Question

s=t^3 - 6t^2 find the displacement when the body is at rest​

Answer 1

Answer:

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Answer 2

GIVEN :–

• S = t³ - 6t²

TO FIND :–

• Acceleration when body is at rest .

SOLUTION :–

\begin{gathered}\\ \implies \bf S= {t}^{3} - 6 {t}^{2} \\\end{gathered}⟹S=t3−6t2

• Differentiate with respect to 't' –

\begin{gathered}\\ \implies \bf \dfrac{dS}{dt}= \dfrac{d({t}^{3})}{dt} - 6 \dfrac{d({t}^{2})}{dt} \\\end{gathered}⟹dtdS=dtd(t3)−6dtd(t2)

\begin{gathered}\\ \implies \bf \dfrac{dS}{dt}= 3 {t}^{2} - 12t \\\end{gathered}⟹dtdS=3t2−12t

• Body is in rest position –

\begin{gathered}\\ \implies \bf \dfrac{dS}{dt}=0\\\end{gathered}⟹dtdS=0

\begin{gathered}\\ \implies \bf 3 {t}^{2} - 12t=0\\\end{gathered}⟹3t2−12t=0

\begin{gathered}\\ \implies \bf t(3t- 12)=0\\\end{gathered}⟹t(3t−12)=0

\begin{gathered}\\ \implies \bf t = 0,t=4\\\end{gathered}⟹t=0,t=4

• Again Differentiate with respect to 't' –

\begin{gathered}\\ \implies \bf \dfrac{d^{2} S}{dt ^{2} }=3(2)t - 12 \\\end{gathered}⟹dt2d2S=3(2)t−12

\begin{gathered}\\ \implies \bf \dfrac{d^{2} S}{dt^{2} }=6t - 12 \\\end{gathered}⟹dt2d2S=6t−12

▪︎When t = 0 :–

\begin{gathered}\\ \implies \bf \left( \dfrac{d^{2} S}{dt ^{2} } \right) _{t = 0}=6(0) - 12 \\\end{gathered}⟹(dt2d2S)t=0=6(0)−12

\begin{gathered}\\ \implies \bf Acceleration = - 12\: \dfrac{m}{ {s}^{2} } \\\end{gathered}⟹Acceleration=−12s2m

▪︎ When t = 4 :–

\begin{gathered}\\ \implies \bf \left( \dfrac{d^{2} S}{dt ^{2} } \right) _{t =4}=6(4) - 12 \\\end{gathered}⟹(dt2d2S)t=4=6(4)−12

\begin{gathered}\\ \implies \bf \left( \dfrac{d^{2} S}{dt ^{2} } \right) _{t =4} = 24- 12 \\\end{gathered}⟹(dt2d2S)t=4=24−12

\begin{gathered}\\ \implies \bf Acceleration = 12 \: \dfrac{m}{ {s}^{2} } \\\end{gathered}⟹Acceleration=12s2m