Aperson walks first at a constant speed of 5.00 m/s along a straight line from point a to point b and then back along the line from b to a at a constant speed of 2.90 m/s. what is the average speed over the entire trip?

This problem is a very interesting problem.
we know wacth:

Speed = distance / time.

Consider that:
distance from point A to point B=distance from point B to point A=d

We calculate the time to go from point A to point B.
time=distance / speed.
T₁=d / (5 m/s)

We calculate the time to go from point B to point A.
T₂=d / (2.9 m/s)

Therefore; the total time wil be: T₁+T₂
Total time=d/(5 m/s) + d / (2.9 m/s)
Least common multiple=(5 m/s)(2.9 m / s)=14.5 m²/s²
Total time=(2.9 m/s d + 5 m/s d)/ 14.5m²/s²
Total time=[(7.9 d) m/s] / (14.5 m²/s²)
Total time=7.9 d/ (14.5 m/s)

Now, the total distance will be =d+d=2d

Therefore:
Average speed=total distance / total time
Average speed=[(2d)]   /  [(7.9 d)/ (14.5 m/s)]
Average speed≈3.67 m/s.

the average speed will be ≈3.67 m/s.

answer and explanation;

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