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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..........$m/{s^2}$
$28$
$22$
$12$
$10$
Solution
(b) $v = 4{t^3} – 2t$ (given) $\therefore $ $a = \frac{{dv}}{{dt}} = 12{t^2} – 2$
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 ${t^4} – {t^2} = 2$
$⇒$ ${t^4} – {t^2} – 2 = 0$ $({t^2} – 2)\;({t^2} + 1) = 0$ $⇒$ $t = \sqrt 2 \;\sec $
Acceleration at $t = \sqrt {2\;} \;\sec $ given by,
$a = 12{t^2} – 2$$ = 12 \times 2 – 2$= $22\;m/{s^2}$
Similar Questions
For the velocity-time graph shown in the figure, in a time interval from $t=0$ to $t=6\,s$, match the following columns.
| Colum $I$ | Colum $II$ |
| $(A)$ Change in velocity | $(p)$ $-5 / 3\,Sl$ unit |
| $(B)$ Average acceleration | $(q)$ $-20\,SI$ unit |
| $(C)$ Total displacement | $(r)$ $-10\,SI$ unit |
| $(D)$ Acceleration at $t=3\,s$ | $(s)$ $-5\,SI$ unit |
