Answer: \(v=\frac{8t}{4t^2-5}-1\), \(a=-\frac{8(4t^2+5)}{(4t^2-5)^2}\); at rest \(t=\frac52\); acceleration \(=-\frac35\).

We start with the main method. Use \(v=ds/dt\) and \(a=dv/dt\), then substitute the requested value of time.

(a) Since

\(s=\ln(4t^2-5)-t,\)

differentiate with respect to \(t\):

\(v=\frac{\mathrm ds}{\mathrm dt}=\frac{8t}{4t^2-5}-1.\)

Now differentiate \(v\) to find the acceleration.

For \(\frac{8t}{4t^2-5}\), use the quotient rule:

\(\frac{\mathrm d}{\mathrm dt}\left(\frac{8t}{4t^2-5}\right) =\frac{8(4t^2-5)-8t(8t)}{(4t^2-5)^2}.\)

So

\(a=\frac{32t^2-40-64t^2}{(4t^2-5)^2}.\)

Hence

\(a=-\frac{32t^2+40}{(4t^2-5)^2} =-\frac{8(4t^2+5)}{(4t^2-5)^2}.\)

(b) The particle is at rest when \(v=0\):

\(\frac{8t}{4t^2-5}-1=0.\)

Thus

\(8t=4t^2-5,\)

so

\(4t^2-8t-5=0.\)

Factorising,

\((2t-5)(2t+1)=0.\)

Since \(t\gt2\),

\(t=\frac52.\)

(c) Substitute \(t=\frac52\) into the expression for acceleration:

\(a=-\frac{8\left(4\left(\frac52\right)^2+5\right)}{\left(4\left(\frac52\right)^2-5\right)^2}.\)

Now

\(4\left(\frac52\right)^2=25.\)

So

\(a=-\frac{8(25+5)}{(25-5)^2} =-\frac{240}{400}=-\frac35.\)

This completes the solution and gives the required result.