A Motion Exercise

KerimF · 2023-11-25 20:58:57

An ordinary ship and a warship were moving, side by side, on two parallel straight lines. The distance between them was (D). Their speed was constant and equal to (S1).

At time zero, t=0, the captain of the warship was instructed to shut down the other ship.

Let us assume that the new constant speed of the warship is (S2), S2>S1 (while moving on a straight line towards the side of the other ship).

The captain can launch a missile whose constant speed is (S3) which is much higher than (S2). The maximum range of the missile is (R), R<D.

Find the minimum time (T) to hit the other ship, that is:
t_min = T = f(D,R,S1,S2,S3)

Kerim

Note1:
I heard of this exercise when I was in the final year of high school (Baccalauréat). At that time, math teachers used to skip it. So, it was out of question to me not to solve it, and naturally I did it (because I was very interested in doing it). Later, I heard that it was removed in the printings of newer math books of the final year.

Note2:
I am afraid that its solution needs to be done in two steps:
(1) Logical
(2) Mathematical
If the logical step is skipped, the mathematical solution will be somehow undetermined, instead of being obvious.

Last edited by KerimF (2023-11-25 21:09:06)

Bob · 2023-11-26 01:47:37

hi KerimF

I have drawn a diagram. The question says "(while moving on a straight line towards the side of the other ship)."

That seems contradictory. I'm marking the two vessel's positions with S for ship and W for warship. If W heads directly towards S while S is moving forwards then W must continuously change the direction of travel so that it is once more heading at S. If W doesn't do this it will end up where S was rather than where S is. But that means W's movement is not in a straight line.

So it seems that the best course for W is to try and close on S following a diagonal line. I'll need to check but I think that the optimal path is so that W's component of velocity in S's direction is S1 but using the extra velocity that S2 allows to move closer to S's path, reducing D until it reaches a certain value*. Then, since S3 is not infinite, the missile must similarly follow a diagonal track so that it closes on where S will be when the missile crosses S's path.

* Initially I though R here, but that won't work as the missile's diagonal path will be greater than R. So I need some more trig. to calculate the position of W at launch.

Sorry if that sounds complicated. I'll have a go at a making and posting a diagram for all the above.

Bob

ps. Now I look at the diagram, I've realised that WX and XY should not be part of the same straight line. As S3 is bigger than S2 the missile should be on a track that is less acute to the ship's direction. The angle between S's direction and the diagonal lines is determined by adjacent = S1, hypotenuse = S2 and then S3.

later edit: I have now changed the diagram.

KerimF · 2023-11-26 02:21:41

Bob wrote:

hi KerimF
I have drawn a diagram. The question says "(while moving on a straight line towards the side of the other ship)."
That seems contradictory. I'm marking the two vessel's positions with S for ship and W for warship. If W heads directly towards S while S is moving forwards then W must continuously change the direction of travel so that it is once more heading at S. If W doesn't do this it will end up where S was rather than where S is. But that means W's movement is not in a straight line.

Sorry for the confusion. I thought that the meaning of 'towards the side of the other ship' is totally different from "towards the other ship'.

Let us assume that the two parallel straights of S and W are drawn horizontally.
The ordinary ship, S, was on the northern line and the warship, W, was on the southern one. Both were moving from west to east.
At t=0, the captain would have to change the direction of his warship, W, towards north-east. The angle formed by the new direction (straight line) of W and the horizontal line could be called θ for example. In theory, the angle θ could be any value between 0 and 90 degrees but the captain had to decide its exact value to achieve the minimum time for his mission. The trajectory of the missile is also a straight line (between the position of the warship, W, when the missile was launched and the position of the other ship, S, when it was hit by the missile).

Please don't hesitate to notify me if I missed something to clarify about the exercise.

Kerim

Last edited by KerimF (2023-11-26 02:26:07)

KerimF · 2023-11-26 02:35:26

Bob wrote:

I've realised that WX and XY should not be part of the same straight line.

I couldn't get well the relation between this sentence and your diagram.
On your diagram, they do have the same direction, unless 'same straight line' and 'same direction' have different meaning.

Bob · 2023-11-26 05:45:02

I have changed my diagram in the earlier post to try and show what I mean. As S3 is more than S2 the missile can take a shorter path to hit the ship. So the direction changes at X from warship's direction to missile's direction.

Starting with XY, you can resolve S3 into a component parallel with the ship's direction and equal to S1, and a component at right angles to the ship's direction.

The distance XY = R, and angle XYS is INV COS (S1/S3) = θ

Time for missile = R/S3

Time for warship from W to X is WX/S2 but what is WX?

Draw a perpendicular from X to SY meeting SY at U. Then XU = R . SIN (θ)

Draw a perpendicular from X to WS meeting WS at V.

angle WXV = INV COS (S1/S2) =Δ

WV = D - XU. hence WX = WV / SIN (Δ)

That's enough to complete the calculation.

Note about my initial comment.

If an object (A) is moving and another object (B) must aim to travel straight towards A then B's path is continuously changing direction. This is illustrated by the Guardian's Four Dogs Puzzle:

https://www.theguardian.com/science/201 … in-pursuit

Bob

KerimF · 2023-11-26 16:40:10

Bob wrote:

If an object (A) is moving and another object (B) must aim to travel straight towards A then B's path is continuously changing direction. This is illustrated by the Guardian's Four Dogs Puzzle:

This may be applied in real life. But, as you know, the situation here is just a math exercise.

For instance, when I saw your first diagram, I thought you got quickly the right logical answer of the situation related to the 2 directions of W and the missile.
Unfortunately, you changed your mind

As we will see, proving that your first diagram is right, and it gives indeed the minimum time of the mission is somehow simple if one discovers how to start it.

I believe you can prove it, right?

KerimF · 2023-11-27 20:19:13

I will refer here to your original diagram (I guess you still have it).

I used to tell my students that, in general, finding how to solve a problem has two main directions, forwards and backwards.
To solve this exercise, walking backwards makes it a simple one.

Let us assume first that the full range (R) of the missile was used.
In this case, the warship had to be on the circle whose center is ‘Y’ and radius is (R).
If we draw this circle, it becomes clear that WX is the shortest distance from W to this circle.
Therefore, the mathematical solution of T=f(D,R,S1,S2,S3) could take advantage of Pythagoras theorem of the right triangle SWY.

One may think that it is not necessarily to use the full range of the missile, say (k*R) instead of (R), where 0<k<1.
The same reasoning above also applies in case (k*R). Only the radius of the circle becomes (k*R) instead of (R).
This means that T_new = f(D,k*R,S1,S2,S3). That is (R) is replaced with (k*R) in the final formula of (T).
And by analyzing the variation of the function (T_new) with the variable parameter (k), we will find out that T_new > T for all values of (k).

Last edited by KerimF (2023-11-29 03:18:46)

KerimF · 2023-11-29 03:18:26

To those who got the logical step and to complete the exercise, here is the mathematical step:
From the right triangle (SWY ), we can write:

[(S2*T1) + R]^2 = D^2 + (S1*T)^2 , //T1 = time from W to X//

(S2*T1)^2 + 2*S2*T1*R + R^2 = D^2 + (S1*T)^2

(S2*T1)^2 + 2*S2*T1*R + R^2 - D^2 - (S1*T)^2 = 0

If T2 = time from X to Y, T2=R/S3
T = T1 + T2
T1 = T – T2
T1 = T – R/S3

By replacing T1:
S2^2*(T – R/S3)^2 + 2*S2*(T – R/S3)*R + R^2 - D^2 - (S1*T)^2 = 0

S2^2*(T^2 – 2*R/S3*T + R^2/S3^2) + 2*S2*(T – R/S3)*R + R^2 - D^2 - (S1*T)^2 = 0

S2^2*T^2 – 2* S2^2*R/S3*T + S2^2*R^2/S3^2 + 2*S2*R*T – 2*S2*R^2/S3 + R^2 - D^2 - S1^2*T^2 = 0

(S2^2 - S1^2)*T^2 – 2* S2*R*(S2/S3 - 1)*T + S2^2*R^2/S3^2 – 2*S2*R^2/S3 + R^2 - D^2 = 0

(S2^2 - S1^2)*T^2 – 2* S2*R*(S2/S3 - 1)*T + R^2*(S2/S3 – 1)^2 - D^2 = 0

This is a quadratic equation whose general parameters (a), (b) and (c) are:
a = S2^2 - S1^2
b= –2* S2*R*(S2/S3 - 1)
c= R^2*(S2/S3 – 1)^2 - D^2

A numeric example:
D = 24.11 km
S1 = 30 km/h
S2 = 45 km/h
S3 = 150 km/h
R = 20 km

The mission time is T = 15 min.

Kerim

Math Is Fun Forum

#1 2023-11-25 20:58:57

A Motion Exercise

#2 2023-11-26 01:47:37

Re: A Motion Exercise

#3 2023-11-26 02:21:41

Re: A Motion Exercise

#4 2023-11-26 02:35:26

Re: A Motion Exercise

#5 2023-11-26 05:45:02

Re: A Motion Exercise

#6 2023-11-26 16:40:10

Re: A Motion Exercise

#7 2023-11-27 20:19:13

Re: A Motion Exercise

#8 2023-11-29 03:18:26

Re: A Motion Exercise

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