"x_cm = \\dfrac{sum(i = 1, N) m_i*x_i}{sum(i=1,N) m_i}"
We have N = 2 in this case. Let's sit on car 2 and call our position x = 0 despite the fact that car 2 is moving with respect to the "outside" world. Then car 1 appears to be moving away from car 2 and moving in the negative x direction. We can describe this by:
"x(t) = (v_1 - v_2)*t - x_0" and since "v1 < v2, x(t) < 0"
Now "x_{cm}" can be computed taking "x_2" = 0 and "x_{cm}= \\dfrac{m1*((v1-v2)*t - x0)}{(m1 + m2)}"
Set "t=0" and "x_{cm}(0)= \\dfrac{m1*(v1-v2)}{m_1+m_2}" -
"x_{cm} = -3.2m" relative to "m_2"
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