A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by `(4t^3-2t)`, where t is in sec and velocity in m/s. what is the acceleration of the particle when it is 2 m from the origin?

Options

A. `28 m//s^(2) `
B. ` 22 m//s^(2) `
C. ` 12 m//s^(2) `
D. ` 10 m//s^(2)`

Answer

B

Explanation

` " "` given `" " a= (dv)/( dt) = 12 t^(2) -2 and x =int _0^(t) v dt =int _0^(t)( 4t^(3) -2t ) dt = t^(4) -t^(2) `
When particle is at 2m from the origin ` t^(4) -t^(2) =2, rArr t^(4) -t^(2) -2= 0rArr (t^(2) -2) (t^(2)+1) =0`
` rArr t= sqrt 2 sec ` Acceleration at t ` =sqrt 2 sec ` is given by ,` a=12t^(2) -2 =12 xx 2 -2=22 m//s^(2) `