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A person walks first at a constant speed of 4.90 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 3.05 m/s.A) What is her average speed over the entire trip?

B) What is her average velocity over the entire trip?

User Ajay Reddy
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1 Answer

5 votes

Answer:

A.) 3.975 m/s

B.) 0.925 m/s

Step-by-step explanation:

Given that a person walks first at a constant speed of 4.90 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 3.05 m/s.

A) What is her average speed over the entire trip?

The average speed = (4.90 + 3.05)/2

Average speed = 7.95/2

Average speed = 3.975 m/s

B) What is her average velocity over the entire trip?

Since velocity is a vector quantity, that is, we consider both the magnitude and direction

Average velocity = ( 4.90 - 3.05)/2

Average velocity = 1.85/2

Average velocity = 0.925 m/s

User Boris WM
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