The displacement of a particle is given by y | vedclass

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(c) $y = a + bt + c{t^2} - d{t^4}$

$	herefore \;v = \frac{{dy}}{{dt}} = b + 2ct - 4d{t^3}$ and $a = \frac{{dv}}{{dt}} = 2c - 12d{t^2}$

Hence, a

Vedclass · Accepted Answer

$b,\,2c$

Vedclass · Answer

(c) $y = a + bt + c{t^2} - d{t^4}$

$	herefore \;v = \frac{{dy}}{{dt}} = b + 2ct - 4d{t^3}$ and $a = \frac{{dv}}{{dt}} = 2c - 12d{t^2}$

Hence, at $t = 0$, $v_{initial} \,= b$ and $a_{initial}$ $= 2c$.

The displacement of a particle is given by $y = a + bt + c{t^2} - d{t^4}$. The initial velocity and acceleration are respectively

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