Question #102697
Three point masses of 2.0 kg each, have the following position vectors:

r^(t) =( t+ 4t²)miˆ + tmk^ ;
r2^(t) = 2t²mj^ -3 mk^ ;
r3^ = (t-1)mi^ + 4t²mj^

Determine the velocity and acceleration of the centre of mass of the system.
1
Expert's answer
2020-02-24T10:49:51-0500
r^=1m+m+m(((t+4t2)mi^+tmk^)+(2t2mj^3mk^)+((t1)mi^+4t2mj^))\hat{r}=\frac{1}{m+m+m}((( t+ 4t^2)m\hat{i} + tm\hat{k})+\\ (2t^2m\hat{j} -3 m\hat{k})+((t-1)m\hat{i}+ 4t^2m\hat{j}))

r^=13((4t2+2t1)i^+6t2j^+(t3)k^))\hat{r}=\frac{1}{3}((4t^2+2t-1)\hat{i} + 6t^2\hat{j} +(t-3)\hat{k}))

The velocity of the centre of mass of the system



v^=13((8t+2)i^+12tj^+k^))\hat{v}=\frac{1}{3}((8t+2)\hat{i} + 12t\hat{j} +\hat{k}))

The acceleration of the centre of mass of the system


a^=13(8i^+12j^))\hat{a}=\frac{1}{3}(8\hat{i} + 12\hat{j} ))


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