Question

If displacement of a particle is given as x= 3t^2 -4t, in
where x is in cms & t is in second find out (i) Time when velocity of the article is zero, (ii) Initial velocity (ii) velocity after 5 second (iv) Acceleration of the particle.

Answer 1

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Answer 2

The displacement of the particle is,

$\longrightarrow\sf{x=3t^2-4t}$

Then the velocity of the particle is,

$\longrightarrow\sf{v=\dfrac{dx}{dt}}$

$\longrightarrow\sf{v=\dfrac{d}{dt}\,\left[3t^2-4t\right]}$

$\longrightarrow\sf{v=6t-4}$

The time at which the velocity becomes zero is given by,

$\longrightarrow\sf{v=0}$

$\longrightarrow\sf{6t-4=0}$

$\longrightarrow\sf{\underline{\underline{t=\dfrac{2}{3}\ s}}}$

$\longrightarrow\sf{\underline{\underline{t=0.667\ s}}}$

Initial velocity is attained at $\sf{t=0.}$ Then initial velocity is,

$\longrightarrow\sf{u=v(0)}$

$\longrightarrow\sf{u=6(0)-4}$

$\longrightarrow\sf{\underline{\underline{u=-4\ cm\,s^{-1}}}}$

Velocity after 5 seconds is,

$\longrightarrow\sf{v(5)=6(5)-4}$

$\longrightarrow\sf{\underline{\underline{v(5)=26\ cm\,s^{-1}}}}$

Acceleration of the particle is,

$\longrightarrow\sf{a=\dfrac{dv}{dt}}$

$\longrightarrow\sf{a=\dfrac{d}{dt}\,[6t-4]}$

$\longrightarrow\sf{\underline{\underline{a=6\ cm\,s^{-2}}}}$