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