Search any question & find its solution
Question:
Answered & Verified by Expert
A particle moves along $x$ -axis and its displacement at any time is given by $x(t)=2 t^{3}-3 t^{2}+4 t$ in SI units. The velocity of the particle when its acceleration is zero, is
Options:
Solution:
1194 Upvotes
Verified Answer
The correct answer is:
$2.5 \mathrm{ms}^{-1}$
Given, $x(t)=2 t^{3}-3 t^{2}+4 t$
So, velocity, $v=\frac{d x}{d t}=\left(6 t^{2}-6 t+4\right)$
and acceleration $a=\frac{d v}{d t}=(12 t-6)$
when acceleration is zero, $i . e . .(12 t-6)=0$
or
$$
t=\frac{6}{12}=\frac{1}{2} \mathrm{s}
$$
$\therefore$ Velocity of the particle at zero acceleration is
$$
v=6\left(\frac{1}{2}\right)^{2}-6\left(\frac{1}{2}\right) 4=2.5 \mathrm{m} / \mathrm{s}
$$
So, velocity, $v=\frac{d x}{d t}=\left(6 t^{2}-6 t+4\right)$
and acceleration $a=\frac{d v}{d t}=(12 t-6)$
when acceleration is zero, $i . e . .(12 t-6)=0$
or
$$
t=\frac{6}{12}=\frac{1}{2} \mathrm{s}
$$
$\therefore$ Velocity of the particle at zero acceleration is
$$
v=6\left(\frac{1}{2}\right)^{2}-6\left(\frac{1}{2}\right) 4=2.5 \mathrm{m} / \mathrm{s}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.