Explain in relative velocity please?

  • Thread starter PiRsq
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  • #1
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Can someone please explain in relative velocity please?
 

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  • #2
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.
 
  • #3
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*THIS IS NOT A HOMEWORK QUESTION* I just dont 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?
 
  • #4
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?
 
  • #5
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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.
 
  • #6
HallsofIvy
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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.
 
  • #7
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Ill have to chew on that info for a bit, thanks guys
 

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