Velocity of a particle in x-y plane at any time t is `v=(2t hati+3t^2 hatj) m//s` At `t=0,` particle starts from the co-ordinates `(2m,4m).` Find
(a) acceleration of the particle at `t=1s.`
(b) position vector and co-ordinates of the particle at `t=2s.`

Answer

A, B, C, D

Explanation

(a) `a=(dv)/dt=d/dt(2t hati+3t^2hatj)`
`=(2hati++6thatj) m//s^2`
At `t=1s,`
`a=(2hati+6 hatj) m//s^2`
(b) intds=intv dt
or `s=intv dt=int(2 thati+3t^2 hatj)dt`
`:. r_f-r_i=int_(i ntitial)^(fi nal) (2t hati+3t^2 hatj)dt`
`or r_(2 sec)-r_(0 sec)=int_0^2(2t hati+3t^2 hatj)dt`
`:. r_(2 sec)=r_(0 sec)+[t^2 hati+t^3 hatj]_0^2`
`=(2 hati+4 hatj)+(4 hati+8 hatj)`
`=(6 hati+12 hatj) m`
Therefore, coordinates of the particle at `t=2s` are `(6m,12m)`