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 $ 12 \,ms^{−2} $
• B. $ x- t $ curve has maximum at $ 4\,s $
• C. Both (a) and (b) are wrong
• D. Both (a) and (b) are correct

Answer 1

Answer: (D)

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