Question

The position ${}^{\prime }{x}^{\prime }$ of the particle as a function of time ${}^{\prime }{t}^{\prime }$ is given by, $x=2t^3-6t^2+12t+6$ ($t$ is in seconds). The acceleration of the particle is zero at $t=$ _____ $\text{s}$.

• A. 1
• B. 2
• C. 3
• D. 0.5

Answer 1

The correct option is A 1

Velocity of particle $=v=\frac{dx}{dt}=6t^2-12t+12$
Acceleration of particle $=a=\frac{dv}{dt}=12t-12$
If acceleration $=0\Rightarrow 12t-12=0$
$\Rightarrow t=\frac{12}{12}=1\text{s}$