Question

A cruise ship maintains a speed of 10 knots ​(nautical miles per​ hour) sailing from San Juan to​ Barbados, a distance of 600 nautical miles. To avoid a tropical​ storm, the captain heads out of San Juan at a direction of 25 degrees off a direct heading to Barbados. The captain maintains the 10 dash knot speed for 7 ​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 10 dash knot speed is​ maintained?

Answer 1

Answer:

a) θ=151.84°, b) t=53.73hr

Step-by-step explanation:

a)

Ok, so the very first thing we need to do when solving this problem is to draw a diagram of what the situation looks like. (See attached picture). This will help us visualize the problem better and determine what to do in order to solve it.

Notice that the ship will take a triangular path, so we can analyze it as if we were talking about a triangle. So the very first thing we need to do is find the length of side b.

We know the ship is traveling at 10knots=10 mi/hr, so we can use the following ratio to find the distance:

$V=\frac{distance}{time}$

when solving for the distance we get that:

distance=Velocity*time

since the ship will travel for 7 hours in that direction, we get that the distance it travels is:

b=10mi/hr*7hr=70mi

Once we found that distance, we can calculate the distance for side c of the triangle by using the law of cosines

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

which can be solved for c, so we get:

$c=\sqrt{a^{2}+b^{2}-2ab cos \gamma}$

since we know all those values, we can directly plug them in, so we get:

$c=\sqrt{(600mi)^{2}+(70mi)^{2}-2(600mi)(70mi) cos (25^{0})}$

which yields:

c=537.37mi

Once we know the length of c, we can use the law of sines to find angle α.

Like this:

$\frac{sin \alpha}{600}=\frac{sin 25^{o}}{537.37}$

which we can solve for α:

$sin \alpha = 600 \frac{sin 25^{o}}{537.37}$

so

$\alpha=sin^{-1}(600\frac{sin 25^{o}}{537.37})$

so we get the angle to be:

α=28.16°

now, since we are interested in finding the angle it has to turn from its origional course, we can now subtract that angle from 180° to get.

θ=180°-28.16°=151.84°

b) in order to find the time it takes to reach Barbados after the final turn is made we just need to use the velocity ratio again, but this time solve for t:

$V=\frac{c}{t}$

when solving for t we get:

$t=\frac{c}{V}$

so when substituting the values we get:

$t=\frac{537.37mi}{10mi/hr}$

so

t=53.73 hr