Q115 / 160 · WBJEE 2009 · Mathematics

A particle is moving in a straight line. At time $t$, the distance between the particle from its starting point is given by $x=t-6 t+z$. Its acceleration will be zero at

A
$t=1$ unit time
Correct answer
B
$t=2$ unit time
C
$t=3$ unit time
D
$t=4$ unit time

Solution

Hints : $x=t-6 t^2+t^3$ $$ \begin{aligned} & \frac{d x}{d t}=1-12 t+3 t^2 \\ & \frac{d^2 x}{d t^2}=-12+6 t \end{aligned} $$ Acceleration $=\frac{d^2 x}{d t^2}$ $\therefore$ Acceleration $=0 \Rightarrow 6 t-12=0 \Rightarrow t=2$

Hence, the correct answer is option: 'B'.

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