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Given, a={\frac  {dv}{dt}}=3\cos ti+4\sin tj+t^{2}k\, [Equation 1]

Integrating (1),v=3\sin ti-4\cos tj+{\frac  {1}{3}}t^{3}k+c\,

[Equation 2]where c is an arbitrary constant.

Given thatv=0 when t=0. Hence,from (2),we obtain

0=-4j+c\, so that c=4j\,. So,(2) becomes,

v=3\sin ti-4\cos tj+{\frac  {1}{3}}t^{3}k+4j\, or

v={\frac  {dr}{dt}}=3\sin ti+4(1-\cos t)j+{\frac  {1}{3}}t^{3}k\, [Equation 3]

Integrating (3), r=-3\cos ti+4(t-\sin t)j+{\frac  {1}{12}}t^{4}k+d\, [4]

where d is a constant.

Given that r=0 when t=0. Henc efrom (4),we get

0=-3i+d,d=3i\,. Now,(4) becomes

r=-3\cos ti+4(t-\sin t)j+{\frac  {1}{12}}t^{4}k+3i\,

On simplification,we get

r=3(1-\cos t)i+4(t-\sin t)j+{\frac  {1}{12}}t^{4}k\,

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