Question

A person walks first at 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 3.00 m/s. What is her average speed over the entire trip?

Answer 1

Let the distance between A and B be d.
Speed s=d/t
5=d/t1
t1=d/5
3=d/t2
t2=d/3
Average speed=(d+d)/(t1+t2)
Hope that helps:)

Answer 2

aveg. speed = total distance / total time

let x be the distance from A to B. The distance from B to A is also x. So the total distance is 2x

time = distance / speed
time from A to B:
t = x/5

time from B to A
t = x/3

the total time is x/5 + x/3 = 8x/15

averg. speed = 2x / (8x/15)
averg. speed = 3.75 m/s


aveg. velocity = total displacement / total time

to find displacement, take final postion and subtract the initial position

the inital position is 0m.
the finial position is also 0m, because the person goes back to here he startedd

total displacement is 0 - 0 = 0m

aveg. velocity = 0 / (8x/15) = 0m/s

hope this helps