$P$、$Q$、$R$ 和 $S$ 是位于一个岛上的四个村庄，其中 $Q$ 位于 $R$ 的正北方。
$PQ = 924\text{ m}$，$QR = 640\text{ m}$，$PR = 1124\text{ m}$，$RS = 583\text{ m}$。
角 $RPS = 26^\circ$ 且角 $PSR$ 为钝角。

1. Calculate the sum of the squares of the two shorter sides of triangle PQR.

$PQ^2 + QR^2 = 924^2 + 640^2$
$= 853776 + 409600$
$= 1263376$

2. Calculate the square of the longest side and compare.

$PR^2 = 1124^2 = 1263376$
Since $PQ^2 + QR^2 = PR^2$, by the converse of Pythagoras' theorem, $\triangle PQR$ is a right-angled triangle with $\angle PQR = 90^\circ$.

3. Use the bearing information to conclude the relative position of P.

We are given that $Q$ is due north of $R$, which means the line segment $QR$ lies on the North-South axis.
Since $\angle PQR = 90^\circ$ and from the given layout $P$ is to the right of the North-South line $QR$, $P$ must be due east of $Q$.

1. Find angle PRQ using right-angled trigonometry

In $\triangle PQR$, since $P$ is due east of $Q$ and $Q$ is due north of $R$, $\angle PQR = 90^\circ$.
$\tan(\angle PRQ) = \frac{PQ}{QR} = \frac{924}{640}$
$\angle PRQ = \tan^{-1}\left(\frac{924}{640}\right) \approx 55.291^\circ$

2. Find the obtuse angle PSR using the Sine Rule

In $\triangle PRS$, using the Sine Rule:
$\frac{\sin(\angle PSR)}{1124} = \frac{\sin 26^\circ}{583}$
$\sin(\angle PSR) = \frac{1124 \sin 26^\circ}{583} \approx 0.84516$
Since $\angle PSR$ is obtuse:
$\angle PSR = 180^\circ - \sin^{-1}(0.84516) \approx 180^\circ - 57.689^\circ = 122.311^\circ$

3. Find angle PRS

Using the sum of angles in $\triangle PRS$:
$\angle PRS = 180^\circ - 26^\circ - 122.311^\circ = 31.689^\circ$

4. Calculate the bearing of S from R

Since $Q$ is due north of $R$, the bearing of $S$ from $R$ is the angle $QRS$.
$\angle QRS = \angle PRQ + \angle PRS = 55.291^\circ + 31.689^\circ = 86.980^\circ$
Bearing $= 087.0^\circ$ (to 1 d.p.)

1. Find angle $PRS$ using the sine rule in triangle $PRS$.

Using the sine rule in $\triangle PRS$:
$\frac{\sin \angle PSR}{1124} = \frac{\sin 26^\circ}{583}$
$\sin \angle PSR = \frac{1124 \sin 26^\circ}{583} \approx 0.8451$
Since $\angle PSR$ is obtuse:
$\angle PSR = 180^\circ - \sin^{-1}(0.8451) \approx 122.317^\circ$
$\angle PRS = 180^\circ - 26^\circ - 122.317^\circ = 31.683^\circ$

2. Find angle $QRS$ by adding angle $PRQ$ and angle $PRS$.

In right-angled $\triangle PQR$ (since $P$ is due east of $Q$ and $Q$ is due north of $R$):
$\tan \angle PRQ = \frac{924}{640}$
$\angle PRQ = \tan^{-1}\left(\frac{924}{640}\right) \approx 55.292^\circ$
$\angle QRS = \angle PRQ + \angle PRS = 55.292^\circ + 31.683^\circ = 86.975^\circ$

3. Calculate the distance $QS$ using the cosine rule in triangle $QRS$.

Using the cosine rule in $\triangle QRS$:
$QS^2 = QR^2 + RS^2 - 2(QR)(RS)\cos \angle QRS$
$QS^2 = 640^2 + 583^2 - 2(640)(583)\cos 86.975^\circ$
$QS^2 = 409600 + 339889 - 746240(0.05276) \approx 710115$
$QS = \sqrt{710115} \approx 842.68 \text{ m}$

4. Calculate the time taken to jog from $Q$ to $S$ and round to the nearest ten seconds.

Speed $= 9.5 \text{ km/h} = \frac{9500}{3600} \text{ m/s} \approx 2.6389 \text{ m/s}$
Time taken $= \frac{\text{Distance}}{\text{Speed}} = \frac{842.68}{2.6389} \approx 319.33 \text{ s}$
$319.33 \text{ s} = 5 \text{ min } 19.33 \text{ s}$
Correct to the nearest ten seconds, the time is $5 \text{ min } 20 \text{ s}$.