How Do Relative Velocities Determine Collision Courses?

In summary, the problem involves a destroyer and a frigate, with the destroyer traveling due west and the frigate traveling in a northwest direction. The velocity of the frigate relative to the destroyer is found to be -13.66i km/h. This means that the frigate is on a collision course with the destroyer, as the distance between them is diminishing. The time of collision is calculated to be 36 hours and 36 minutes later. The constant of proportionality, k, can be placed on either side of the equation, as long as the ratio between the distance and velocity remains positive.
  • #1
Woolyabyss
143
1
This isn't really homework its an example, But I need help understanding it to do my homework

Homework Statement


A destroyer is 500 km due west of a frigate. The destroyer is traveling at 10 km/h in a direction of 30 degrees north of east.The frigate is traveling at 5(2)^(1/2) in a NW direction.

(i) Find the velocity of the frigate relative to the destroyer.

(ii) Show that they are on a collision course.

(iii) When will they collide?

Homework Equations


Vab = Va -Vb (V is a vector I don't know how to write vector notation on a computer)
d is the destroyer
f is the frigate

The Attempt at a Solution


(solution of example as in the book)
(i)Vd = 10cos30i + 10sin30j
= 8.66i + 5j

Vf = -5(2)^(1/2)cos45i + 5(2)^(1/2)sin45j = -5i +5j

Vfd = (-5i +5j) - (8.66i +5j)
=-13.66i

(ii)
Position of frigate relative to the destroyer is at 500i km
we write this as Rfd = 500i km

the velocity of the frigate relative to the destroyer is
Vfr = -13.66i km/h (I think this is an error and it should Vfd)

since Vfd = -k(Rfd) where k is a positive constant , they must be on a collision course.

(iii)
the time of the collision is given by relative distance/ relative speed
500/13.66 = 36 hours and 36 minutes later.I understand (i) completely my main problem lies with (ii) where it says
"since Vfd = -k(Rfd) where k is a positive constant , they must be on a collision course."
The only way I can see they could make them both equal is if k were the inverse of time and if it is then why not just put k on the left and let it equal to time. I also don't get why the negative sign is needed.If anybody could help explain this one line to me it would be greatly appreciated.
 
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  • #2
Woolyabyss said:
"since Vfd = -k(Rfd) where k is a positive constant , they must be on a collision course."
The only way I can see they could make them both equal is if k were the inverse of time and if it is then why not just put k on the left and let it equal to time. I also don't get why the negative sign is needed.
It doesn't matter whether you put k on the right and have it as an inverse of time, or put it on the left and have it represent (the more obvious) collision time. The point is merely that (with the minus sign) the ratio is positive: i.e. the distance is diminishing in magnitude, so the time at which they will be at the same point is in the future.
 
  • #3
Thanks I get it now its just that generally when I see constants of proportionality there usually put on the side of the equation where they wouldn't be inverted.
 
  • #4
haruspex said:
It doesn't matter whether you put k on the right and have it as an inverse of time, or put it on the left and have it represent (the more obvious) collision time. The point is merely that (with the minus sign) the ratio is positive: i.e. the distance is diminishing in magnitude, so the time at which they will be at the same point is in the future.

Sorry to bother you again but it would be fine if I said -k(Vfd) = Rfd wouldn't it?
since k would now mean time and the ratio between them would still be positive.
 
  • #5


I can help clarify this for you. Let's break down the equation Vfd = -k(Rfd) and understand what it means.

Vfd represents the velocity of the frigate relative to the destroyer. This means it is the velocity of the frigate as seen by someone on the destroyer, taking into account the motion of the destroyer.

-k(Rfd) represents a constant (k) multiplied by the relative position of the frigate with respect to the destroyer (Rfd). This means that the magnitude and direction of Vfd is determined by the constant k and the relative position of the frigate.

Now, since the velocity of the frigate relative to the destroyer (Vfd) is equal to -k(Rfd), this means that the magnitude and direction of Vfd is completely determined by k and Rfd. In other words, no matter what the value of k is, as long as the relative position (Rfd) stays the same, the velocity of the frigate relative to the destroyer (Vfd) will also stay the same.

So, if the relative position (Rfd) is constant, this means that the velocity of the frigate relative to the destroyer (Vfd) is also constant. This is why they must be on a collision course - because their relative velocity (Vfd) is not changing, and they are getting closer and closer to each other as time passes.

The negative sign is needed because it represents the direction of the relative velocity. Since the frigate is moving towards the destroyer, the relative velocity (Vfd) is in the opposite direction, which is why it is represented as negative.

I hope this helps clarify the equation and why they must be on a collision course. Let me know if you have any other questions.
 

Related to How Do Relative Velocities Determine Collision Courses?

1. What is relative velocity collision?

Relative velocity collision is a type of collision that occurs between two objects in motion. It takes into account the velocities of both objects and how they affect the outcome of the collision.

2. How is relative velocity calculated in a collision?

Relative velocity is calculated by taking the difference between the velocities of the two objects and considering their direction. This can be represented by a vector quantity and can be used to determine the final velocities of the objects after the collision.

3. What are the factors that can affect relative velocity collision?

The factors that can affect relative velocity collision include the velocities of the objects, their masses, and the angle at which they collide. External forces, such as friction, can also have an impact on the outcome of the collision.

4. How does relative velocity collision differ from elastic and inelastic collisions?

Relative velocity collision differs from elastic and inelastic collisions in that it takes into account the velocities of both objects, rather than just the total momentum. In elastic collisions, the objects bounce off each other with no loss of energy, while in inelastic collisions, some energy is lost to heat or deformation.

5. What are some real-world examples of relative velocity collision?

Some real-world examples of relative velocity collision include car accidents, billiard balls colliding, and objects falling from a height. These collisions can be analyzed using the principles of relative velocity to determine the outcomes and potential damage.

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