A particle is moving on a straight line. The distance \(\mathrm{S}\) travelled in time \(t\) is given by \(S=a t^2+b t+6\). If the particle comes to rest after 4 seconds at a distance of \(16 \mathrm{~m}\). from the starting point, then the acceleration of the particle is.

\(
\begin{aligned}
& \mathrm{S}=\mathrm{at} \mathrm{t}^2+\mathrm{bt}+6 \\
& \therefore \mathrm{V}=\frac{\mathrm{dS}}{\mathrm{dt}}=2 \mathrm{at}+\mathrm{b} \text { and } \mathrm{A}=\frac{\mathrm{d}^2 \mathrm{~S}}{\mathrm{dt}^2}=2 \mathrm{a}
\end{aligned}
\)
When particle comes to rest,
\(
\begin{aligned}
& S=16, \mathrm{t}=4, \mathrm{~V}=0 \\
& \therefore 16=\mathrm{a}(4)^2+\mathrm{b}(4)+6 \\
& \Rightarrow 16 \mathrm{a}+4 \mathrm{~b}=10 \ldots(1)
\end{aligned}
\)
Also \(0=2 \mathrm{a}(1)(2)+4\)
\(
\Rightarrow \mathrm{b}=-8 \mathrm{a}
\)
From (1) and (2), we get
\(
16 a+4(8 a)=10 \Rightarrow-16=10 \Rightarrow a=\frac{-5}{9}
\)
We have acceleration \(\mathrm{A}=2 \mathrm{a}=2\left(\frac{-5}{8}\right)=\frac{-5}{4} \mathrm{~m} / \mathrm{sec}^2\)