(a) Since

\[ v=t-7t^{\frac12}+12, \]

differentiate to find the acceleration:

\[ a=\frac{dv}{dt}=1-\frac72t^{-\frac12}. \]

At \(t=4\),

\[ a=1-\frac72(4^{-\frac12}). \]

\[ a=1-\frac72\cdot\frac12=1-\frac74=-\frac34. \]

Answer: \(-0.75\text{ m s}^{-2}\).

(b) \(P\) is instantaneously at rest when \(v=0\).

\[ t-7t^{\frac12}+12=0. \]

Let \(u=\sqrt t\). Then \(t=u^2\), so

\[ u^2-7u+12=0. \]

\[ (u-3)(u-4)=0. \]

So

\[ u=3\quad\text{or}\quad u=4. \]

Therefore

\[ t=9\quad\text{or}\quad t=16. \]

Answer: \(t=9\text{ s}\) and \(t=16\text{ s}\).

(c) The velocity changes sign at \(t=9\) and \(t=16\).

First find the displacement function:

\[ s=\int \left(t-7t^{\frac12}+12\right)\,dt. \]

\[ s=\frac12t^2-\frac{14}{3}t^{\frac32}+12t. \]

Evaluate:

\[ s(0)=0, \]

\[ s(9)=\frac{45}{2}, \]

\[ s(16)=\frac{64}{3}, \]

\[ s(25)=\frac{175}{6}. \]

Total distance is

\[ \left(s(9)-s(0)\right)+\left(s(9)-s(16)\right)+\left(s(25)-s(16)\right). \]

\[ =\frac{45}{2}+\left(\frac{45}{2}-\frac{64}{3}\right)+\left(\frac{175}{6}-\frac{64}{3}\right). \]

\[ =\frac{63}{2}=31.5. \]

Answer: \(31.5\text{ m}\).