Posted by Jack on October 01, 2002 at 10:25:39:
In Reply to: Physics acceleration problem posted by Jimmu on September 30, 2002 at 20:17:01:
: Okay, here it goes,
: A skier moves downslope, starting from rest, with an acceleration of 2 m/s^2. At the end of 6 seconds, he slows down to stop, which takes 1 more second.
: I need to know:
: 1. How to graph it
:Let y-axis be speed, x-axis be time
: 2. Skier's max speed
Vf=Vi+at where Vi=0, a=2 m/s/s, t=6 s
Vf=0+2*6 = 12 m/s
: 3. Skier's acceleration while stopping
:Vf=Vi+at or a=(Vf-Vi)/t where Vf=0, Vi=12 m/s,
:t=1 s So a=(0-12)/1 = -12 m/s/s
: 4. Total distance traveled by the skier
:Use the formula S=Vit+(1/2)at^2
:Let S1=distance traveled during the 6 s interval
:Let S2=distance traveled while stopping
:S1=(0)(6)+(1/2)(2)(6)^2 = 36 m
:S2=(12)(1)+(1/2)(-12)(1)^2 = 6 m
:Total distance traveled S=S1+S2=36+6=42 m
: Thank you very much.