Matlab中的相对运动分析速度

Question

我想找到一个方程，用于确定在t = 0（显示的形状）到t = 10 s的时间间隔内的速度A和B和A / B

http://uupload.ir/files/o27p_1.png

http://uupload.ir/files/i4qz_2.png

http://uupload.ir/files/s0st_3.png

   clc;
    clear;
    close all ;
    w = 2 ;
    rA = 100/1000 ; 
    rB = 75/1000 ;
    vB = rB .* w ;
    rAB = sqrt((100-80).^1/2 + (250-75).^2) / 1000 ;
    n= 1000 ;
    Teta0 = atan((100-50) / (250-75)) ;
    t = linspace(0,10,n) ;
    Teta = Teta0 + w .* t ;
    Beta  = asin((rAB/rA) .* sin(Teta)) ; 
    % vAB = (vB -( vB * tan(Teta) .* cos(Beta))) ./ cos(Teta) ;
    vAB =vB ./ cos(Teta) ;
    vA = sqrt(vB.^2 + vAB .^ 2 - (2 .* vB .*vAB .* cos(Teta)))  ;
    wOA = vA ./ rA ;
    wAB = vAB ./ rAB ;
    Number_Of_experience = (1:n)' ;
    Time = reshape(t,n,1) ;
    Velocity_Of_B = repmat(vB,n,1);
    Velocity_Of_A = reshape(vA,n,1);
    Velocity_Of_AB = reshape(vAB,n,1);
    Teta = reshape(Teta,n,1);
    Beta = reshape(Beta,n,1);
    Omega_of_OA = reshape(wOA,n,1) ;
    Omega_of_AB = reshape(wAB,n,1) ;

    T = table(Number_Of_experience,Time,Velocity_Of_B,Velocity_Of_A,Omega_of_OA,Velocity_Of_AB,Omega_of_AB,Teta,Beta) ;
    disp(T);