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To cross the river directly to the opposite bank (along the shortest path):

Given:

• Speed of the boat, $v = 14\text{ km/h}$
• Speed of the current, $u = 7\text{ km/h}$
• Width of the river, $d = 12.125\text{ km}$

1. Direction:
Let the boat be steered at an angle $\alpha$ with the direction of the river current so that the resultant motion is perpendicular to the bank.
$\cos\alpha = -\frac{u}{v} = -\frac{7}{14} = -0.5$
$\alpha = \cos^{-1}(-0.5) = 120^\circ$
Therefore, the boat should be steered at an angle of $120^\circ$ with the direction of the current (or $30^\circ$ upstream from the perpendicular to the bank).

2. Time taken:
The component of velocity across the river is:
$v_y = v \sin\alpha = 14 \sin(120^\circ) = 14 \times \frac{\sqrt{3}}{2} = 7\sqrt{3}\text{ km/h} \approx 12.12436\text{ km/h}$

Thus, the time required to cross the river is:
$t = \frac{d}{v_y} = \frac{12.125}{7\sqrt{3}} \approx 1.00005\text{ hours} \approx 1\text{ hour}$

(Note: If crossing in the minimum time is intended, the boat should be steered perpendicular to the bank, i.e., $\alpha = 90^\circ$, and the time taken would be $t = \frac{12.125}{14} \approx 0.866\text{ hours} \approx 52\text{ minutes}$).

Answer:

• Direction: $120^\circ$ to the river current
• Time required: $1\text{ hour}$