Question

A particle moves along x axis and its displacement at any time is given by x is equal to 2t^3-3t^2+4t . The velocity of the particle when its acceleration is zero is

Answer 1

Answer:

explanation is in image

Answer 2

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GiveN :-

• A particle moves along x axis and its displacement at any time is given by x is equal to 2t³- 3t² + 4t .

To FinD :-

• The velocity of the particle when its acceleration is 0 .

AnsweR :-

$\underline{\red{\sf \mathscr{A}ccording \ \mathscr{T}o\ \mathscr{Q} uestion :- }}$

$\tt:\implies x(t) = 2t^3 - 3t^2 + 4t \\\\\tt:\implies \dfrac{d}{dt}x(t) = \dfrac{d}{dt}(2t^3-3t^2+4t) \:\:\bigg\lgroup \bf {Differnciating \ both \ sides \ wrt \ t .}\bigg\rgroup \\\\\tt:\implies \dfrac{d}{dt}x(t)= 2\times 3t^{(3-1)}-2\times 3t^{(2-1)}+4t^{(1-1)}\\\\\boxed{\green{\tt \leadsto v(t) = 6t^2-6t+4 }}$

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$\underline{\underline{\pink{\sf Again \ differenciating \ both \ sides \ w.r.t \ t }}}$

$\tt:\implies v(t) = 6t^2-6t+4\\\\\tt:\implies \dfrac{d}{dt}v(t) = \dfrac{d}{dt}(6t^2-6t+4) \\\\\tt:\implies a(t) = 12t^{(2-1)}-6t^{(-1-1)}+0 \\\\\boxed{\orange{\tt\leadsto a = 12t - 6 }}$

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$\underline{\underline{\blue{\sf Putting \ a \ = \ 0 :- }}}$

$\tt:\implies a = 12t - 6 \\\\\tt:\implies 12t - 6 = 0 \\\\\tt:\implies 12t = 0+6 \\\\\tt:\implies 12t = 6 \\\\\tt:\implies t = \dfrac{6}{12} \\\\\underline{\boxed{\red{\tt\longmapsto Time \:\:=\:\:\dfrac{1}{2}s }}}$

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$\underline{\underline{\blue{\sf Putting\ this \ in \ v :- }}}$

$\tt:\implies v = 6t^2-6t+4 \\\\\tt:\implies v = 6\times \bigg( \dfrac{1}{2}\bigg)^2-6\times \dfrac{1}{2}+4 \\\\\tt:\implies v = \bigg(6\times \dfrac{1}{4}\bigg) - 3 + 4 \\\\\tt:\implies v= \dfrac{3}{2}+1 \\\\\underline{\boxed{\orange{\tt \longmapsto Velocity = 2.5 ms^{-1} }}}$

$\boxed{\green{\pink{\dag}\:\: \bf Hence \ the \ Velocity \ will \ be \ 2.5 m/s . }}$

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