A person 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 over the entire trip? What is her average speed over the entire trip? How do I calculate this without knowing the distance traveled? I know that average speed equals the total distance over the total time...
2007-06-04 17:54:15 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
speed coming back is 3.00 m/sec...sorry about that...
2007-06-04 17:58:38 · update #1
I actually did that and got 4.00m/s. However, the answer in the back of the book is 3.75 m/s. I understand why-b/c one would spend a longer period of time going back at 3.00 m/s than at 5.00 m/s, when traversing the same distance. Thanks for taking the time to answer.
2007-06-04 18:16:43 · update #2
I'm just wondering how to come up with the number 3.75, without knowing the exact distance traveled. I'm sure it has to do with setting certain equations equal to each other, blah, blah, but I just can't seem to figure it out...
2007-06-04 18:18:54 · update #3
Let s = distance.
let t = total time.
s/5.0 + s/3.0 = t
3s/15 + 5s/15 = t
8s/15 = t
2s/t = average
2s * 15/(8s) = average
15/4 = 3.75 m/s average
2007-06-04 18:32:59 · answer #1 · answered by J C 5 · 1⤊ 0⤋
just add both the velocites and divide it by two to get the average.
(5+3)/2 = 4m/s
2007-06-05 01:11:40 · answer #2 · answered by lilmaninbigpants 3 · 0⤊ 1⤋