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: C
Explanation
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\(^{-2}\)
