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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: 2,530 Thanks: 1390 What have you tried? February 18th, 2019, 07:49 PM   #3
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Thanks: 1390 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

$20 t = x_0 - 15t \cos(\theta)$

$0 = y_0 + 15t \sin(\theta)$

$(x_0, y_0) = 50(\cos(22.5^\circ), -\sin(22.5^\circ))$

$-\dfrac{20t-x_0}{15t} = \cos(\theta)$

$-\dfrac{y_0}{15t} = \sin(\theta)$

$\left(\dfrac{20t-x_0}{15t}\right)^2 + \left(\dfrac{y_0}{15t}\right)^2 = 1$

$(20t-x_0)^2 + y_0^2 = 225 t^2$

and this is a simple quadratic which I leave to you
Attached Images Clipboard01.jpg (12.0 KB, 40 views) February 21st, 2019, 11:40 AM   #4
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Quote:
 Originally Posted by romsek $(20t-x_0)^2 + y_0^2 = 225 t^2$ and this is a simple quadratic which I leave to you
Thanks for the help!

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 right-angled 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
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Quote:
 Originally Posted by RachelGreen Thanks for the help! I'm assuming the reason there are 2 answers is because of $\theta$'s value which can change?
Imagine ship B heading due north.

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
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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.
Attached Images Maneuver1.jpg (92.7 KB, 7 views) Tags meet, ships, time Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Tarata12 Linear Algebra 3 November 12th, 2017 12:56 PM Monox D. I-Fly Elementary Math 10 February 27th, 2015 12:17 PM rapshin New Users 2 August 8th, 2014 08:08 AM Dacu Math Events 16 July 10th, 2013 11:13 AM jazker Calculus 1 March 28th, 2010 07:00 PM

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