Question

The position$\left(x\right)$ of a particle at any time$\left(t\right)$ is given by
$x\left(t\right)=4t^3-3t^2+2$

The acceleration and velocity of the particle at any time $t=2sec$ are respectively

• A. $16{ms}^{-2}$ and $22{ms}^{-1}$
• B. $42{ms}^{-2}$ and $36{ms}^{-1}$
• C. $36{ms}^{-2}$ and $42{ms}^{-1}$
• D. $12{ms}^{-2}$ and $25{ms}^{-1}$

Answer 1

Answer: (B)

$42{ms}^{-2}$ and $36{ms}^{-1}$

We have $x\left(t\right)=4t^3-3t^2+2$

$\Rightarrow v=\frac{dx}{dt}=12t^2-6t$

and $a=\frac{dv}{dt}=24t-6$

$\therefore v$ at $t=2s$ is $12{\left(2\right)}^2-6\left(2\right)$ i.e., $36{ms}^{-1}$

and $a$ at $t=2s$ is $24\times 2-6$ i.e., $42{ms}^{-2}$