2.Motion in Straight Line
hard

A particle is moving in a straight line. The variation of position $x$ as a function of time $t$ is given as $x=\left(t^3-6 t^2+20 t+15\right) m$. The velocity of the body when its acceleration becomes zero is ........... $m/s$

A $4$
B$8$
C$10$
D$6$
(JEE MAIN-2024)

Solution

$ x=t^3-6 t^2+20 t+15 $
$ \frac{d x}{d t}=v=3 t^2-12 t+20$
$ \frac{d v}{d t}=a=6 t-12$
When $\mathrm{a}=0$
$6 \mathrm{t}-12=0 ; \mathrm{t}=2 \mathrm{sec}$
At $\mathrm{t}=2 \mathrm{sec}$
$\mathrm{v}=3(2)^2-12(2)+20$
$\mathrm{v}=8 \mathrm{~m} / \mathrm{s}$
Standard 11
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.