Question

position(x) of an object is related to time as x=t^3-12t where (x) is in metre and (t) in second. find the acceleration of the object when it's velocity is zero​

Answer 1

Answer:

12 $\frac{m}{s^{2} }$

Explanation:

x = $t^{3}$ - 12t

$\frac{dx}{dt}$ = $\frac{d}{dt}$ ($t^{3}$ - 12t)

v = 3$t^{2}$ - 12

$\frac{dv}{dt}$ = $\frac{d}{dt}$ (3$t^{2}$ - 12)

a = 6t

v = 3$t^{2}$ - 12 = 0

t = 2s

a = 6t = 12 $\frac{m}{s^{2} }$