Question

A person walks first at a constant speed of 4.70 m/s along a straight line from point circled A to point circled B and then back along the line from circled B to circled A at a constant speed of 3.15 m/s.(a) What is her average speed over the entire trip?m/s(b) What is her average velocity over the entire trip?m/s

Answer 1

Answer:

(a) 3.925 m/s

(b) 0 m/s

Explanation:

(a) As the distance from point A to point B is the same as the distance from point B to point A, the total distance is twice the distance of each. The average speed would be the literally the average of the 2 speeds

$\frac{4.7 + 3.15}{2} = 3.925m/s$

(b) Since she's back to her starting point, her net displacement would be 0, given the total travel time as t, her average velocity would be the net displacement divided by t, which is

0 / t = 0 m/s