91Ó°ÊÓ

The initial revolutions per minute are\({\omega _1} = 3500\,{\rm{rpm}}\).

The final revolutions per minute are\({\omega _2} = 1200\,{\rm{rpm}}\).

The time taken is\(t = 2.5\;{\rm{s}}\).

In order to calculate the angular acceleration, divide the change in angular speed of the automobile engine with the time taken by the engine. Consider the constant angular acceleration for determining the angular displacement.

The relation to find the angular acceleration is given by:

\(\alpha = \frac{{{\omega _2} - {\omega _1}}}{t}\)

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

\(\begin{aligned}{l}\alpha &= \left( {\frac{{1200\,{\rm{rpm}} - 3500\,{\rm{rpm}}}}{{2.5\;{\rm{s}}}}} \right)\\\alpha &= \left( {\frac{{ - 1200\,{\rm{rpm}}}}{{2.5\;{\rm{s}}}} \times \frac{{2\pi \,{\rm{rad}}}}{{1\;{\rm{rev}}}} \times \frac{{1\,{\rm{min}}}}{{60\;{\rm{s}}}}} \right)\\\alpha &= - 96.3\;{\rm{rad/}}{{\rm{s}}^2}\end{aligned}\)

Thus, \(\alpha = - 96.3\;{\rm{rad/}}{{\rm{s}}^2}\) is the required angular acceleration.

The relation to find the angular displacement is given by:

\(\theta = \frac{1}{2}\left( {{\omega _1} + {\omega _2}} \right)t\)

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

\(\begin{aligned}{l}\theta &= \frac{1}{2}\left( {1200\,{\rm{rpm}} + 3500\,{\rm{rpm}}} \right)\left( {2.5\;{\rm{s}} \times \frac{{1\,{\rm{min}}}}{{60\;{\rm{s}}}}} \right)\\\theta &= 97.9\;{\rm{rev}}\end{aligned}\)

Thus, \(\theta = 97.9\;{\rm{rev}}\) is the required angular displacement.