her average speed over the entire trip?

Explanation / Answer

(a) Let's assume that the distance from point A to point B is 1612 meters (it's the common denominator of 52 and 31 for simplicity of calculations). Given that d = r * t, we can calculate the times for traveling from A to B and back from B to A: A --> B: 1612 = 5.2 * t, t = 310 seconds B --> A: 1612 = 3.1 * t, t = 520 seconds The average speed over the entire trip = (total distance) / (total time): Average speed = (1612 + 1612) m / (310 + 520) s Average speed = 3224 m / 830 s Average speed = 3.88 m/s (b) Please note that velocity is a vector quantity and the rate at which the position changes per time. This is different than the average speed, which is a scalar quantity and not associated with direction. Average velocity = (displacement from original position) / total time So, if the person started at point A and ended up at point A (even though he went to B and back), the displacement is 0 and average velocity over the entire trip is 0 m/s.

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