Question

The motion of a particle in rectilinear motion is defined by the relation s=2t^3 -9t^2+12t - 10 where s is expressed in metres and t in seconds, Find the acceleration of the particle when the velocity is zero?​

Answer 1

Answer:

$\large\boxed{\sf{-6\:m{s}^{-2}\:,\:+6\:m{s}^{-2}}}$

Explanation:

The motion of particle in rectilinear motion is defined by the relation,

$\sf{s = 2 {t}^{3} - 9 {t}^{2} + 12t - 10}$

where, s is in metres and t in seconds.

Differentiating both sides wrt 't',

$\sf{= > \dfrac{ds}{dt} = \dfrac{d}{dt} (2 {t}^{3} - 9 {t}^{2} + 12t - 10) }\\ \\ \sf{= > v = 6 {t}^{2} - 18t + 12}$

where v is velocity in $m{s}^{-1}$

It's given that, Velocity is Zero.

$\sf{= > v = 0 }\\ \\ \sf{= > 6 {t}^{2} - 18t + 12 = 0}\\ \\ \sf{ = > 6 {t}^{2} - 6t - 12t + 12 = 0 }\\ \\ \sf{ = > 6t(t - 1) - 12(t - 1) = 0 }\\ \\ \sf{ = > (t - 1)(6t - 12) = 0} \\ \\ \sf{= > t = 1 \: \: and \: \: t = 2}$

Thus, Velocity will be zero at 1 sec. and 2 sec.

Again differentiating wrt 't',

$\sf{= > \dfrac{dv}{dt} = \dfrac{d}{dt} (6 {t}^{2} - 18t + 12) }\\ \\ \sf{= > a = 12t - 18}$

Therefore, acceleration will be,

$\sf{= > a =(( 12 \times 1 )- 18 )\: \: and \: \:( (12 \times 2) - 18)} \\ \\ \sf{ = > a = (12 - 18) \: \: and \: \: (24 - 18) }\\ \\ \sf{ = > a = - 6 \: m {s}^{ - 2} \: \: and \: \: 6 \: m {s}^{ - 2} }$