The position vector of a particle is given as `vecr=(t^2-4t+6)hati+(t^2)hatj`. The time after which the velocity vector and acceleration vector becomes perpendicular to each other is equal to

Options

A. 2 sec
B. 1 sec
C. 1.5 sec
D. 5 sec

Answer

B

Explanation

`vec r =(t^2 - 4t + 6) hat i + t^2 hatj , vec v = (d vec r)/(dt)`
=`(2t - 4)hat i + 2t hat j, vec a = (d vec v)/(dt) = 2 hat i + 2 hat j`
If `vec a` and `vec v` are perpendicular, `vec a. vec v = 0`
`(2 hat i + 2 hat j).((2 t - 4) hat i + 2t hat j) = 0`
`8t - 8 = 0, t = 1 sec`.