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.

 

[Nickie]

I think I need to use the equations q=ut and v=dq/dt, but I am not sure how to apply them in this situation.

First, let's define our variables:
q = position of the ant's image
u = velocity of the ant's image
t = time
f = distance from -2f to -f
v = constant speed of the ant

To find the position (q) as a function of time, we can use the equation q=ut, where u is the velocity and t is the time. Since the ant is walking from -2f to -f, the total distance traveled is f, and the time it takes is t. Therefore, we can rewrite the equation as q=(f/t)t.

Now, we know that velocity is defined as the rate of change of position with respect to time, or v=dq/dt. So, we can also rewrite the equation as q=(f/t)dt. This means that we can find the position of the ant's image at any given time by multiplying the constant f/t by the time (t).

Next, we can use the given equation for velocity (u = f2v/(f-vt)2) to find the velocity of the ant's image as a function of time. This equation is derived from the formula for the derivative of a quotient, d(u/v)/dx = [V(du/dx)-U(dv/dx)]/v2. In this case, u is the velocity and v is the position, so we can substitute them into the equation to get u = [f(dv/dt) - v(du/dt)]/f2.

Since we know that v=dq/dt, we can substitute this into the equation to get u = [f(d2q/dt2) - v(du/dt)]/f2. From the first equation we found (q=(f/t)t), we know that dq/dt = f/t, so we can substitute this into the equation to get u = [f(d2q/dt2) - v(f/t)]/f2.

We can simplify this further by recognizing that the ant is moving at a constant speed, so the derivative of its velocity (du/dt) is 0. This means that u = f(d2q/dt2)/f2, or u = (d2q/dt2)/f