A bar magnet is released from rest and falls through a very long...

As the magnet falls through the copper tube, the changing magnetic flux induces eddy currents in the tube.

According to Faraday's law of induction, the induced electromotive force (EMF) is directly proportional to the velocity of the magnet $v$. The induced eddy currents are $I = \dfrac{\text{EMF}}{R}$, which means $I \propto v$.

By Lenz's law, these currents produce an upward magnetic drag force $F_{drag}$ that opposes the motion. Since the magnetic force is proportional to the induced current, $F_{drag} \propto v$. We can write $F_{drag} = kv$, where $k$ is a positive constant.

Applying Newton's second law to the falling magnet of mass $m$:
$mg - F_{drag} = ma$
$mg - kv = ma$
$a = g - \left(\dfrac{k}{m}\right)v$

This equation represents a straight line with a negative slope \left(-\dfrac{k}{m}\right)$. At$v = 0$, the acceleration is$a = g$. As$v$increases,$a$decreases linearly until it reaches zero at the terminal velocity$v = \dfrac{mg}{k}$, intersecting the positive$v$-axis.<br><br>Answer: A straight line with a negative slope, starting at$a = g$on the$a$-axis and intersecting the positive$v-axis.