Question

the displacement of a particle moving along x-axis is given as as if x is equal to t^3 - 12tsquare + 4t+ 5 square x is and calculate time at which acceleration is zero​

Answer 1

Explanation:

$find \: time \: when \: acceleration \: will \: \\ be \: zero \\ \\ x = > ( {t}^{3} + 12 {t}^{2} + 4t + 5) \\ \\ if \: you \: double \: differentiate \: it \: \\ we \: will \: get \: acceleration.. \\ \frac{ {d}^{2} x}{ {dt}^{2} } = > \\ 6t + 24 \\ for \: acceleration \: to \: be \: zero = > \\ t = 4sec \\ \\ \\ hope \: it \: helps \: uh....$