a) Look at Ball C in the figure above. Imagine that Ball C is sitting still (so we are in the reference frame of Ball C). What is the distance to Ball D after time \(\Delta t\)? What about Ball B?
If the initial distance between Ball C and Ball B (or D) is \(x_0\), the distance after the time interval should be \(d=x_0+\Delta x\).
b) What are the distances from Ball C to Ball A and Ball E?
If we assume that the balls are evenly spaced to begin with, the distance to Ball A should be \(d=2(x_0+\Delta x)\), to account for both the distance that Ball B moves from Ball C and the distance that Ball A moves from Ball B. The same is true for Ball E.
c) Write a general expression for the distance to a ball N balls away from Ball C after time \(\Delta t\). Interpret your finding.
\(d=N(x_0+\Delta x)\)
\(d(t)=N\frac{\Delta x}{\Delta t}\Delta t+Nx_0\)
This means that the distance between a ball N balls away from Ball C increases linearly with time.
d) Write the velocity of a ball N balls away from Ball C during \(\Delta t\). Interpret your finding.
The velocity would be the slope of the above equation: \(N\frac{\Delta x}{\Delta t}\). This means that the velocity is not only dependent on the rate of change of the distance, but also on how many balls are between that ball and Ball C. Farther balls will move away from Ball C at higher velocities than balls close to Ball C.
Great! You’re thinking like a Hubble!
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