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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