Explain in relative velocity please?

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Discussion Overview

The discussion centers around the concept of relative velocity, particularly in the context of vector analysis. Participants explore how to calculate relative velocity when two objects are moving at different angles and speeds, with a focus on understanding the vector components involved.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant asks for clarification on relative velocity, indicating a lack of understanding of the concept in vector terms.
  • Another participant explains that if one car is traveling at 60 km/h and another at 40 km/h, the relative velocity between the two is 20 km/h.
  • A participant presents a scenario involving two cars, one traveling at 15 mph at a 45° angle and another at 8 mph due East, and seeks to determine the relative velocity between them.
  • One response provides a breakdown of the components of the velocities, suggesting that the relative velocity could be calculated by subtracting the horizontal components.
  • Another participant introduces a formula for calculating the magnitude of the relative velocity vector, emphasizing the vector nature of the problem.
  • One participant prefers to distinguish between "velocity" as a vector and "speed" as its magnitude, and provides a detailed method for calculating the relative velocity vector using vector subtraction.
  • A later reply expresses a need to reflect on the information shared, indicating ongoing contemplation of the concepts discussed.

Areas of Agreement / Disagreement

Participants present various methods and interpretations for calculating relative velocity, indicating that multiple approaches exist. There is no consensus on a single method or conclusion, as participants explore different aspects of the problem.

Contextual Notes

The discussion includes various assumptions about the coordinate system and the interpretation of vector components, which may not be universally agreed upon. The calculations involve approximations and specific angle considerations that could affect the results.

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Can someone please explain in relative velocity please?
 
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If you are in a car and traveling 60 Km/h you are moving 60 Km/h relative to the ground.

If another car is moving 40 Km/h, you are moving 20 Km/h relative to the other car.
 
*THIS IS NOT A HOMEWORK QUESTION* I just don't understand the concept of relative velocity dealing within vectors...

What if you are going 45°North of East at 15 mph and you see another car is going due East at 8mph. What is your velocity compared to the other car?
 
Lets see,

15 mph at 45º

15 cos 45º = 10.6 mph
15 sin 45º = 10.6 mph

8 mph at 0º

8 cos 0º = 8 mph
8 sin 0º = 0 mph

Wouldnt your relative velocity be 10.6 mph - 8 mph?

= 2.6 mph faster horizontally?
 
Not quite.
vrel=√(v12+v22-2v1v2cos45)
This is the magnitude of the vector with the origin in the head of v1 and the head in the head of v2. it's a sliding vector. It's the speed you see the other one moving if you are in one of them. So if you're moving at speed v1 with respect to the origin you see the other one moving at v1-v2 and the vector's origin must be in the other one's position.
 
I prefer to use the term "velocity" to mean the vector, "speed" to mean the magnitude of that vector.

"What if you are going 45°North of East at 15 mph and you see another car is going due East at 8mph. What is your velocity compared to the other car?"

The basic idea is to treat the other car as if it were standing still. If you are going 45 degrees North of East at 15 mph, then you "components" (setting up a coordinates system with x to the east and y to the north) are (15 cos(45),15 sin(45))= (15[sqrt](2)/2, 15[sqrt](2)/2). The other car is going due east a 8 mph so it components are (8, 0). "Compare" the two by subtracting the two vectors:
(15[sqrt](2)/2,15[sqrt](2)/2)- (8,0)=(15[sqrt](2)/2-8,15[sqrt](2)/2)
Since [sqrt](2)/2 is about .7, 15[sqrt](2)/2 is about 10.5 and 15[sqrt](2)/2- 8 is about 2.5. The occupants of the second car would see you going mostly east with a little bit of north component.

Since the problem asked for the vector relative to the second car, there is no reason to calculate it "length" or speed. The answer is
(15[sqrt](2)/2-8,15[sqrt](2)/2).

If you are required to give the vector in the same form as the two velocities were originally given, speed and direction, you could use the Pythagorean theorem to find the speed and
tan(theta)=(15[sqrt](2)/2-8)/(15[sqrt](2)/2) to find the direction.
 
Ill have to chew on that info for a bit, thanks guys
 

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