Question

Two sea trawlers left a sea port simultaneously in two mutually perpendicular directions. Half an hour later, the shortest distance between them was 17 km and another 15 minutes later, one sea trawler was 10.5 km farther from the original than the other. Find the speed of each sea trawler

• A. 16 km/hr, 30 km/hr
• B. 18 km/hr, 24 km/hr
• C. 20 km/hr, 22 km/hr
• D. 18 km/hr, 36 km/hr

Answer 1

The Correct Option is A

The problem involves two trawlers leaving a port in perpendicular directions, creating a right triangle with their paths as legs. We'll denote the speed of the first trawler as \( x \) km/hr and the second trawler as \( y \) km/hr.

After 0.5 hours, their distances from the port are \( 0.5x \) and \( 0.5y \) km, forming a right triangle. The hypotenuse, or the shortest distance between them, is 17 km:

\(\text{Distance} = \sqrt{(0.5x)^2 + (0.5y)^2} = 17\)

\(0.25x^2 + 0.25y^2 = 289\)

Multiplying through by 4 gives:

\(x^2 + y^2 = 1156\)

After an additional 0.25 hours, the distances are \( 0.75x \) and \( 0.75y \). If one trawler is 10.5 km farther than the other, we obtain:

\(|0.75x - 0.75y| = 10.5\)

Simplifying, we find:

\(0.75|x - y| = 10.5\)

\(|x - y| = 14\)

We have two equations:

• \(x^2 + y^2 = 1156\)
• \(|x - y| = 14\)

Case 1: \(x - y = 14\)

Using \(x = y + 14\) in \(x^2 + y^2 = 1156\):

\((y + 14)^2 + y^2 = 1156\)

\(y^2 + 28y + 196 + y^2 = 1156\)

\(2y^2 + 28y + 196 = 1156\)

\(2y^2 + 28y - 960 = 0\)

Dividing by 2:

\(y^2 + 14y - 480 = 0\)

Applying the quadratic formula:

\(y = \frac{-14 \pm \sqrt{14^2 + 4 \times 480}}{2} = \frac{-14 \pm \sqrt{2916}}{2}\)

\(y = \frac{-14 \pm 54}{2}\)

Solutions are \(y = 20\) or \(-34\), but only positive values are valid, so \(y = 20\).

Thus, \(x = y + 14 = 34\).

Case 2: \(y - x = 14\)

Using \(y = x + 14\) yields identical speed values. Both trawlers' speeds match incorrect options, implying a miscalculation. Reevaluation shows:

Speeds must adhere to specific criteria. Out of the given speeds, \(x = 16\), \(y = 30\) aligns with derived equations, ensuring:\(16^2 + 30^2 = 1156 \text{ is valid}\). Hence, the correct speeds are:

16 km/hr, 30 km/hr