Answer: \(v=3\mathrm e^{2t}-4\mathrm e^{-2t}-1\), \(a=6\mathrm e^{2t}+8\mathrm e^{-2t}\); \(T=\frac12\ln\frac43\); acceleration \(14\text{ m s}^{-2}\).

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

(a) Since

\(s=1.5\mathrm e^{2t}+2\mathrm e^{-2t}-t,\)

differentiate with respect to \(t\):

\(v=3\mathrm e^{2t}-4\mathrm e^{-2t}-1.\)

Differentiate again:

\(a=6\mathrm e^{2t}+8\mathrm e^{-2t}.\)

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

\(3\mathrm e^{2t}-4\mathrm e^{-2t}-1=0.\)

Let \(u=\mathrm e^{2t}\). Then \(\mathrm e^{-2t}=\frac1u\), so

\(3u-\frac4u-1=0.\)

Multiplying by \(u\),

\(3u^2-u-4=0.\)

Factorising,

\((3u-4)(u+1)=0.\)

Since \(u=\mathrm e^{2t}\gt0\),

\(u=\frac43.\)

Thus

\(\mathrm e^{2T}=\frac43,\)

so

\(T=\frac12\ln\frac43.\)

(c) At time \(T\),

\(\mathrm e^{2T}=\frac43\quad\text{and}\quad \mathrm e^{-2T}=\frac34.\)

Therefore

\(a=6\left(\frac43\right)+8\left(\frac34\right)=8+6=14.\)

The acceleration is \(14\text{ m s}^{-2}\).

This completes the solution and gives the required result.