Question

S= t³-6t² . Find the acceleration when body is at rest .

{ kinematics}

Answer 1

$velocity = \frac{ds}{dt} \\ = \frac{d( {t}^{3} - 6 {t}^{2}) }{dt} \\ v= 3 {t}^{2} - 12t$

when body is at rest v=0

then,

t-4=0

t=4


$accleration = \frac{velocity}{time} \\ = \frac{dv}{dt} \\ = \frac{d(3 {t}^{2} - 12t)}{dt} \\ = > 6t - 12$


acceleration=6(4)-12

=>24-12=12