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: B
Explanation
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
