A person walks first at a constant speed of 5.20 m/s along a straight line from

Question

A person walks first at a constant speed of 5.20 m/s along a straight line from point to point and then back along the line from to at a constant speed of 3.20 m/s.

a) What is her average speed over the entire trip? in m/s

What is her average velocity over the entire trip?
m/s

Explanation / Answer

(a) Let's assume that the distance from point A to point B is 1664 meters (it's the common denominator of 52 and 32 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: 1664 = 5.2 * t, t = 320 seconds B --> A: 1664 = 3.2 * t, t = 520 seconds The average speed over the entire trip = (total distance) / (total time): Average speed = (1664 + 1664) m / (320 + 520) s Average speed = 3224 m / 830 s Average speed = 3.96 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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