Question

a particle moving in straight line such that it's displacement at any time t is given by S =t³-6t²+3t +4 m, the velocity when acceleration is zero.

Answer 1

Put second derivative of s equals 0 that is the time when accelaration equals 0
And then put t calculated in first derivative of s. You will get your answer

Answer 2

Correct Question

A particle moves along a straight line such that its displacement at any time t is given by s=t³-6t²+3t+4 find the velocity when acceleration is zero.

Given:

Displacement of particle, s= t³-6t²+3t+4

To Find:

Velocity of given particle when acceleration is 0

Solution:

We know that,

On differentiating displacement with respect to time once, we get the velocity

i.e. $\bf\pink{\dfrac{ds}{dt}=v}$

On differentiating displacement with respect to time twice, we get the acceleration

i.e. $\bf\purple{\dfrac{d^{2}s}{dt^{2}}=a}$

Now,

On differentiating displacement of given particle with respect to time, we get

$\longrightarrow\mathrm{\dfrac{ds}{dt}=3t^{2}-12t+3+0}$

$\longrightarrow\mathrm{\dfrac{ds}{dt}=3t^{2}-12t+3}-----(1)$

$\longrightarrow\mathrm{v=3t^{2}-12t+3}-------(2)$

On differentiating (1) with respect to time again, we get

$\longrightarrow\mathrm{\dfrac{d^{2}s}{dt^{2}}=6t-12+0}$

$\longrightarrow\mathrm{\dfrac{d^{2}s}{dt^{2}}=6t-12}$

$\longrightarrow\mathrm{a=6t-12}-------(3)$

Now,

On putting a=0 in (3), we get

$\longrightarrow\mathrm{0=6t-12}$

$\longrightarrow\mathrm{6t-12=0}$

$\longrightarrow\mathrm{6t=12}$

$\longrightarrow\mathrm{t=\dfrac{12}{6}}$

$\longrightarrow\mathrm{t=2\:s}$

This means, at t=2, acceleration is 0

On putting t=2 in (2), we get

$\longrightarrow\mathrm{v=3(2)^{2}}$

$\longrightarrow\mathrm{v=3(4)-24+3}$

$\longrightarrow\mathrm{v=12-21}$

$\longrightarrow\mathrm{v=-9\:m/s}$

This means, at t=2 or a=0, v is -9 m/s

Hence, required velocity is -9 m/s.