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February 18th, 2019, 02:46 PM  #1 
Newbie Joined: Feb 2019 From: Scotland Posts: 6 Thanks: 0  Time taken for 2 ships to meet
2 ships A and B are 50 nautical miles apart. Ship B, which is sailing at 15 knots, is ESE of ship A, which continues to sail due East at 20 knots. Ship B sets course to meet ship A. How long will it take them to meet? 
February 18th, 2019, 04:13 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,427 Thanks: 1314 
What have you tried?

February 18th, 2019, 07:49 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,427 Thanks: 1314  The position of ship A (in blue) is given by $p_A(t) = (20t, 0)$ And likewise ship B $p_B(t) = 50(\cos(22.5^\circ), \sin(22.5^\circ))+15t(\cos(\theta),\sin(\theta))$ so we end up with two equations $20 t = x_0  15t \cos(\theta)$ $0 = y_0 + 15t \sin(\theta)$ $(x_0, y_0) = 50(\cos(22.5^\circ), \sin(22.5^\circ))$ $\dfrac{20tx_0}{15t} = \cos(\theta)$ $\dfrac{y_0}{15t} = \sin(\theta)$ $\left(\dfrac{20tx_0}{15t}\right)^2 + \left(\dfrac{y_0}{15t}\right)^2 = 1$ $(20tx_0)^2 + y_0^2 = 225 t^2$ and this is a simple quadratic which I leave to you 
February 21st, 2019, 11:40 AM  #4  
Newbie Joined: Feb 2019 From: Scotland Posts: 6 Thanks: 0  Quote:
I'm assuming the reason there are 2 answers is because of $\theta$'s value which can change? My original answer was 2 hours and I'm pretty sure it is one of the right answers, but it wasn't part of the 2 solutions from the quadratic... (1.59 and 8.96) I got my answer by drawing a perpendicular line from Ship B's position to Ship A's direction (due East). With the rightangled triangle's hypotenuse being 50 knots: $(20t)^2 + (15t)^2 = 50^2$ $400t^2 + 225t^2 = 2500$ $625t^2 = 2500$ $t^2 = 4$ $t = 2$ Is this wrong? Last edited by RachelGreen; February 21st, 2019 at 11:46 AM.  
February 21st, 2019, 12:54 PM  #5  
Senior Member Joined: Sep 2015 From: USA Posts: 2,427 Thanks: 1314  Quote:
In one solution ship B catches ship A to the west of that due north line. In the other solution ship B catches ship A to the east of it. The time to intercept is shorter in the first solution. I'm getting a solution of $t \approx 1.59347hr,~\theta \approx 53.18^\circ$ The two ships meet at the point $(31.8694, 0)$ The other solution is $t\approx 8.96515hr,~\theta \approx 171.82^\circ$ and the two ships meet at $(179.303, 0)$ Last edited by romsek; February 21st, 2019 at 01:14 PM.  
February 21st, 2019, 03:10 PM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 2,923 Thanks: 1518 
This problem makes me recall doing intercept problems onboard ship using pencil, a set of parallel rulers, a pair of dividers, and a maneuvering board chart. (strictly an analog world ) Referring to the attached diagram, start by plotting ship A's initial position 50 nm at a bearing of 157.5 deg (each ring on the plot is 5 nm) Since an intercept is desired, plot a relative motion position vector from point $\color{red}{A_0}$ to the center of the plot, which represents ship B's position at all times. The plotted relative motion position vector is the dashed red line showing a constant bearing / decreasing range. Plot ship A's course and speed in blue. Each ring represents 5 kts of speed. With the parallel rulers, construct a line parallel from the head of ship A's course/speed vector. It is the solid red line which is in the direction of the relative motion speed vector. (I eyeballed this parallel) Plot ship B's course/speed vector (using just the 15 kts) such that it intersects the relative motion course/speed vector. Note the two possible points of intersection for $\color{green}{B_1}$ and $\color{green}{B_2}$. $\color{green}{B_1}$ looks to be in the direction of about 55 deg N of W. $\color{red}{R_1}$ is the relative course/speed vector from the head of $\color{green}{B_1}$ to the head of $\color{blue}A$ The length of $\color{red}{R_1}$ is about 31 kts ... (50 nm)/(31 kts) is about 1.6 hrs. $\color{green}{B_2}$ looks to be in the direction of about 9 deg N of E. $\color{red}{R_2}$ is the relative course/speed vector from the head of $\color{green}{B_2}$ to the head of $\color{blue}A$ The length of $\color{red}{R_2}$ is about 5.7 kts ... (50 nm)/(5.7 kts) is about 8.8 hrs. 

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