Question

An automobile engine slows down from 4500 rpm to 1600 rpm in 6.0 s. (a) calculate its angular acceleration, assumed constant. 3036.87 incorrect: your answer is incorrect. rad/s2 (b) calculate the total number of revolutions the engine makes in this time. 8700 incorrect: your answer is incorrect. rev

Answer 1

1) The angular acceleration of the engine is given by:
$\alpha = \frac{\omega _f - \omega _i}{\Delta t}$
where $\omega _f$ and $\omega _i$ are the final and initial angular velocity of the engine, while $\Delta t$ is the time interval considered.
We need first to convert the velocities from rpm (revolutions per minute) into rad/s. Keeping in mind that 
$1 rev=2\pi rad$
$1min = 60 s$
The factor of conversion is
$1 \frac{rev}{min} = \frac{2 \pi rad}{60 s}$
So, the two velocities become
$\omega _i = 4500 rpm \cdot ( \frac{2 \pi rad}{60 s} )=471 rad/s$
$\omega _f = 1600 rpm \cdot ( \frac{2 \pi rad}{60 s} )=167 rad/s$
And using $\Delta t=6.0 s$ we can find the angular acceleration of  the engine:
$\alpha = \frac{167 rad/s - 471 rad/s}{6.0 s}=-50.7 rad/s^2$
where the negative sign means the engine is decelerating.

2) The total number of revolutions is given by the law of angular motion:
$S(\Delta t)= \omega _i \Delta t + \frac{1}{2} \alpha (\Delta t)^2 = (471 rad/s)(6.0 s) + \frac{1}{2}(-50.7 rad/s^2)(6.0 s)=$
$=1915 rad$
And keeping in mind that $1 rev=2 \pi rad$, the number of revolutions is
$1915 rad \cdot \frac{1}{2 \pi}=305 rev$