Question

The displacement of a particle starting from rest at t=o is given by s=6t*2-t*3. the time at which the particle will attain zero velocity again is

Answer 1

displacement of particle is given by...

s=6t^2-t^3
differentiate wrt time..

ds/dt =12t-3t^2
v=12t-3t^2

velocity will be zero..

12t-3t^2=0
t(12-3t)=0
t=0 or 12-3t=0
t=0 or t=4

Answer 2

$\underline{\large\bf{\mathfrak{Bonjour!}}}$

Relation between s and t is given as;

s = 6t² - t³

To find the time at which the velocity of particle.
First we need to do differentiation of s and t relation.
Then we will get velocity and by placing the velocity = 0 we will the time at which the velocity will be 0.

$= > v = \frac{ds}{dt} = \frac{d(6 {t}^{2} - {t}^{3}) }{dt} \\ \\ = > v = 6\frac{d {t}^{2} }{dt} - \frac{d {t}^{3} }{dt} \\ \\ = > v = 6 {t}^{2 - 1} - {t}^{3 - 1} \\ \\ = > v = 6t - {t}^{2} \\ \\ \\ = > 6t - {t}^{2} = 0 \\ \\ = > t(6 - t) = 0 \\ \\ = > t = 6 sec \: or \: 0sec$

So, the time at which the velocity will be zero is 6 sec(if there is some displacement) but the time can be 0 sec if the body is at rest.

$\bf{\mathfrak{Hope \: this \: helps...:)}}$