The speed after t seconds of acceleration is $\omega = \alpha t,$ giving a final angular velocity (speed) of

$$\omega = \alpha t = 0.75 \, \frac{rad}{s^2} (60 \, s) = 45 \, \frac{rad}{s}$$

For (b), in order to find the number of rotations, we'll need the average angular velocity over those 60 seconds, which is just the average of the starting and ending angular velocities:

$$\bar \omega = \frac{0 + 45}{2} = 22.5 \, \frac{rad}{s}$$

Now the number of radians is

$$ \require{cancel} 22.5 \frac{rad}{\cancel{s}} (60 \, \cancel{s}) = 1350 \, rad$$

If we divide 2π radians into that, we get the number of full rotations:

$$1350 \, \cancel{rad} \left( \frac{1 \, rev}{2\pi \cancel{rad}} \right) = 215 \, rev.$$

Now for the linear velocity at the rotor tips:

$$v = r\omega = (7.25 \, m)\left( 45 \frac{rad}{s} \right) = 326 \, \frac{m}{s}$$

That's just below the speed of sound (~340 m/s). Generally, helicopters keep the blade-end speed below the speed of sound, where strange things happen and the machine can lose a lot of power.

The inital angular velocity in rad/s is

$$ \begin{align} 258 \, &\frac{\cancel{rev}}{\cancel{min}} \left( \frac{2\pi \, rad}{1 \cancel{rev}} \right) \left( \frac{1 \cancel{min}}{60 \, s} \right) \\[5pt] &= 27 \, \frac{rad}{s} \end{align}$$

Now the definition of angular acceleration can be rearranged to find the time:

$$\alpha = \frac{\omega_f - \omega_i}{\Delta t} \: \color{magenta}{\longrightarrow} \Delta t = \frac{\omega_f - \omega_i}{\alpha}$$

Then

$$\Delta t = \frac{(0 - 27)\frac{rad}{s}}{0.13 \frac{rad}{s^2}} = 208 \, s$$

We'll need the average angular velocity, which is just the numerical average of 20 and 7 rad/s:

$$\bar{\omega} = \frac{20 + 7}{2} = 13.5 \, \frac{rad}{s}$$

We can convert this to revolutions per second:

$$\bar{\omega} = 13.5 \, \frac{\cancel{rad}}{s} \left( \frac{1 \, rev}{2 \pi \, \cancel{rad}} \right) = 2.1486 \, \frac{rev}{s}$$

Now if we divide 500 revolutions by that number of revolutions per second (just pay attention to the units), we get the time that it takes for 500 revolutions:

$$\frac{500 \cancel{rev}}{2.1486 \frac{\cancel{rev}}{s}} = 232.7 \, s$$

That's 3:53 in minutes:seconds. Now we have all we need to calculate the angular accleration:

$$\alpha = \frac{(20 - 7) \frac{rev}{s}}{232.7 \, s} = 0.056 \, \frac{rad}{s^2}$$

First convert revolutions per minute (rpm) into radians per second:

$$ \begin{align} \frac{30,000 \cancel{rev}}{\cancel{min}} &\left( \frac{2 \pi \, rad}{1 \cancel{rev}} \right) \left( \frac{1 \cancel{min}}{60 \, s} \right) \\[5pt] \omega &= 3141 \, \frac{rad}{s} \end{align}$$

Now calculate the angular accleration:

$$\alpha = \frac{(3141 - 0) \frac{rev}{s}}{5 \, s} = 628 \, \frac{rad}{s^2}$$

The linear velocity is $v = \omega r,$

$$v = 0.01 \, m \left( 3141 \frac{rad}{s} \right) = 31.41 \, \frac{m}{s}$$

Finally, convert to mi./h, where 1 mile = 1609 meters and 1 hour = 3600 s.

$$ \begin{align} 31.41 &\frac{\cancel{m}}{\cancel{s}} \left( \frac{mi}{\cancel{m}} \right)\left( \frac{\cancel{s}}{h} \right) \\[5pt] &= 70 \, \frac{mi.}{h} \end{align}$$

Recall that for linear motion, $d = \frac{1}{2} at^2.$ We can make the same analogy for rotational motion, and rearranging for time gives:

$$\theta = \frac{1}{2} \alpha t^2 \: \color{magenta}{\longrightarrow} \: t = \sqrt{\frac{2 \theta}{\alpha}}$$

The time to spin up to the final angular velocity is then

$$t = \sqrt{\frac{2 \theta}{\alpha}} = \sqrt{\frac{2 (10 \pi)}{2}} = 5.605 \, s,$$

where $10 \pi$ is the angle measure of 5 revolutions. Now by further analogy with the linear world, we know that $\omega_f = \alpha t + \omega_i,$ so we have our final angular velocity:

$$ \begin{align} \omega_f &= 2 \frac{rad}{s^{\cancel{2}}} (5.605 \, \cancel{s}) + 2 \, \frac{rad}{s} \\[5pt] &= 13.2 \, \frac{rad}{s} \end{align}$$