a jogger runs 100 meters at a constant speed of 5 m/s and then becoming tired , walks 100 meteres of a constance speed of 1.00m/s. what is the jogger's average speed over the 200.0 meters?
2006-09-26 17:46:53 · 3 answers · asked by Julie Kim 1 in Education & Reference ➔ Homework Help
That is easy.
5 m/s for 100 meters = 20seconds.
1 m/s for 100 meters = 100 seconds.
200 meters / 120 seconds = 1.667 m/s
2006-09-26 17:52:07 · answer #1 · answered by jbtascam 5 · 0⤊ 0⤋
That is easy.
5 m/s for 100 meters = 20seconds.
1 m/s for 100 meters = 100 seconds.
200 meters / 120 seconds = 1.667 m/s
2006-09-27 01:07:19 · answer #2 · answered by kahoonas 1 · 0⤊ 0⤋
First answer is correct and the second one is only a copy of the first one but I want to give you a formula for this kind of question:
V(average)=(x1+x2+....+xn)/
[(x1/v1)+(x2/v2)+...+(xn/vn)]
x1=distance that body move at a constant speed of v1
x2=distance that body move at a constant speed of v2
.
.
.
xn=distance that body move at a constant speed of vn
2006-09-27 04:49:48 · answer #3 · answered by Mamad 3 · 0⤊ 0⤋