OK, but it doesn't matter. It's just a convention. ie. People do it that way by agreement; not because it doesn't otherwise work.

Wike wrote:

The sign convention used here is that the focal length is positive for concave mirrors and negative for convex ones, and are positive when the object and image are in front of the mirror, respectively. (They are positive when the object or image is real.)[3]

For convex mirrors, if one moves the term to the right side of the equation to solve for the result is always a negative number, meaning that the image distance is negativeĀthe image is virtual, located "behind" the mirror. This is consistent with the behavior described above.

So they have a different convention.

With a concave mirror, the object, image and focal length are all on the same side of the mirror so it makes sense to maintain same signs throughout.

A convex mirror has a virtual image on the opposite side to the object.

Note that your source and Wiki have positive and negative swapped arouind.

Changing the signs in my post does not alter the rest of the analysis.

Bob

]]>a convex mirror is considered to have a negative focal length as the image is virtual

I do not think so.

That is the New Cartesian Sign Convention

...................some thinking later......................

I've not seen this done so I'm exploring new ground. What follows may not be right; comments welcome.

The object distance, do, and the image distance, di, are related by this formula, where f is the focal length:

According to Wiki, a convex mirror is considered to have a negative focal length as the image is virtual, so I'm going to keep f positive by reversing the sign in the formula:

And therefore

Differentiating with respect to t :

where vi and vo are the velocities.

From the original formula :

Substituting for di

When do is zero the speeds are the same. As do tends to infinity vi tends to zero.

This graph shows vi (x) with f = 10 and vo = 1

So, to answer the original question: As the object approaches the mirror the image accelerates from zero (starting at the focus and travelling 'towards the mirror' albeit virtually), to a maximum when the object hits the mirror. Only at this moment are the two speeds equal (but opposite in sign of course!).

Bob

]]>As the object moves away, the image diminishes in size and gets gradually closer to the focus, until it is reduced to a point in the focus when the object is at an infinite distance

So it seems that the speed of movement is not linear. Your answer is good. If you want a formula it will take a little more thought.

Bob

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I'm sure you'd like to work this out for yourself, wouldn't you?

Look at

http://en.wikipedia.org/wiki/Curved_mirror

: the section on 'image'. The paragraph and the diagram should enable you to deduce what happens.

Bob

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