A particle is moving in a straight line. At time $t$, the distance between the particle from its starting point is given by $x = t - 6t^{2} + t^{3}.Its$ acceleration will be zero at

Question

Dr. Abubacker · Accepted Answer

(b)
Given, $x = t - 6t^{2} + t^{3}$
On differentiating, w. r. t. x, we get
$\frac{dx}{dt} = 1 - 12t + 3t^{2}$
Again, differentiating, we get
$\ \ \frac{d^{2}x}{dt^{2}} = - 12 + 6t\$
When $\ \ \ \frac{d^{2}x}{dt^{2}} = 0 \Rightarrow t = 2$units time

Q. A particle is moving in a straight line. At time $t$, the distance between the particle from its starting point is given by $x = t - 6t^{2} + t^{3}.Its$ acceleration will be zero at

Explanation

Reviewed by Dr. Abubacker

Want more practice?