Relative velocity of two ships

  • Thread starter Aramatheis
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  • #1
Hi, I'm having difficulty with this problem, and was hoping whether someone could lend me a hand?

After being provided with this info;

Ship A is located 4.0 km north and 2.5 km east of ship B. Ship A has a velocity of 19 km/h toward the south and ship B has a velocity of 40 km/h in a direction 36° north of east.
I have found the vector component equation of the velocity (in km/h), which was;

V = (-32.36)i + (-42.51)j

The rest of the question demands at what point in time is the separation between the ships the least; and what is this separation, in km?

The first part of the question demanded the x and y components of the vector (for which I have found the correct values). My problem stems from the fact that the minimum distance would normally be found by using the derivative of the vector equation, and solving for when it equals 0, but my equation has no variable for time (t). This leaves my derivative lacking in variables (the derivative should still have at least one t variable, should it not?)
I would greatly appreciate any hints/information provided on how to go about resolving this dilemma.
Thanks in advance!

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Answers and Replies

  • #2
alphysicist
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Hi Aramatheis,

Hi, I'm having difficulty with this problem, and was hoping whether someone could lend me a hand?

After being provided with this info;

Ship A is located 4.0 km north and 2.5 km east of ship B. Ship A has a velocity of 19 km/h toward the south and ship B has a velocity of 40 km/h in a direction 36° north of east.
I have found the vector component equation of the velocity (in km/h), which was;

V = (-32.36)i + (-42.51)j

The rest of the question demands at what point in time is the separation between the ships the least; and what is this separation, in km?

The first part of the question demanded the x and y components of the vector (for which I have found the correct values). My problem stems from the fact that the minimum distance would normally be found by using the derivative of the vector equation

Which vector equation do you mean here? If you take the time derivative of V, you'll get zero because the relative velocity is constant.

, and solving for when it equals 0, but my equation has no variable for time (t).

They are asking for the minimum distance; what equation can you get for the distance between these two ships at any time t?

This leaves my derivative lacking in variables (the derivative should still have at least one t variable, should it not?)
I would greatly appreciate any hints/information provided on how to go about resolving this dilemma.
Thanks in advance!

Homework Statement





Homework Equations





The Attempt at a Solution

 
  • #3
Hi Aramatheis,



Which vector equation do you mean here? If you take the time derivative of V, you'll get zero because the relative velocity is constant.

The vector equation I am referring to is the one received once you subtract the x & y-components of ship B from ship A. For this question, it becomes;

Va= (19cos270)i + (19sin270)j
Vb= (40cos36)i + (40sin36)j

V= Va - Vb
V= (19cos270 - 40cos36)i + (19sin270 - 40sin36)j
v= (-32.36)i + (-42.51)J


They are asking for the minimum distance; what equation can you get for the distance between these two ships at any time t?

This is the part I am not understanding. How do I formulate an equation for the distance between the ships as a function of time? I have no idea where to begin for this part.
Thank you for your aid.
 
  • #4
alphysicist
Homework Helper
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This is the part I am not understanding. How do I formulate an equation for the distance between the ships as a function of time? I have no idea where to begin for this part.
Thank you for your aid.

First, wow would you find the distance between the two ships at t=0?

Now, ship A starts 4 km north of ship B. How far north (or south) is it at any later time t? Remember you already have found the north-south component of the velocity of ship A relative to B.

How far east (or west) is it any any later time t?

From these, what is the distance at any time t?
 
  • #5
First, wow would you find the distance between the two ships at t=0?

Now, ship A starts 4 km north of ship B. How far north (or south) is it at any later time t? Remember you already have found the north-south component of the velocity of ship A relative to B.

How far east (or west) is it any any later time t?

From these, what is the distance at any time t?


1. at t=0, the distance between ships is d= sqrt [(4)^2 + (2.5)^2]
which equals 4.72 km

2. The north/south component at any time would be (4 - 19t) km/h

3. Would the east/west component be 0 km/h for ship A, or (2.5 - 40cos36)km/h ?
I am not sure about this one.

4. would the distance at any time just be the sum of the north/south & east/west components?
 
  • #6
alphysicist
Homework Helper
2,238
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1. at t=0, the distance between ships is d= sqrt [(4)^2 + (2.5)^2]
which equals 4.72 km

2. The north/south component at any time would be (4 - 19t) km/h

That would be the north/south component of ship A relative to the origin (the place where ship B started at); that's because -19 km/h is the velocity of ship A relative to the water.

But you want the north/south component of ship A's displacement relative to ship B, which is moving. Since you've already found the velocity of ship A relative to ship B, you can just use the components of the relative velocity (for both this part and the next).

3. Would the east/west component be 0 km/h for ship A, or (2.5 - 40cos36)km/h ?
I am not sure about this one.

4. would the distance at any time just be the sum of the north/south & east/west components?

It's not just a simple addition; you've already done the right thing in finding the distance at t=0; the north/south and east/west components are perpendicular, and so you have to use the Pythagorean theorem.
 

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