Answer to Question #13371 in Mechanics | Relativity for Shetal Varghese
For particle 1,a1=3t^2+4 b1=0
For particle 2,a2=7t+5 b2=2
For particle 3,a3=2t b3=3t+4
Determine the position and velocity of the centre of mass as functions of time
ac = (a1+a2+a3)/3 = (3t²+4+7t+5+2t)/3 = (3t²+9t+9)/3 = t²+3t+3
bc = (b1+b2+b3)/3 = (0+2+3t+4)/3 = t+2
The velocity& of the center of mass:
vca = ac'(t) = (t²+3t+3)' = 2t+3
vcb = bc'(t) = (t+2)' = 1
vc = √(vca²+vcb²) = √((2t+3)²+1²) = √(4t²+12t+9+1) = √(4t²+12t+10).
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