(a) Differentiate \(v\) to find the acceleration:
\[
a=\frac{dv}{dt}
=\frac{2k_1}{(4t+1)^{\frac12}}-(2t+1).
\]
When \(t=1.25\),
\[
4t+1=6
\]
and
\[
2t+1=3.5.
\]
The deceleration is \(0.5\text{ m s}^{-2}\), so
\[
a=-0.5.
\]
Hence
\[
\frac{2k_1}{\sqrt6}-3.5=-0.5.
\]
\[
\frac{2k_1}{\sqrt6}=3.
\]
Therefore
\[
k_1=\frac{3\sqrt6}{2}.
\]
Answer: \(k_1=\frac{3\sqrt6}{2}\).
(b) Since \(P\) is at instantaneous rest when \(t=3.5\),
\[
v=0
\]
when \(t=3.5\).
So
\[
0=k_1(15)^{\frac12}-\frac14(8)^2+k_2.
\]
Using \(k_1=\frac{3\sqrt6}{2}\),
\[
0=\frac{3\sqrt6}{2}\sqrt{15}-16+k_2.
\]
\[
0=\frac{9\sqrt{10}}{2}-16+k_2.
\]
Therefore
\[
k_2=16-\frac{9\sqrt{10}}{2}.
\]
The particle is not at instantaneous rest in \(0\leq t\leq1\), so the distance travelled is
\[
\int_0^1 v\,dt.
\]
Thus
\[
\int_0^1\left(k_1(4t+1)^{\frac12}-\frac14(2t+1)^2+k_2\right)\,dt.
\]
\[
=\left[\frac{k_1}{6}(4t+1)^{\frac32}-\frac1{24}(2t+1)^3+k_2t\right]_0^1.
\]
\[
=\frac{k_1}{6}\left(5^{\frac32}-1\right)-\frac{13}{12}+k_2.
\]
Substitute \(k_1=\frac{3\sqrt6}{2}\) and \(k_2=16-\frac{9\sqrt{10}}{2}\):
\[
\text{distance}
=
\frac{\sqrt6}{4}(5\sqrt5-1)-\frac{13}{12}+16-\frac{9\sqrt{10}}{2}.
\]
\[
=6.9205\ldots
\]
Answer: \(6.92\text{ m}\).