Question

The position $x$ of a body as a function of time $t$ is given by the equation:
$x=2t^3-6t^2+12t+16$
The acceleration of the body is zero at time $t$ is equal to

• A. $1\text{s}$
• B. $2\text{s}$
• C. $3\text{s}$
• D. $0.5\text{s}$

Answer 1

The correct option is A $1\text{s}$

$x=2t^3-6t^2+12t+6$
$v=\frac{dx}{dt}$
$v=6t^2-12t+12$
Acceleration $=\frac{d}{dt}\left(v\right)=12t-12$
$0=12t-12$
$t=1\text{s}$