Position and Velocity as a function of time

[Wilkins]

Homework Statement

At time t=0 an ant starts to walk from -2f to -f at a constant speed v. Show that the position (q) and velocity (u) of the ants image as a function of time is given by:

q=(2f-vt)f/(f-vt) AND u = f²v/(f-vt)²

Homework Equations

d(u/v)/dx = [V(du/dx)-U(dv/dx)]/v²

The Attempt at a Solution

I am finding it hard to start the solution.

 

[haruspex]

When the ant is at distance x from the lens, where is the image? You surely have a standard equation for that.

 

[Wilkins]

When the ant is at distance x from the lens, where is the image? You surely have a standard equation for that.

1/q=1/f+1/x
where x is the ants distance
q is the image distance and
f is the focal length

OR In Newtonian terms:
X₁ = S₁ - f
X₂ = S₂ - f

 

[haruspex]

1/q=1/f+1/x
where x is the ants distance
q is the image distance and
f is the focal length

OK, so if you can express the ant's position as a function of time, you can use the above equation to get the position of the image as a function of time.