WAEC - Further Mathematics (2019 - No. 21)
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.
12m
16m
64m
96m
توضیح
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