Dee G.

asked • 12/06/19# Word Problem Solving

A ship's bearing is 240°. If a submarine, due east of the ship, shoots a torpedo traveling 5 times as fast as the ship, what angle should the submarine fire the torpedo in order to hit the ship?

## 2 Answers By Expert Tutors

Michael H. answered • 12/06/19

High School Math, Physics, Computer Science & SAT/GRE/AP/PRAXIS Prep

This problem appears to be unsolvable without knowledge of the initial distance between the ship and the sub. However, when the relative speeds are given, then it becomes solvable.

Bearings are measured from due North, and they run counterclockwise. Trig functions assume angles are measured relative to the x-axis, and run clockwise.

We start by realizing that if the torpedo travels at a bearing of 240 degrees, then it will travel along a parallel path towards the ship, overtaking it at some point, then passing it without ever making contact. Hence, the bearing to aim the torpedo is somewhat greater than 240 degrees.

The solution will be found by noting that regardless of how far apart the ship and sub are to begin with, when the ship and torpedo meet, their distance South of the x-axis will be the same. Let t be the common time of travel. Traveling at speed v at an angle of -30 degrees, the Ship's distance South will be

v*t*sin(-30 deg)

Traveling at speed 5v at an angle of θ degrees, the torpedo's distance South will be

5*v*t*sin( θ )

When they meet, these two expressions will be equal:

5*v*t*sin(θ) = v*t*sin(-30 deg)

Solving, we get

sin(θ) = sin(-30 deg) / 5

θ = arcsin [ sin(-30 deg) / 5 ]

= -5.74 deg.

Converting to Bearing, -5.74 deg + 270 deg = 264.26 deg, ans.

Note that when the torpedo's speed is only 1% higher than the Ship's, we would expect to get a nearly parallel track:

θ = arcsin [ sin(-30 deg) / 1.01 ] = -29.67 deg, which is very close to -30 deg, as expected.

Shaun M.

12/11/19

Shaun M. answered • 12/06/19

Doctoral candidate teaching and using Calculus since freshman year

Draw the ship at the origin of an XY-axis. As time progresses it sails in the direction of 240° (down-left). Now rewind and place the submarine on the x-axis to the right of the origin where the ship is. As time progresses, the launched torpedo must also travel down and to the left, but at some angle less than 240° or it will never meet with the ship.

So extend the trajectory of the ship then intersect it with the trajectory of the torpedo. Since the same amount of time passed for both the ship and torpedo to meet, the torpedo's distance should be 5 times that of the ship (it's okay if your drawing isn't to scale!). The two paths should now make a scalene triangle with the x-axis. The inner angle at the ship is the remainder from 360°, which is 120°. The inner angle at the submarine is θ-180°, where θ is the absolute angle at which the torpedo is fired.

We can relate these angles and sides of the triangle using the law of sines:

sin(120°)/(5*ship travel distance) = sin(θ-180°)/(ship travel distance)

Negating the common "ship travel distance" from both sides, theta can be found with an inverse sine function:

θ = 180° + sin^{-1}(sin(120°)/5)

You should find a value less than 240° (parallel trajectory), but greater than 180° (firing directly at the ship).

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Mark M.

12/06/19