A ship leaves port at 1:00 P.M. and travels S35°E at the rate of 28 mi/hr. Another ship leaves the same port at 1:30 P.M. and travels S20°W at 22 mi/h

Question

A ship leaves port at 1:00 P.M. and travels S35°E at the rate of 28 mi/hr. Another ship leaves the same port at 1:30 P.M. and travels S20°W at 22 mi/hr. Approximately how far apart are the ships at 3:00 P.M.? (Round your answer to the nearest mile.)

Answer 1

A sketch of this situation could look like this:

The distance the first ship travels (d_A) is its rate (R_A) multiplied by the time it traveled (T_A2 - T_A1) or:

d_A = R_A•(T_A2 - T_A1)

d_A = 28•(3 - 1)

d_A = 28•(2)

d_A = 56 miles

The distance the second ship travels (d_B) is its rate (R_B) multiplied by the time it traveled (T_B2 - T_B1) or:

d_B = R_B•(T_B2 - T_B1)

d_B = 22•(3 - 1.5)

d_B = 22•(1.5)

d_B = 33 miles

But since they are in different directions we must solve this considering that.

Vector Component Method:

Break each vector into "x" and "y" components then use the distance formula to calculate the distance

Vector d_A can be broken into x and y components as follows:

d_AX = 56sin(35°) = 32.1203

d_AY = -56cos(35°) = -45.8725 (from the sketch we can see that the vector goes in the negative y direction)

Vector d_B can be broken into x and y components as follows:

d_BX = -33sin(20°) = -11.2867

d_BY = -33cos(20°) = -31.0099

The distance is then:

d_A-B = √[(32.1203 - -11.2867)² + (-45.8725 - -31.0099)²]

d_A-B = √[43.4069² + (-14.8627)²]

d_A-B = √[1884.1629 + 220.8986]

d_A-B = √2105.0615

d_A-B = 46 miles

Law of Cosines Method:

Using the Law of Cosines, plug and chug:

(d_A-B)² = 33² + 56² - 2•33•56•cos(55°)

(d_A-B)² = 1089 + 3136 - 2119.9385

(d_A-B)² = 2105.0615

d_A-B = √2105.0615

d_A-B = 46 miles