Question #fb12b

1 Answer
Dec 19, 2014

Since the dimensions involved here are much smaller than the radius of curvature of the Earth we can treat this problem as set in a two dimensional flat surface.

Approach : To solve this problem we are going to take the check point as located at the coordinate origin and represent the positions of the coast guard ship and the distressed ship which are given in the polar form. Remember in polar representation of vectors the angle is always measures from the positive X axis (East) along the counter-clock wise direction.

#\vec{A] # - Position vector of the coast guard ship as measured from the checkpoint (coordinate origin),
#\vec{B}# - Position vector of the distressed ship, as measured from the check point (coordinate origin)
#\vec{C}# - Position vector of the distressed ship, as measured from the coast guard ship.

If you illustrate the problem in a simple two-dimensional rectangular grid system it is easy to see that,

#\vec{A}+\vec{C}=\vec{B}; \qquad => \qquad \vec{C} = \vec{B} - \vec{A}. \qquad #

#\vec{A}# and #\vec{B}# are given in their polar forms. To find #\vec{C}# it is better to convert them to cartesian forms. Because, in their cartesian form, the operation of vector addition/subtraction is simply a question of adding/subtracting their components.

#\vec{A} = (r_A,\theta_A) = (X_A, Y_A); #
#r_A=35km, \qquad \theta_A=180^{\circ}-42^{\circ}=138^{\circ}#
#X_A=r_A\cos\theta_A = -26.01 km,#
#Y_A=r_A\sin\theta_A = +23.42 km#

#\vec{B} = (r_B,\theta_B) = (X_B, Y_B); #
#r_B=20km, \qquad \theta_B=360^{\circ}-36^{\circ}=324^{\circ}#
#X_B=r_B\cos\theta_B = 16.18 km#
#Y_B=r_B\sin\theta_B = -11.75 km#

#\vec{C} = (r_C,\theta_C) = (X_C, Y_C); \qquad \vec{C}=\vec{B} -\vec{A};#
#X_C=X_B-X_A=+42.19 km#
#Y_C=Y_B-Y_A=-35.17 km#
This gives us the position of the distressed ship as measured from the coast guard ship in its cartesian form. This shows that the coast guard ship has to travel #42.19 km# to the East and #35.17 km# to the south . To find this in terms of the straight line distance and angle, convert it to its polar form,

#r_C=\sqrt{X_C^2+Y_C^2} = 54.93 km;#
#\theta_C=\tan^{-1}(Y_C/X_C)=-39.81^{\circ}#.

So the coast guard must travel #54.93 km# along a direction that is #39.81^{\circ}# to the South of East .