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.
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Correct Answer: Option 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\(^{-2}\)
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\(^{-2}\)