VC2.12

Given, $\frac{d^2 r}{dt^2}=6ti-24t^2j+4\sin t k\,$ [Equation 1]

with $\frac{dr}{dt}=-i-3k\,$ at t=0 [Equation 2]

and $r=2i+j\,$ at t=0.[Equation 3]

Integrating (1), $\frac{dr}{dt}=3t^2i-8t^3j-4\cos t k+c\,$ [4] where c is an arbitrary constant vector.

Putting t=0 in (4) and using (2),we have

$-i-3k=-4k+c \,$ so that $c=-i+k\,$ [5]

Using (5),(4) reduces to $\frac{dr}{dt}=3t^2i-8t^3j-4\cos t k-i+k=(3t^2-1)i-8t^3j+(1-4\cos t)k\,$ [6]

Integrating (6), $r=(t^3-t)i-2t^4j+(t-4\sin t)k+d\,$ [7] where d is a constant vector.

Putting t=0 in (7) and using(3),we have $d=2i+j\,$ [8]

Using (8),(7)reduces to $r=(t^3-t)i-(2t^4j+(t-4\sin t)k+2i+j\,$

On simplification,we get

$r=(t^3-t+2)i+(1-2t^4)j+(t-4\sin t)k\,$