Solve Physics Problem: Speed of Image Relative to Dubbie

[Mekiel]

Hello everyone, I'm knew to this forum. I was wondering if someone could help me solve this physics problem. I've searched my textbook, the internet, and asked tons of people, but I still can't find an answer. Here's the question:

What is the speed of the image, relative to Dubbie, if Dubbie walk away from the mirror surface at 3.5 m/s at an angle of 30 degrees to the mirror surface?

Obviously there's a formula I need to use, but which formula is it?

 

[ice109]

why wouldn't it be -3.5m/s

 

[Mekiel]

My apologies, I didn't notice the sticky about homework questions. I'll post this in the appropriate forum.

My apologies again.

 

[Mekiel]

Wouldn't the 30 degree angle make a difference in the speed? I mean, if it was just a flat/plane mirror, then I could see the speed being equal.

 

[Danger]

Welcome to PF, Mekiel. The 'speed' of the image will be the speed of light through whatever refractive indices are provided by the atmosphere and the material of the mirror.

 

[Mekiel]

Okay, so if the material is made of glass, and the atmosphere was oxygen, I would need to find the index of refraction for O2?

 

[ice109]

Welcome to PF, Mekiel. The 'speed' of the image will be the speed of light through whatever refractive indices are provided by the atmosphere and the material of the mirror.

i really think that has nothing to do with the problem

upon rereading the problem i think by symmetry alone you could reason that it's -7m/s

 

[Danger]

i really think that has nothing to do with the problem

upon rereading the problem i think by symmetry alone you could reason that it's -7m/s

So you have somehow managed to alter the speed of light from 300,000Km/sec to -7m/s? How'd you manage that? There might be a paper in the future.

 

[ice109]

So you have somehow managed to alter the speed of light from 300,000Km/sec to -7m/s? How'd you manage that? There might be a paper in the future.

the question asks what is the speed of the image's recession relative to the object not how fast does the movement's information propagate. obviously if i pull something along the normal away from a mirror at 3.5m/s its image is not receding from at c.

 

[Danger]

i really think that has nothing to do with the problem

upon rereading the problem i think by symmetry alone you could reason that it's -7m/s

edit: Somehow, my post pointing out that only 23% or so of the atmosphere is oxygen, and it's primarily nitrogen, disappeared.

edit #2: Ice, I think that I just realized the difference in our approach to the problem. I was considering how fast the image responds to movement of the original; you were dealing with how fast the image moves in relation to its environment. A subtle, yet critical, difference. By the way that you're approaching the problem, I agree with you.

 

[Mekiel]

3x10^5km/second eh? I reckon my car can go faster =p. It's a '92 honda accord by the way.

So, I've been further researching and have come across various formulae such as snell's law, critical angle, lens-marker's formula, but I don't know how to implement them or if they are even useful for solving this.

Could someone shed some "light" on this... Pardon the pun.

 

[Danger]

We tried to, from 2 different perspectives. Depending upon how you meant the queston, one of them answered it.
And a '92 Honda is not a car; it's a roller skate with delusions of grandeur.

 

[Integral]

Danger, Danger But please remember this is now in the Homework help forum.

Ok, now for the problem. First start by drawing a picture.

Decompose D's velocity into components. One parallel to the mirror the other normal to the mirror. From the diagram you should see that the image of D will have the same velocity parallel to the mirror but will be receding "into" the mirror at the same rate but in the opposite direction that D is receding from the mirror.

I have given to much already.

 

[andrevdh]

If two objects are receeding from each other with a speed v their relative speed of separation would be 2v.

 

[belliott4488]

If two objects are receeding from each other with a speed v their relative speed of separation would be 2v.

That's where the 30 deg. angle comes in; v is a vector, not a speed (scalar). You need to consider vector arithmetic, as someone already hinted at above.

 

[ice109]

That's where the 30 deg. angle comes in; v is a vector, not a speed (scalar). You need to consider vector arithmetic, as someone already hinted at above.

no you don't

edit

maybe you do, the question is very vague