Chapter 8.3, Problem 45AYU

a.

To determine

What angle should the captain turn to head directly to Barbados?

Answer to Problem 45AYU

  $\boxed{26.4°}$

Explanation of Solution

Given information:

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.

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

  

Calculation:

For finding the angle $\theta$ at which the ship has to be turned, first the distance from the point of tuming to . Let the distance be represented as $x$ .

  

Since the ship is moving at a constant speed of 15 knots for 10 hours, the distance travelled by the ship up to the point of turning will be $15.10=150$ nautical miles.

We know two sides and the included angle of the triangle. By using the Law of Cosines,we can find the third side $x.$

  $x^2={600}^2+{150}^2-\mathrm{2.600.150.}\cos 20°$

  $=213355.3$

Take square root on both the sides.

  $\sqrt{x^2}=\sqrt{213355.3}$

  $x\approx 462$

For finding the angle $\theta$ , we have to find angle A opposite the side of length 600 nautical miles.

We know the three sides of the triangle. So , use the Law of Cosines to find A.

  ${600}^2={462}^2+{150}^2-\mathrm{2.462.150}\cos A$

  $A={\cos }^{-1}\left(\frac{{462}^2+{150}^2-{600}^2}{\mathrm{2.462.150}}\right)$

  $={\cos }^{-1}\left(-0.895\right)$

  $=153.5°$

From the figure, $A+\theta =180°$

  $\theta =180°-A$

  $\approx 180°-153.5°$

  $\approx 26.4°$

Hence, the ship should be turned through an angle of about $26.4°$ .

b.

To determine

How much time the ship will take to reach after the turn is made?

Answer to Problem 45AYU

  $\boxed{t=30.8hr}$

Explanation of Solution

Given information:

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.

Once the turn is made, how will it be before the ship reaches Barbados if the same 15-knot speed is maintained? After taking for

Calculation: travelling is the distance divided by the speed.

  $t=\frac{Dis\tan ce}{speed}$

  $t=\frac{462}{15}$

  $t=30.8$

Hence, the ship will take 30.8 hr to reach after the turn is made.