《数学建模第三次作业——追击问题》由会员分享，可在线阅读，更多相关《数学建模第三次作业——追击问题（8页珍藏版）》请在金锄头文库上搜索。
1、数学建模（机自75班）丁鑫07011114 8数学建模实验报告机械工程及自动化75班07011114丁鑫四人追击问题问题：在一个边长为1的正方形跑道的四个顶点上各站有一人，他们同时开始以等速顺时针追逐下一人，在追逐过程中，每个人时刻对准目标，试模拟追击路线。并讨论：（1） 四个人能否追到一起？（2）若能追到一起，则每个人跑过多少路程？（3）追到一起所需要的时间（设速率为1）？（4）如果四个人追逐的速度不一样，情况又如何呢分析：先建立坐标系，设计程序使从A,B,C,D四个点同时出发，画出图形并判断。程序设计流程：四个人追击的速度相等，则有。针对这种情形，可有以下的程序。hold onaxis(0
2、 2 0 2);gridA=0,0;B=0,1;C=1,1;D=1,0;k=0;s1=0;s2=0;s3=0;s4=0; %四个人分别走过的路程t=0; v=1;dt=0.002;while k10000 k=k+1; plot(A(1),A(2),r.,markersize,15); plot(B(1),B(2),b.,markersize,15); plot(C(1),C(2),m.,markersize,15); plot(D(1),D(2),k.,markersize,15); e1=B-A;d1=norm(e1); e2=C-B;d2=norm(e2); e3=D-C;d3=norm
3、(e3); e4=A-D;d4=norm(e4); fprintf(k=%.0f ,k) fprintf(A(%.2f,%.2f) d1=%.2f ,A(1),A(2),d1) fprintf(B(%.2f,%.2f) d2=%.2f ,B(1),B(2),d2) fprintf(C(%.2f,%.2f) d3=%.2f ,C(1),C(2),d3) fprintf(D(%.2f,%.2f) d4=%.2fn,D(1),D(2),d4) A=A+v*dt*e1/d1; B=B+v*dt*e2/d2; C=C+v*dt*e3/d3; D=D+v*dt*e4/d4; t=t+dt; s1=s1+v
4、*dt; s2=s2+v*dt; s3=s3+v*dt; s4=s4+v*dt; if norm(A-C)=5.0e-3&norm(B-D)=5.0e-3 break endendts1s2s3s4部分运行结果：k=481 A(0.52,0.52) d1=0.04 B(0.52,0.48) d2=0.04 C(0.48,0.48) d3=0.04 D(0.48,0.52) d4=0.04k=482 A(0.52,0.52) d1=0.04 B(0.52,0.48) d2=0.04 C(0.48,0.48) d3=0.04 D(0.48,0.52) d4=0.04k=483 A(0.52,0.5
5、2) d1=0.04 B(0.52,0.48) d2=0.04 C(0.48,0.48) d3=0.04 D(0.48,0.52) d4=0.04k=484 A(0.52,0.51) d1=0.04 B(0.51,0.48) d2=0.04 C(0.48,0.49) d3=0.04 D(0.49,0.52) d4=0.04k=485 A(0.52,0.51) d1=0.04 B(0.51,0.48) d2=0.04 C(0.48,0.49) d3=0.04 D(0.49,0.52) d4=0.04k=486 A(0.52,0.51) d1=0.03 B(0.51,0.48) d2=0.03 C
6、(0.48,0.49) d3=0.03 D(0.49,0.52) d4=0.03k=487 A(0.52,0.51) d1=0.03 B(0.51,0.48) d2=0.03 C(0.48,0.49) d3=0.03 D(0.49,0.52) d4=0.03k=488 A(0.52,0.51) d1=0.03 B(0.51,0.48) d2=0.03 C(0.48,0.49) d3=0.03 D(0.49,0.52) d4=0.03k=489 A(0.52,0.51) d1=0.03 B(0.51,0.48) d2=0.03 C(0.48,0.49) d3=0.03 D(0.49,0.52)
7、d4=0.03k=490 A(0.52,0.50) d1=0.03 B(0.50,0.48) d2=0.03 C(0.48,0.50) d3=0.03 D(0.50,0.52) d4=0.03k=491 A(0.52,0.50) d1=0.02 B(0.50,0.48) d2=0.02 C(0.48,0.50) d3=0.02 D(0.50,0.52) d4=0.02k=492 A(0.52,0.50) d1=0.02 B(0.50,0.48) d2=0.02 C(0.48,0.50) d3=0.02 D(0.50,0.52) d4=0.02k=493 A(0.51,0.50) d1=0.02
8、 B(0.50,0.49) d2=0.02 C(0.49,0.50) d3=0.02 D(0.50,0.51) d4=0.02k=494 A(0.51,0.50) d1=0.02 B(0.50,0.49) d2=0.02 C(0.49,0.50) d3=0.02 D(0.50,0.51) d4=0.02k=495 A(0.51,0.50) d1=0.02 B(0.50,0.49) d2=0.02 C(0.49,0.50) d3=0.02 D(0.50,0.51) d4=0.02k=496 A(0.51,0.50) d1=0.01 B(0.50,0.49) d2=0.01 C(0.49,0.50
9、) d3=0.01 D(0.50,0.51) d4=0.01k=497 A(0.51,0.49) d1=0.01 B(0.49,0.49) d2=0.01 C(0.49,0.51) d3=0.01 D(0.51,0.51) d4=0.01k=498 A(0.51,0.49) d1=0.01 B(0.49,0.49) d2=0.01 C(0.49,0.51) d3=0.01 D(0.51,0.51) d4=0.01k=499 A(0.50,0.49) d1=0.01 B(0.49,0.50) d2=0.01 C(0.50,0.51) d3=0.01 D(0.51,0.50) d4=0.01k=5
10、00 A(0.50,0.50) d1=0.01 B(0.50,0.50) d2=0.01 C(0.50,0.50) d3=0.01 D(0.50,0.50) d4=0.01k=501 A(0.50,0.50) d1=0.01 B(0.50,0.50) d2=0.01 C(0.50,0.50) d3=0.01 D(0.50,0.50) d4=0.01k=502 A(0.50,0.50) d1=0.00 B(0.50,0.50) d2=0.00 C(0.50,0.50) d3=0.00 D(0.50,0.50) d4=0.00t = 1.0040s1 = 1.0040s2 = 1.0040s3 =
11、 1.0040s4 =1.0040从运行的结果来看，如果四个人的追击速度相同，均为1，可有以下的结果：（1） 四人最后可以追到一起。（2） 每个人跑过相等的路程，均为1.0040.（3） 追到一起的时间为1.0040秒。如果四个人追击的速度不一样，可取，运行程序。运行程序：hold onaxis(0 2 0 2);gridA=0,0;B=0,1;C=1,1;D=1,0;k=0;s1=0;s2=0;s3=0;s4=0; %四个人分别走过的路程t=0; v=1;dt=0.001;while k10000 k=k+1; plot(A(1),A(2),r.,markersize,15); plot(B
12、(1),B(2),b.,markersize,15); plot(C(1),C(2),m.,markersize,15); plot(D(1),D(2),k.,markersize,15); e1=B-A;d1=norm(e1); e2=C-B;d2=norm(e2); e3=D-C;d3=norm(e3); e4=A-D;d4=norm(e4); fprintf(k=%.0f ,k) fprintf(A(%.2f,%.2f) d1=%.2f ,A(1),A(2),d1) fprintf(B(%.2f,%.2f) d2=%.2f ,B(1),B(2),d2) fprintf(C(%.2f,%.
13、2f) d3=%.2f ,C(1),C(2),d3) fprintf(D(%.2f,%.2f) d4=%.2fn,D(1),D(2),d4) A=A+v*dt*e1/d1; B=B+2*v*dt*e2/d2; C=C+2*v*dt*e3/d3; D=D+v*dt*e4/d4; t=t+dt; s1=s1+v*dt; s2=s2+v*dt; s3=s3+v*dt; s4=s4+v*dt; if norm(A-C)=1.0e-3&norm(B-D)=1.0e-3 break endendts1s2s3s4部分运行结果：k=662 A(0.41,0.32) d1=0.06 B(0.43,0.27)
14、d2=0.00 C(0.43,0.27) d3=0.00 D(0.43,0.27) d4=0.05k=663 A(0.41,0.32) d1=0.05 B(0.43,0.27) d2=0.00 C(0.43,0.27) d3=0.00 D(0.43,0.27) d4=0.05k=664 A(0.41,0.32) d1=0.05 B(0.43,0.27) d2=0.00 C(0.43,0.27) d3=0.00 D(0.43,0.27) d4=0.05k=665 A(0.41,0.32) d1=0.05 B(0.43,0.27) d2=0.00 C(0.43,0.27) d3=0.00 D(0.43,0.28) d4=0.05k=666 A(0.41,0.31) d1=0.05 B(0.43,0.28) d2=0.00 C(0.43,0.28) d3=0.00 D(0.43,0.28) d4=0.04k=667 A(0.41,0.31) d1=0.04 B(0.43,0.28) d2=0.00 C(
