已知OABC为同一直线上的四点,AB间的距离为L1,BC间的距离为L2,一物体自O点由静止出发,沿

1个回答

  • 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)]