A Coast Guard cutter detects an unidentified ship at a distance of 18.0 km in th
ID: 1527062 • Letter: A
Question
A Coast Guard cutter detects an unidentified ship at a distance of 18.0 km in the direction 15.0° east of north. The ship is traveling at 30.0 km/h on a course at 40.0° east of north. The Coast Guard wishes to send a speedboat to intercept and investigate the vessel.
(a) If the speedboat travels at 44.0 km/h, in what direction should it head? Express the direction as a compass bearing with respect to due north.
__________ ° east of north
(b) Find the time required for the cutter to intercept the ship.
_____________ min
Explanation / Answer
a)
given
v_ship = 30 km/h
v_boat = 44 km/h
let East be +x axis.
let t is the time taken for the speed boat to catch the ship.
let theta is the angle made by the speed boat with North towards East.
in North direction
v_boat*cos(theta)*t = 18*cos(15) + v_ship*cos(40)*t
44*cos(theta)*t = 17.38 + 30*cos(40)*t
44*cos(theta)*t = 17.38 + 22.98*t
t(44*cos(theta) - 22.98) = 17.38 ----(1)
in East direction
v_boat*sin(theta)*t = 18*sin(15) + v_ship*sin(40)*t
44*sin(theta)*t = 4.66 + 30*sin(40)*t
44*sin(theta)*t = 4.66 + 19.3*t
t(44*sin(theta) - 19.3) = 4.66 ----(2)
take equation(2)/equation(1)
(44*sin(theta) - 19.3)/(44*cos(theta) - 22.98) = 4.66/17.38
on solving the above equation we get
theta = 31.8 degrees East of North. <<<<<<<<<------------Answer
b) from equation 2
t = 4.66/(44*sin(31.8) - 19.3)
= 1.2 hours
= 72.0 <<<<<<<<<------------Answer
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