一火车沿直线轨道从静止出发由A地驶向B地并停在B地,AB两地相距为s,火车先做匀加速运动,其加速度最大为a1,做匀减速运

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  • 运动包含二部分:

    1.速度从0,以a1做匀加速运动,达到速度V,用时间为t1,路程s1

    s1=1/2a1t1^2 , v=a1t1

    2.速度从V,以a2的加速度减速运动到停车,所用时间t2,路程s2

    s2=vt2-1/2 a2t2^2 ,v=a2t2

    最短时间:T=t1+t2 S=s1+s2

    v=a1t1=a2t2=a2(T-t1)==>t1=a2T/(a1+a2), t2=a1t1/a2

    s2=vt2-1/2 a2t2^2

    =(a1t1)(a1t1/a2)-1/2 a2(a1t1/a2)^2

    =(a1t1)^2/a2-1/2(a1t1)^2/a2

    =1/2(a1t1)^2/a2

    S=s1+s2=1/2a1t1^2+1/2(a1t1)^2/a2

    =1/2a1(a1+a2)t1^2/a2

    t1=a2T/(a1+a2),

    可以求出 T=√2S(a1+a2)/a1a2

    V=a1t1=a1a2T/(a1+a2)=√2Sa1a2/(a1+a2)