Question

find position X velocity V and acceleration a with help of derivation

Answer 1

By differentiation of "x" we get "v" similarly when we differentiate "v" we get "a" when we substitute t = 2 we get the following results

Answer 2

$x = {t}^{4} - 12{t}^{2} - 40 \\ At \: t=2s,\\ x={2}^{4}-12{(2)}^2-40 \\ x=-72m \\ v = \frac{dx}{dt} \\ v = \frac{d}{dt}( {t}^{4} - 12{t}^{2} - 40) \\ v = 4 {t }^{3} - 24t \: \: \: \: \: \: (i)\\ at \: t = 2s \\ v = 4( {2)}^{3} - 24t \\ v = -16 \frac{m}{s}$
Using (i),
$a = \frac{dv}{dt} \\ a = \frac{d}{dt} (4 {t}^{3} - 24t) \\ a = 12{t}^{2}-24 \\ at \: t = 2s \\ a = 12{(2)}^{2}-24 \\ a = 24 \frac{m}{ {s}^{2} }$