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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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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
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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.
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 Maneuver1.jpg (92.7 KB, 7 views)

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