How fast is the distance changing? (FODE, I think)

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In summary, the speed of distance change is calculated by dividing the change in distance by the time it takes for the change to occur. Speed and velocity are both measures of distance change, but velocity takes into account direction while speed does not. Acceleration affects the rate of distance change by changing the velocity over time. The rate of distance change can be negative, indicating deceleration, and is affected by external factors such as friction, air resistance, and gravity which can either increase or decrease the speed of distance change.
  • #1
PhysicsMark
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Homework Statement


2 cars start from the same point. Car A travels a constant 50 mph due west. Car B travels a constant 27 mph due south. After 3 hours, how fast is the distance changing between them?


Homework Equations





The Attempt at a Solution



I saw this problem online have been stumped trying to solve it. To start I drew a right triangle according to the problem. I have car A traveling west and car B traveling south. I labeled the hypotenuse side C. I tried to write an equation relating the three sides and (using phythagoras) I said that:

[tex]C=\sqrt{A^2+B^2}[/tex]

My thought was to take the derivative of this equation and then find it's value at 3 hours. I don't think this is correct because I would be taking the derivative with respect to time (I think) and I am not sure how to handle both A and B.

I would like a small hint in the correct direction. I am anxious to solve this one myself, if possible please hold off on presenting the solution.

Thanks for taking the time to read and help.
 
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  • #2
PhysicsMark said:

Homework Statement


2 cars start from the same point. Car A travels a constant 50 mph due west. Car B travels a constant 27 mph due south. After 3 hours, how fast is the distance changing between them?


Homework Equations





The Attempt at a Solution



I saw this problem online have been stumped trying to solve it. To start I drew a right triangle according to the problem. I have car A traveling west and car B traveling south. I labeled the hypotenuse side C. I tried to write an equation relating the three sides and (using phythagoras) I said that:

[tex]C=\sqrt{A^2+B^2}[/tex]

My thought was to take the derivative of this equation and then find it's value at 3 hours. I don't think this is correct because I would be taking the derivative with respect to time (I think) and I am not sure how to handle both A and B.

I would like a small hint in the correct direction. I am anxious to solve this one myself, if possible please hold off on presenting the solution.

Thanks for taking the time to read and help.

Instead of using A and B, you need expressions that represent the distance each car travels, as a function of time.

dA = 50t
dB = 27t

Now, what expression gives you the distance between the two cars as a function of time t? (What you had is a step in the right direction.)
 
  • #3
BTW, what does FODE stand for?
 
  • #4
Mark44 said:
BTW, what does FODE stand for?

Sorry about the abbreviation. In my DiffEq book, they use FODE to mean First Order ordinary Differential Equation.

This problem does not come from my DiffEq book, but it did remind me of a similar problem in that book that also stumped me.

Thank you for the hint. As always your posts lend insight. I will work on this a bit more and then reply.
 
  • #5
Hmm, ok. Using your suggestions here is what I did:

I re-wrote my equations (Thanks, organization is crucial) as:

[tex]C(t)=\sqrt{(50t)^2+(27t)^2}[/tex]

I then simplified as follows:

[tex]C(t)=\sqrt{t^2(50^2+27^2)}[/tex]

...then:

[tex]C(t)=\sqrt{t^2(3229)}[/tex]

...and finally:

[tex]C(t)=t{\sqrt{3229}}[/tex]

I stopped here, because it implies that the distance grows at a constant rate. I originally thought this makes sense, since the cars are traveling at constant velocities . I then thought that this couldn't be correct because I was under the impression this problem involved changing rates (which is why I thought it was a FODE).

I assumed it would be a DiffEq problem because it was similar to a previous DiffEq problem I have encountered.

Does this look correct?
 
  • #6
Looks fine to me. As a sanity check, notice that C(0) = 0, C(1) = sqrt(3229), and C(2) = 2 sqrt(3229). After 2 hours, the cars are twice as far apart as they were after 1 hour.
 
  • #7
Thanks. This makes sense to me also. I would like to point out something about this problem. I am not sure how relevant this is because the "correct" answer could have come from anywhere. I found this problem on another forum and could not solve it. The original poster came up with the same answer as I now have (about 57 mph) and they stated it was not correct. The poster said the "correct" answer is about 68mph and that they (the poster) did not understand how to get that answer.

Again, I have no idea where the "correct" answer comes from and I have no idea whether or not the original poster posted the problem correctly.

I wanted to add that bit of information to see if it struck a chord.

Thanks for the help! Unless that last bit changes anything, I am satisfied with the answer.
 
  • #8
I can't find anything wrong with what you did, so an answer of about 57 mi/hr looks good to me. I don't see how the other person came up with an answer of 68.
 

1. How is the speed of distance change calculated?

The speed of distance change is calculated by dividing the change in distance by the time it takes for the change to occur. This is known as average speed and is represented by the formula: speed = distance / time.

2. What is the difference between speed and velocity in relation to distance change?

Speed refers to the rate of change of distance without taking direction into account, while velocity takes into account both the speed and direction of the change in distance. In other words, velocity is a vector quantity while speed is a scalar quantity.

3. How does acceleration affect the rate of distance change?

Acceleration is a measure of the change in velocity over time. Therefore, if an object is accelerating, its velocity is changing and as a result, the rate of distance change will also change. If an object is accelerating at a constant rate, the speed of distance change will increase or decrease at a constant rate as well.

4. Can the rate of distance change be negative?

Yes, the rate of distance change can be negative. This indicates that the distance is decreasing over time, which is known as deceleration. It is important to note that negative speed does not necessarily mean the object is moving backwards, as it depends on the direction of the change in distance.

5. How is the rate of distance change affected by external factors?

The rate of distance change can be affected by various external factors such as friction, air resistance, and gravity. These forces can either increase or decrease the speed of distance change. For example, friction can slow down an object's speed of distance change, while gravity can increase it.

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