圈上泽充大峰 Example solution ▣Rise segment 6 boundary conditions requires 6 terms in equation >Thus use 5th degree polynomial Use 0/B as independent variable(x),Differentiate polynomial to get v,a .c.aa-cfa)cca =c+c[xacj+c ac..c)1c.( 圆上泽夫道大学 Example solution ▣Boundary conditions >When 0=0,s=0 Thus Co=0 s=C。+C e9ccc >When 0=0,v=0 Thus C,=0 +ccc 2424  Rise segment  6 boundary conditions requires 6 terms in equation  Thus use 5th degree polynomial  Use  as independent variable (x), Differentiate polynomial to get v, a Example solution 5 5 4 4 3 3 2 0 1 2                                                        s C C C C C C                                             4 5 3 4 2 1 2 2 3 3 4 5 1          v C C C C C                                    3 5 2 2 2 2 6 3 12 4 20 1        a C C C C  Boundary conditions  When =0, s=0 Thus C0=0 5 5 4 4 3 3 2 0 1 2                                                        s C C C C C C  When =0, v=0 Thus C1=0                                             4 5 3 4 2 1 2 3 2 3 4 5 1          v C C C C C  When q=0, a=0 Thus C2=0                                    3 5 2 2 2 3 4 2 6 12 20 1        a C C C C Example solution