Question

The position of the particle is given by $x=3t^2-6t+4$ in $\text{m}$. The acceleration of the particle at $10\text{sec}$ is

• A. $5m/s^2$
• B. $6m/s^2$
• C. $4m/s^2$
• D. $10m/s^2$

Answer 1

The correct option is B $6m/s^2$

As we know that the acceleration of the particle is given by
$a=\frac{dv}{dt}$ and also, $v=\frac{dx}{dt}$
Given, $x=3t^2-6t+4$
So, $v=\frac{dx}{dt}=\frac{d(3t^2-6t+4)}{dt}$
$\Rightarrow v=6t-6$
also, $a=\frac{dv}{dt}=\frac{d(6t-6)}{dt}=6$
Thus, the acceleration of the particle is constant i.e $6m/s^2$