Question

A particle moves along a straight line such that its position $x$ at any time $t$ is $x=6t^2-t^3$ . Where $x$ is in metre and $t$ is in second, then

• A. at $t=0$ acceleration is $12m{s}^{-2}$
• B. $x-t$ curve has maximum at $4s$
• C. Both (a) and (b) are wrong
• D. Both (a) and (b) are correct

Answer 1

Answer: (D)

Given, $x=6t^2-t^3$
$\frac{dx}{dt}=12t-3t^2...(i)$
$\frac{dx}{dt}=0$
$\Rightarrow t=4s$
Now, again differentiating Eq. (i), we get
$\frac{d^{2}x}{d{t}^2}=12-6t$
$=12-6(4)=-12$
Since, $\frac{d^{2}x}{d{t}^2}$ is negative, hence $t=4s$ gives the maximum value for $x-t$ curve.
Moreover, acceleration $a=\frac{d^{2}x}{d{t}^2}$, at $t=0$, $\frac{d^{2}x}{d{t}^2}=12m{s}^{-2}$