A particle moves along 1 ositive branch of the curve `y=(x)/(2)` where `x=(t^(3))/(3)`, x and y are measured in metres and t in seconds, then-

Options

A. the velocity of particle at t- 1 s is `hati+(1)/(2)hatj`
B. the velocity of particle at t= 1 s is `(1)/(2)hati+hatj`
C. the acceleration of particle at t=2s is `4hati+2hatj`
D. the acceleration of particle at t=2s is `hati+2hatj`

Explanation

`(dx)/(dt)=t^(2)` . . . ..(i)
`y=(1)/(2)(t^(3))/(3)`
`(dy)/(dt)=(t^(2))/(2)`
`t=1,v_(x)=1,v_(y)=(1)/(2)`
`vecv=hati+(1)/(2)hatj`
`(d^(2)x)/(dt^(2))=2t` ltrbgt `(d^(2)y)/(dt^(2))=t`
`"at" t=1s a_(x)=2 "and" a_(y)=1`
`veca=2hati+hatj`