Question

(a) If ` s = 2t^(3)+ 3t^(2) + 2t + 8` then find time at which acceleration is zero.
(b) Velocity of a particle (starting at t =0) varies with time as v= 4t. Calculate the displacement of part between t = 2 to t = 4 s

Answer 1

(a) `v = (ds)/(dt) = 6t^(2) + 6t+ 2 rArr a = (dv)/(dt) = 12t = 6 =0 rArr t =-(1)/(2)` which is impossible. There acceleration can never be zero.
(b) `because (dx)/(dt) = v therefore x = int v dt = overset4 underset2 int 4tdt = [2t^(2) ] _(2)^(4) = 2(4)^(2)- 2(2)^(2) = 32 - 8 = 24` m