A concave mirror forms a real image twice as large as the object. The object is then moved such that the new real image produced is three times the size of the object. If the image was moved 94.1 cm from its initial position, how far was the object moved?

I have no clue how to start this...

this should help

http://farside.ph.utexas.edu/teaching/302l/lectures/node137.html

To solve this question, we can use the mirror equation and magnification formula for concave mirrors.

First, let's assign some variables:
- Distance of the object from the mirror = u
- Distance of the image from the mirror = v
- Focal length of the mirror = f
- Height of the object = h
- Height of the image = h'

According to the problem, the mirror forms a real image twice as large as the object. This means the magnification (M) is given by:
M = h'/h = 2

We also know that:
M = -v/u (negative sign because concave mirrors produce real images)

Using the magnification formula, we have:
2 = -v/u

Now, let's determine the values of u and v when the object is moved to the new position.

When the object is moved, the image is also moved by 94.1 cm. Therefore, the new image distance (v') can be expressed as:
v' = v + 94.1

The magnification for the new position is given as three times the size of the object:
M' = h'/h = 3

Using the magnification formula, we have:
3 = -(v')/u

Substituting v' = v + 94.1, we get:
3 = -(v + 94.1)/u

Now, we have two equations:
1) 2 = -v/u
2) 3 = -(v + 94.1)/u

To find the values of u and v, we can solve these two equations simultaneously.

By cross multiplication, we can rewrite equation 2 as:
3u = -(v + 94.1)

Multiplying both sides by -1, we get:
-3u = v + 94.1

Now, we can substitute v from equation 1 into equation 3:
-3u = -(2u) + 94.1

Expanding the equation:
-3u = -2u + 94.1

Combining similar terms, we have:
-u = 94.1

Finally, by multiplying both sides by -1, we get the distance moved by the object:
u = -94.1 cm

Therefore, the object was moved 94.1 cm.