Velocity of a particle is given by` v = (3t ^(2) +2t) m/s.` Find its average velocity between t = 0 to t=3s and also find its acceleration at t = 3s. Motion of the particle is in one dimension.

Options

A. `11 m /s, 10 (m)/(s ^(2))`
B. `12 (m)/(s), 20 (m)/(s ^(2))`
C. `11 (m)/(s) , 20 (m)/(s^(2))`
D. None of the given

Answer

B

Explanation

`v = (3t ^(2) + 2t)`
`therefore int v dx = int (3t ^(2) + 2r ) dt`
`therefore x = 3 (t ^(3))/(3) + 2 (t ^(2))/( 2) = t ^(3) + t ^(2)`
At `t =0` time ,` x _(0) = (0) ^(2) + (0) ^(2) =0`
At t =3 time, `x _(3) = (3) ^(2) + (3) ^(2) = 36 m`
`therefore` Average velocity `= (x _(3) -x _(0))/(3-0)=(36-0)/(3-0)`
`=12 ms ^(-1)` Acceleration of particle,
`a = (dv)/(dt) = (d)/(dt) (3t ^(2) + 2t) = 6t +2`
Put `t =3s,`
`a =6(3) +2=20 ms ^(-2)`