Question

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

Answer 1

✴ Correct Question :

A particle moves along a straight line such that its displacement at any time t is given by $\bf{s=t^3-6t^2+3t+4}$ find the velocity when acceleration is zero.

✴ Solution :

✏ We know that,

$\bigstar\bf\:v=\dfrac{ds}{dt}\\ \\ \bigstar\bf\:a=\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2}$

▪ Acceleration of particle :

$\longrightarrow\sf\:a=\dfrac{d^2s}{dt^2}\\ \\ \longrightarrow\sf\:a=\dfrac{d^2(t^3-6t^2+3t+4)}{dt^2}\\ \\ \longrightarrow\sf\:a=6t-12\\ \\ \longrightarrow\sf\:ATQ,\:a=0\\ \\ \longrightarrow\sf\:a=6t-12=0\\ \\ \longrightarrow\sf\:6t=12\\ \\ \longrightarrow\boxed{\bf{\large{t=2\:s}}}$

▪ Velocity of particle at t = 2s

$\longrightarrow\sf\:v=\dfrac{ds}{dt}\\ \\ \longrightarrow\sf\:v=\dfrac{d(t^3-6t^2+3t+4)}{dt}\\ \\ \longrightarrow\sf\:v=3t^2-12t+3\\ \\ \longrightarrow\sf\:v=3(2)^2-12(2)+3\\ \\ \longrightarrow\sf\:v=12-24+3\\ \\ \longrightarrow\boxed{\bf{\red{v=-9\:mps}}}$

▶ Additional information :

⏭ Velocity is a vector quantity.

⏭ It can be positive, negative and zero.

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

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

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

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

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

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

Hence, required velocity is -9 m/s.