# Relative Velocity of Two Ships: When Is Their Separation Least?

• Aramatheis
In summary: Pythagorean theorem to find the distance between the two ships at any time t. The Pythagorean theorem says that the square of the hypotenuse (the distance between the two ships) is equal to the sum of the squares of the other two sides (the north/south and east/west components). So the equation for the distance at any time t would be:d = sqrt[(4 - 19t)^2 + (2.5 - 40cos36t)^2] To find the minimum distance, you would need to take the derivative of this equation with respect to time and set it equal to 0. However, since you have two variables (t and theta),
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, 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.

Hi Aramatheis,

Aramatheis said:
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.

alphysicist said:
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.

Aramatheis said:
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.

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?

alphysicist said:
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 ?

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

Aramatheis said:
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 ?

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.

## 1. What is relative velocity?

Relative velocity is the measurement of the motion of one object in relation to another object. It takes into account both the speed and direction of the objects.

## 2. How is relative velocity calculated?

Relative velocity is calculated by taking the difference between the velocities of the two objects and considering their direction. This can be represented using vector notation.

## 3. What factors affect relative velocity?

The relative velocity of two ships is affected by the speed and direction of each ship, as well as any external forces such as wind or currents.

## 4. How does relative velocity affect collisions between ships?

Relative velocity plays a crucial role in determining the outcome of collisions between ships. If two ships are moving in the same direction, their relative velocity will be lower and the impact will be less severe compared to two ships moving towards each other with higher relative velocity.

## 5. Can relative velocity change during a ship's journey?

Yes, relative velocity can change during a ship's journey as the speed and direction of the ships may change due to external factors such as wind or currents. It is important for ships to consider and adjust their relative velocity to avoid collisions.

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