Question

The motion of a particle is represented by the equation $ S = 8t^3 - 2t^2+ 6t + 7 $ . Find the velocity of the particle at the end of $ 2 $ seconds in $ m s^{-1} $

• A. $ 108 $
• B. $ 57 $
• C. $ 94 $
• D. $ 41 $

Answer 1

Answer: (C)

$S=8t^{3}-2t^{2}+6t+7$
$x=n \frac{\lambda}{t}=8 \left(3t^{2}\right)-2\left(2t\right)+6\left(1\right)+7\left(0\right)$
$=24t^{2}-4t+6$
At $t=2\,s$,
$v=24\left(2\right)^{2}-4\left(2\right)+6$
$=96-8+6$
$=94\,m\,s^{-1}$