Answer: \(v=2e^{2t}-10e^t-12\), \(a=4e^{2t}-10e^t\); \(t=\ln6=1.79\); acceleration \(84\).

(a) Velocity is

\(v=\frac{ds}{dt}.\)

Differentiate

\(s=e^{2t}-10e^t-12t+9.\)

This gives

\(v=2e^{2t}-10e^t-12.\)

Acceleration is

\(a=\frac{dv}{dt}.\)

So

\(a=4e^{2t}-10e^t.\)

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

\(2e^{2t}-10e^t-12=0.\)

Divide by \(2\):

\(e^{2t}-5e^t-6=0.\)

Factorise:

\((e^t+1)(e^t-6)=0.\)

Since \(e^t\gt0\),

\(e^t=6.\)

Therefore

\(t=\ln6=1.79\)

to 3 significant figures.

(c) At this time, \(e^t=6\), so \(e^{2t}=36\). Hence

\(a=4(36)-10(6)=144-60=84.\)