VC2.13

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

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

[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,

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

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

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

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

On simplification,we get

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