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.

0 ms\(^{-2}\)

B.

3 ms\(^{-2}\)

C.

6 ms\(^{-2}\)

D.

9 ms\(^{-2}\)

Correct answer is C

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

\(\frac{dx}{dt}\) = 6r - 6

at t = 4

\(\frac{dx}{dt}\) = 6 x 4 - 6 = 24 - 6

= 18

at t = 3

\(\frac{dy}{dt}\) = 6 x 4 - 6 

= 24 - 6

= 18 

at t = 3

\(\frac{dx}{dt}\) = 6 x 3 - 6 = 18 - 6

= 12

\(\frac{dy}{dt}\) = 18 - 12 

= 6ms\(^{-2}\)