A person walks first at a constant speed of 4.99 m/s along a straight line from

Question

A person walks first at a constant speed of 4.99 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?
What is the average velocity over the entire trip? A person walks first at a constant speed of 4.99 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?
What is the average velocity over the entire trip?
What is the average speed over the entire trip?
What is the average velocity over the entire trip?

Explanation / Answer

Avarage speed = (Total distance covered) /( Total time taken)

Avarage velocity = ( Total displacement ) / ( Total time )

Let t be the time taken for the first time journey , then the distance from A to B is

D = 4.99 t meter.

Time taken for the second journey = 4.99 t / 2.90 = 1.72 t

Avarage speed = (2*4.99 t ) /(1.72 t + t ) = 3.67 m/s

But since the total displacement is 0.

So, Avarage velocity = 0.

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