The diameter of the wheel is\({\rm{D}} = 31\;{\rm{cm}}\).

The initial revolutions per minute are\({\omega _1} = 240\;{\rm{rpm}}\).

The final revolutions per minute are\({\omega _2} = 360\;{\rm{rpm}}\).

The time taken is \({\rm{t}} = 6.8\;{\rm{s}}\).

In determining the angular displacement, multiply the average angular speed of the wheel with the time taken. It should be known that each revolution corresponds to a circumference of the distance traveled.

The average angular speed can be calculated as:

\({\omega _{\rm{A}}} = \frac{{{\omega _1} + {\omega _2}}}{2}\)

On plugging the values in the above relation, you get:

\(\begin{aligned}{l}{\omega _{\rm{A}}} = \left( {\frac{{240\;{\rm{rpm}} + 360\;{\rm{rpm}}}}{2}} \right)\\{\omega _{\rm{A}}} = 300\;{\rm{rpm}}\end{aligned}\)

The relation to find the angular displacement is given by:

\(\begin{aligned}{l}\theta &= {\omega _{\rm{A}}}t\\\theta &= \left( {300\;{\rm{rpm}} \times \frac{{1\;{\rm{min}}}}{{60\;{\rm{s}}}}} \right)\left( {6.8\;{\rm{s}}} \right)\\\theta &= 34\;{\rm{rev}}\end{aligned}\)

The relation to find the required distance is given by:

\(\theta = \left( {34\;{\rm{rev}}} \right)\left( {\frac{{\pi D\;{\rm{m}}}}{{1\,{\rm{rev}}}}} \right)\)

On plugging the values in the above relation, you get:

\(\begin{aligned}{l}\theta &= \left( {34\;{\rm{rev}}} \right)\left( {\frac{{\pi \left( {31\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)\;{\rm{m}}}}{{1\,{\rm{rev}}}}} \right)\\\theta &= 33.1\;{\rm{m}}\end{aligned}\)

Thus, \(\theta = 33.1\;{\rm{m}}\) is the required distance.