Question

A particle moves along a straight line such that its displacement at any time is given by s=(t2-3t2+2)m .The displacement when the acceleration becomes zero is​

Answer 1

Correct Question:

A particle moves along a straight line such that its displacement at any time is given by s=(t³ - 3t² + 2)m .What is the displacement of particle when the acceleration becomes zero?

Answer:

$\boxed{\mathfrak{Displacement \ (s) = zero}}$

Explanation:

Displacement of particle w.r.t. time:

$\rm s = {t}^{3} - 3 {t}^{2} + 2$

Double differentiation of displacement-time relation gives accelration:

$\rm \implies a = \dfrac{ {d}^{2} s}{d {t}^{2} } \\ \\ \rm \implies a = \dfrac{ {d}^{2} }{d {t}^{2} } ( {t}^{3} - 3 {t}^{2} + 2) \\ \\ \rm \implies a = 6t - 6 \\ \\ \rm \implies a = 6(t - 1)$

When Acceleration (a) = 0

$\rm \implies 0 = 6(t - 1) \\ \\ \rm \implies t - 1 = 0 \\ \\ \rm \implies t = 1s$

So, displacement (s) when accelration becomes zero:

$\rm \implies s = {1}^{3} - 3( {1}^{2}) + 2 \\ \\ \rm \implies s =1 - 3 + 2 \\ \\ \rm \implies s =3 - 3 \\ \\ \rm \implies s =0$

Note: As the displacement of particle is zero it means that particle have returned to its initial position.

Answer 2

GIVEN :-

• A ᴘᴀʀᴛɪᴄʟᴇ ᴍᴏᴠᴇs ᴀʟᴏɴɢ ᴀ sᴛʀᴀɪɢʜᴛ ʟɪɴᴇ .

• Tʜᴇ ᴅɪsᴘʟᴀᴄᴇᴍᴇɴᴛ ᴏғ ᴛʜᴇ ᴘᴀʀᴛɪᴄʟᴇ ᴀᴛ ᴀɴʏ ᴛɪᴍᴇ ɪs ɢɪᴠᴇɴ ʙʏ "S = [t³ - 3t² + 2] m" .

• Aᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ = 0 m/s² .

TO FIND :-

• Tʜᴇ ᴅɪsᴘʟᴀᴄᴇᴍᴇɴᴛ ᴡʜᴇɴ ᴛʜᴇ ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ ʙᴇᴄᴏᴍᴇs ᴢᴇʀᴏ .

SOLUTION :-

ᴡᴇ ʜᴀᴠᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

$\huge\red\checkmark$ $\bf\purple{\dfrac{dS}{dt}\:=\:Velocity\:(v)\:}$

$\huge\red\checkmark$ $\bf\purple{\dfrac{dv}{dt}\:=\:Acceleration\:(a)\:}$

☯︎ Iᴛ ᴍᴇᴀɴs, ᴛʜᴇ ᴅᴏᴜʙʟᴇ ᴅɪғғᴇʀᴇɴᴄɪᴀᴛɪᴏɴ ᴏғ Dɪsᴘʟᴀᴄᴇᴍᴇɴᴛ ᴇǫᴜᴀᴛɪᴏɴ ɪs ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ .

$\bf{:\implies\:\dfrac{d^2S}{dt^2}\:=\:Acceleration\:}$

$\rm{:\implies\:\dfrac{d}{dt}\:\left[\dfrac{d(t^3\:-\:3t^2\:+\:2)}{dt}\right]\:=\:Acceleration\:}$

$\rm{:\implies\:\dfrac{d}{dt}\:\left[\dfrac{dt^3}{dt}\:-\:3\times{\dfrac{dt^2}{dt}}\:+\:2\times{\dfrac{d1}{dt}}\right]\:=\:Acceleration\:}$

$\rm{:\implies\:\dfrac{d}{dt}\:\left[3t^2\:-\:6t\:+\:0\right]\:=\:Acceleration\:}$

$\rm{:\implies\:3\times{\dfrac{dt^2}{dt}}\:-\:6\times{\dfrac{dt}{dt}}\:=\:Acceleration\:}$

$\bf\pink{:\implies\:6t\:-\:6\:=\:Acceleration\:}$

☯︎ Gɪᴠᴇɴ ᴛʜᴀᴛ, ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ ɪs ᴢᴇʀᴏ .

$\rm{:\implies\:6t\:-\:6\:=\:0\:}$

$\rm{:\implies\:6t\:=\:6\:}$

$\rm{:\implies\:t\:=\:\dfrac{6}{6}\:}$

$\bf\green{:\implies\:t\:=\:1\:second}$

☯︎ Nᴏᴡ, ᴘᴜᴛᴛɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏғ 't' ɪɴ ᴛʜᴇ ɢɪᴠᴇɴ ᴅɪsᴘʟᴀᴄᴇᴍᴇɴᴛ ᴇǫᴜᴀᴛɪᴏɴ .

➳ S = 1³ - (3 × 1²) + 2

➳ S = 1 - (3 × 1) + 2

➳ S = 3 - 3

➳ S = 0

$\huge\red\therefore$ Tʜᴇ ᴅɪsᴘʟᴀᴄᴇᴍᴇɴᴛ ᴡʜᴇɴ ᴛʜᴇ ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ ʙᴇᴄᴏᴍᴇs ᴢᴇʀᴏ is "0 m" .