A Coast Guard cutter detects an unidentified ship at a distance of 19.0 km in th
ID: 2045449 • Letter: A
Question
A Coast Guard cutter detects an unidentified ship at a distance of 19.0 km in the direction 15.0° east of north. The ship is traveling at 20.0 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 54.0 km/h, in what direction should it head? Express the direction as a compass bearing with respect to due north.
(b) Find the time required for the cutter to intercept the ship.
PLEASE EXPLAIN STEP BY STEP. THANK YOU
Explanation / Answer
A Coast guard cutter detects an unidentified ship at a distance of 20.0 km in the direction 15.0 degrees east of north. The ship is traveling at 26.0 km/h on a course at 40.0 degrees east of north. The Coast Guard wishes to send a speedboat to intercept the vessel and investigate it. If the speed boat travels 50.0 km/h, in what direction should it head? Express direction as a compass bearing with respect to due north and using unit vector notation The unidentified ship starts at a distance of 20 km, at 15º east of north, with a velocity of 26 km/h, at 40º east of north. These (r,?) polar coordinates can be transformed (x,y) coordinates by using the following formulas. x = r sin? y = r cos? So the ship unidentified starts at position ( 5.18x + 19.3y ) km with a velocity of ( 16.7 x + 19.9y) km/h The speedboat starts at corrdinates (0,0), and has an initial velocity of (50.0 , f) in polar coordinates. In cartesian, this becomes ( 50sinf , 50cosf ).. At a given time t from the start, the x-coordinate of the ship will be 5.18 + 16.7t, while that of the speedboat will be 50sinf*t. The y-coordinate of the ship will be 19.3 + 19.9t while that of the speedboat will be 50cosf*t. We want the solution for which the equations 5.18 + 16.7t = 50sinf*t 19.3 + 19.9t = 50cosf*t are both satisfied. This is a system of two independdant equations with two unknowns, which can be solved. Taking the first equation, we have t = 5.18 / (50sinf - 16.7) Plugging it into the second one gets us 19.3 = (50cosf - 19.9)t 19.3 = (50cosf - 19.9) * 5.18 / (50sinf - 16.7) 3.732 = (50cosf - 19.9) / (50sinf - 16.7) 3.732 = (cosf - 0.398) / (sinf - 0.334) 3.732 * (sinf - 0.334) = (cosf - 0.398) 3.732 sinf - 1.247 = cosf - 0.398 3.732 sinf = cosf + 0.849 Let's take the square of each side 13.93 sinf^2 = cosf^2 + 1.70 cosf + 0.721 Knowing that [sinf^2 + cosf^2 = 1], we have 13.93 (1 - cosf^2) = cosf^2 + 1.70 cosf + 0.721 13.93 - 13.93 cosf^2 = cosf^2 + 1.70 cosf + 0.721 14.93 cosf^2 + 1.70 cosf - 13.21 = 0 This is a simple quadratic equation in cosf. The solutions are cosf = 0.885 cosf = -0.999 acos (0.885) = 27.7º acos (-0.999) = 177º (clearly not what we're looking for) So it seems the direction in which to head is about 27.7º east of north As in the start, this direction can be expresed in (x,y) coordinates by using x = r sin? ; y = r cos? so it becomes; v = (23.2x + 44.3y) km/h
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.