A particle starts from rest and moves in a straight line such that its velocity, V ms\(^{-1}\), at time t second is given by V = 3t\(^2\) - 6t. Calculate the acceleration in the 3rd second. 

A.

12m

B.

16m

C.

64m

D.

96m

Correct answer is B

V = 3t\(^2\) - 6t

\(\frac{ds}{dt} = 3t^2 - 6t\)

s = \(\int 3t^2 - 6t\)

s = \(\frac{3t^3}{3} - \frac{6t^2}{2} + k\)

s = t\(^3\) - 3t\(^2\) + k

s = 0, t = 0

s = t\(^3\) - 3t\(^2\) 

s = 4\(^3\) - 3t\(^2\) 

s = 4\(^3\) - 3(4)\(^2\) 

= 64 - 48 = 16m