The position vector of a aprticle is given as `vecr=(3t^2-2t+5)hati+(3t^2)hatj`. The time after which the velocity vector and acceleration vector becomes perpendicular to each other is equal to

Answer

A

Explanation

Time to cross 2m is `(2/(v sin theta))`….
To avoid an accident
Displacement `=4+v cos thetaxx2/(v sin theta)`
`8xx2/(v sin theta)=4+2 cot theta`
`v sin theta=(16 sin theta)/(4 sin theta +2 cos theta)`
`v_("min")=16/sqrt(4^(2)+2^(2))=1.6 sqrt(5) m//s`
`[ :' (a cos theta + b sin theta) "has max. value "=sqrt(a^(2)+b^(2))]`