A Coast Guard ship is traveling at a constant velocity of 4.25 m/s, due east, re

Question

A Coast Guard ship is traveling at a constant velocity of 4.25 m/s, due east, relative to the water. On his radar screen the navigator detects an object that is moving at a constant velocity. The object is located at a distance of 2305 m with respect to the ship, in a direction 32.5° south of east. Six minutes later, he notes that the object's position relative to the ship has changed to 1115 m, 56.5° south of west. What are the magnitude and direction of the velocity of the object relative to the water? Express the direction as an angle with respect to due west.

Explanation / Answer

let East be +x axis

let the initial poistion of gaurd is a orogin.

so, initial cordinates of object.

x1 = 2305*cos(32.5) = 1944 m

y1 = -2305*sin(32.5) = -1238.5 m

finalposition of the coardites,

x2 = 6*60*4.25 - 1115*cos(56.5) = 914.6 m

y2 = - 1115*sin(56.5) = -930 m

now diplacemt of object, d = sqrt((x2-x1)^2 + (y2-y1)^2)

= sqrt((914.6 - 1944)^2 + (-930 - (-1238.5))^2)

= 1074.6 m/s

velocity = dispalcemnt/time taken

= 1074.6/(6*60)

= 2.99 m/s

direction, theta = tan^-1((y2-y1)/(x2-x1))

= tan^-1((-930+1238.5)/(914.6 - 1944))

= 16.7 degrees south of west

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