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)

$94$

Answer 2

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