Question

given s= t^3 -6t^2 find int. acceleration when body is at rest​

Answer 1

GIVEN :–

• S = t³ - 6t²

TO FIND :–

• Acceleration when body is at rest .

SOLUTION :–

$\\ \implies \bf S= {t}^{3} - 6 {t}^{2}$

• Differentiate with respect to 't' –

$\\ \implies \bf \dfrac{dS}{dt}= \dfrac{d({t}^{3})}{dt} - 6 \dfrac{d({t}^{2})}{dt}$

$\\ \implies \bf \dfrac{dS}{dt}= 3 {t}^{2} - 12t$

• Body is in rest position –

$\\ \implies \bf \dfrac{dS}{dt}=0$

$\\ \implies \bf 3 {t}^{2} - 12t=0$

$\\ \implies \bf t(3t- 12)=0$

$\\ \implies \bf t = 0,t=4$

• Again Differentiate with respect to 't' –

$\\ \implies \bf \dfrac{d^{2} S}{dt ^{2} }=3(2)t - 12$

$\\ \implies \bf \dfrac{d^{2} S}{dt^{2} }=6t - 12$

▪︎When t = 0 :–

$\\ \implies \bf \left( \dfrac{d^{2} S}{dt ^{2} } \right) _{t = 0}=6(0) - 12$

$\\ \implies \bf Acceleration = - 12\: \dfrac{m}{ {s}^{2} }$

▪︎ When t = 4 :–

$\\ \implies \bf \left( \dfrac{d^{2} S}{dt ^{2} } \right) _{t =4}=6(4) - 12$

$\\ \implies \bf \left( \dfrac{d^{2} S}{dt ^{2} } \right) _{t =4} = 24- 12$

$\\ \implies \bf Acceleration = 12 \: \dfrac{m}{ {s}^{2} }$