Question

一物体从斜坡顶端由静止开始匀加速下滑,下滑的加速度为2m/s若滑到斜坡低端前最后1s下滑的距离为

斜坡长度的9分之5，球斜坡长 多少

Answer 1

设总时间为t,斜坡全长为S
对全程:
S=(1/2)at^2...................................................................................................<1>
对(t-1)s:
(1/2)(t-1)^2=(1-5/9)S=(4/9)S..........................................................................<2>
<1>:<2>得
t=3s
S=(1/2)at^2=(1/2)*2*(3*3)=9m

Answer 2

设斜坡长为x，下落时间为t
则x=at^2/2
4x/9=a(t-1)^2/2
则9/4=t^2/(t-1)^2
得t=3s
x=9m

追问

同样感谢！