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2 किग्रा तथा 3 किग्रा द्रव्यमान के दो कणों के स्थिति सदिश क्रमश: `vecr_(1) = t hati + 2t^(2) hatj + 2t hatk` तथा ` vecr_(2) = hati + 2t^(3) hatj + 5 hatk` है | इस निकाय के लिये ज्ञात कीजिये : (i ) द्रव्यमान केन्द्र का स्थिति सदिश , (ii ) द्रव्यमान केन्द्र का वेग , (iii ) द्रव्यमान केन्द्र का त्वरण |

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(i) द्रव्यमान केन्द्र का स्थिति सदिश
`vecr_(CM) = (m_(1) vecr_(1) + m_(2) vecr_(2))/(m_(1) + m_(2))`
`= (2(t hati + 2 t^(2) hatj + 2t hatk) + 3(hati + 2t^(3) hatj + 5hatk))/(2 + 3)`
`= (1)/(5) [(2t + 3) hati + (4t^(2) + 6t^(3)) hatj + (4t + 15) hatk]`
(ii ) द्रव्यमान केन्द्र का वेग
`vecv_(CM) = (d)/(dt)"" vecr_(CM)`
`= (1)/(5)[(2xx 1 + 0 ) hati + (8t + 18t^(2)) hatj + (4 xx 1 + 0 ) hatk]`
`= (1)/(5) [2hati + (8t + 18t^(2) ) hatj + 4hatk]`
(iii) द्रव्यमान केन्द्र का त्वरण
`veca_(CM) = (dvecv_(CM))/(dt)`
`=(1)/(5) [0 + (8 + 36 t) hatj + 0]`
`= ((8 + 36t)/(5)) hatj`

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