Chapter 8.3, Problem 45AYU

Avoiding a Tropical Storm A cruise ship maintains an average speed of 15 knots in going from San Juan, Puerto Rico, to Barbados, West Indies, a distance of 600 nautical miles. To avoid a tropical storm, the captain heads out of San Juan in a direction of ${20}^{\circ }$ off a direct heading to Barbados. The captain maintains the 15-knot speed for 10 hours, after which time the path to Barbados becomes clear of storms.

a. Through what angle should the captain turn to head directly to Barbados?

b. Once the turn is made, how long will it be before the ship reaches Barbados if the same 15-knot speed is maintained?

To determine

To find:

a. Through what angle should the captain turn to head directly to Barbados?

b. Once the turn is made, how long will it be before the ship reaches Barbados if the same 15–knot speed is maintained?

Answer to Problem 45AYU

Solution:

a. The captain need to turn the ship through the angle of $26.4°$

b. $30.8$ hours are required for the second leg of the trip.

Explanation of Solution

Given:

A cruise ship maintains an average speed of 15 knots in going from San Juan, Puerto Rico, to Barbados, West Indies, a distance of 600 nautical miles. To avoid a tropical storm, the captain heads out of San Juan in a direction of $20°$ off a direct heading to Barbados. The captain maintains the 15–knot speed for 10 hours, after which time the path to Barbados becomes clear of storms.

Formula used:

$c^2 = a^2 + b^2 - 2ab\text{cos}C$

Calculation:

The captain maintains the 15–knot speed for 10 hours,

The distance he travels is $15 * 10 = 150$ Nautical miles.

a. Through what angle should the captain turn to head directly to Barbados, we need to find the angle opposite to the side 600.

using cosine formula.

$c^2 = a^2 + b^2 - 2ab\text{cos}C$

$a = 600, b = 150, C = 20°$

$c^2 = {600}^2 + {150}^2 - 2 * 150 * 600\text{cos}20$

$c \approx 461.9 \text{mi}$

To find angle opposite the side 600

$\text{cos}A = (b^2 + c^2 - a^2)/2bc$

$a = 600, b = 150, c = 461.9$

$\text{cos}A = \frac{{150}^2 + {461.9}^2 - {600}^2}{2 * 150 * 461.9}$

$A = 153.6°$

The captain need to turn the ship through the angle of $180 - 153.6 = 26.4°$

b. To find how long will it be before the ship reaches Barbados if the same 15–knot speed is maintained, we need to find the time.

$t = \frac{461.9}{15} = 30.8 hours.$

$30.8$ hours are required for the second leg of the trip.