A body starts from the origin and moves along | vedclass

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Question

(b) $v = 4{t^3} - 2t$ (given) $	herefore $ $a = \frac{{dv}}{{dt}} = 12{t^2} - 2$

and $x = \int_0^t {v\;dt} = \int_0^t {(4{t^3} - 2t)} \;dt = {t^4}

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(b) $v = 4{t^3} - 2t$ (given) $	herefore $ $a = \frac{{dv}}{{dt}} = 12{t^2} - 2$

and $x = \int_0^t {v\;dt} = \int_0^t {(4{t^3} - 2t)} \;dt = {t^4} - {t^2}$

When particle is at 2m from the origin ${t^4} - {t^2} = 2$

$&rArr;$  ${t^4} - {t^2} - 2 = 0$ $({t^2} - 2)\;({t^2} + 1) = 0$ $&rArr;$ $t = \sqrt 2 \;\sec $

Acceleration at $t = \sqrt {2\;} \;\sec $ given by,

$a = 12{t^2} - 2$$ = 12 	imes 2 - 2$= $22\;m/{s^2}$

A body starts from the origin and moves along the $X-$axis such that the velocity at any instant is given by $(4{t^3} - 2t)$, where $t$ is in sec and velocity in$m/s$. What is the acceleration of the particle, when it is $2\, m$ from the origin..........$m/{s^2}$

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