设在OA段的时间为t1,加速度为a;在AB和BC段的时间为t.则:
OA=(1/2)at1^2
OB=(1/2)a(t1+t)^2
OC=(1/2)a(t1+2t)^2
则,AB=OB-OA=(1/2)a(t1+t)^2-(1/2)at1^2=L1
===> (t1+t)^2-t1^2=2L1/a
===> t1^2+2tt1+t^2-t1^2=2L1/a
===> t^2+2tt1=2L1/a…………………………………………………………(1)
BC=OC-OB=(1/2)a(t1+2t)^2-(1/2)a(t1+t)^2=L2
===> (t1+2t)^2-(t1+t)^2=2L2/a
===> t1^2+4tt1+4t^2-t1^2-2tt1-t^2=2L2/a
===> 3t^2+2tt1=2L2/a…………………………………………………………(2)
(2)-(1)得:2t^2=2(L2-L1)/a
===> t=√(L2-L1)/a
代入(1)得到:(L2-L1)/a+2√[(L2-L1)/a]t1=2L1/a
===> √[(L2-L1)/a]t1=(3L1-L2)/(2a)
===> t1^2=(3L1-L2)^2/(4a^2)*[a/(L2-L1)]=(3L1-L2)^2/[4a(L2-L1)]
所以,OA=(1/2)at1^2
=(1/2)a*(3L1-L2)^2/[4a(L2-L1)]
=(3L2-L1)^2/[8(L2-L1)]