VC1.12

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Given r=at^{2}+bt+c\,

Since a,b,c are constant vectors {\frac  {da}{dt}}={\frac  {db}{dt}}={\frac  {dc}{dt}}=0\,

From the given, {\frac  {dr}{dt}}={\frac  {da}{dt}}t^{2}+2at+{\frac  {db}{dt}}t+b+{\frac  {dc}{dt}}\,

Therefore,{\frac  {dr}{dt}}=2at+b\,


Again differentiating the above, we get {\frac  {d^{2}r}{dt^{2}}}=2[{\frac  {da}{dt}}+a]+{\frac  {db}{dt}}=2a\,

Therefore,the acceleration of the particle={\frac  {d^{2}r}{dt^{2}}}=2a\,=constant,showing that the point whose path is given by the given is moving with constant acceleration.

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