Answer: (a) \(\displaystyle v=\frac{(2t+1)^{1/2}}{(t+1)^2}(t+2)\), so \(a=2\), \(b=1\); (b) \(P\) is never instantaneously at rest for \(t\gt 0\).

Answer: (a) \(\displaystyle v=\frac{(2t+1)^{1/2}}{(t+1)^2}(t+2)\), so \(a=2\), \(b=1\); (b) \(P\) is never instantaneously at rest for \(t\gt 0\).

(a) The displacement is

\(s=\frac{(2t+1)^{3/2}}{t+1}-1.\)

Differentiate using the quotient rule. Let

\(u=(2t+1)^{3/2},\qquad v=t+1.\)

Then

\(u'=\frac32(2t+1)^{1/2}\cdot2=3(2t+1)^{1/2},\)

and

\(v'=1.\)

Therefore the velocity is

\(\frac{(t+1)3(2t+1)^{1/2}-(2t+1)^{3/2}}{(t+1)^2}.\)

Factor out \((2t+1)^{1/2}\):

\(\frac{(2t+1)^{1/2}\left(3(t+1)-(2t+1)\right)}{(t+1)^2}.\)

Simplify the bracket:

\(3(t+1)-(2t+1)=t+2.\)

Hence

\(v=\frac{(2t+1)^{1/2}}{(t+1)^2}(t+2).\)

So

\(a=2,\qquad b=1.\)

(b) After passing through \(O\), \(t\gt 0\). For \(t\gt 0\),

\((2t+1)^{1/2}\gt 0,\qquad (t+1)^2\gt 0,\qquad t+2\gt 0.\)

Therefore

\(v\gt 0.\)

So the velocity is never zero, and \(P\) is never at instantaneous rest after passing through \(O\).