We see that the point on the terminal side is (5,6).  So we know that with this point a right triangle is formed with a base that is 5 units long, and a leg that is 6 units high.  We can use the Pythagorean Theorem to solve for the hypotenuse that is formed by this triangle and this will tell us the distance of the point from the origin. 

\(\displaystyle a^{2}+b^{2}=c^{2}\)

\(\displaystyle 5^{2}+6^{2}=c^{2}\)

\(\displaystyle 25+36=c\)

\(\displaystyle 59=c^{2}\)

\(\displaystyle \sqrt{59}=c\)

Let’s solve for sine first.  When working with right triangles recall that \(\displaystyle \sin = \frac{opposite}{hypotenuse}\)and we are considering the angle formed by the x-axis and the hypotenuse.  So the opposite side is the leg that is 6 units high.

\(\displaystyle \sin = \frac{opposite}{hypotenuse}\)

\(\displaystyle \sin = \frac{6}{\sqrt{59}}\)

\(\displaystyle \sin = \frac{6\sqrt{59}}{{59}}\)

We can solve for cosine if we recall that \(\displaystyle \cos =\frac{adjacent}{hypotenuse}\).  Our adjacent side would be the base that is 5 units long.

\(\displaystyle \cos =\frac{adjacent}{hypotenuse}\)

\(\displaystyle \cos = \frac{5}{\sqrt{59}}\)

\(\displaystyle \cos = \frac{5\sqrt{59}}{{59}}\)

Using the equation,

 \(\displaystyle \omega=\frac{v}{r}\) where 

\(\displaystyle \omega\)=angular velocity, \(\displaystyle v\)=linear velocity, and \(\displaystyle r\)=radius of the circle.

In this case the radius is 5 (half of the diameter) and linear velocity is 20 m/s.  

\(\displaystyle \omega=\frac{20}{5}=4\).

Write the formula for angular velocity.

\(\displaystyle \omega =2\pi f_\)

The frequency of the tire is 8 revolutions per second. The radius is not used.  

Substitute the frequency and solve.

\(\displaystyle \omega =2\pi f= 2\pi(8)=16\pi\)

Write the formula for average velocity.

\(\displaystyle \omega=\frac{\theta}{t}\)

The units of omega is radians per second.

Substitute the givens and solve for omega.

\(\displaystyle \omega=\frac{\theta}{t}=\frac{\pi}{\frac{1}{3}}=\pi*3=3\pi\)

Recall that  \(\displaystyle angular\ velocity=\theta/t\).

Since the tire revolves 9.3 times/second it would seem that the tire would rotate

\(\displaystyle 9.3 (2\pi) radians/second\) or \(\displaystyle 18.6 \pi radians/ second\).

We use \(\displaystyle 2\pi\) to indicate that the tire is rolling 360 degrees or \(\displaystyle 2\pi\) radians each revolution (as it should).

Thus, 

\(\displaystyle 18.6 \pi radians/ second\) is your final answer.

Note that radians is JUST a different way of writing degrees. The higher numbers in the answers above are all measures around the actual linear speed of the tire, not the angular speed.

Angular speed is the same as linear speed, but instead of distance per unit time we use degrees or radians. Any object traveling has both linear and angular speed (though objects only have angular speed when they are rotating).

\(\displaystyle Angular\hspace{1mm}speed= \frac{\Theta }{t}\)

Since our tire completes 8 revolutions per second we multiply by \(\displaystyle 2\pi\) since a full rotation (360°) equals \(\displaystyle 2\pi\).

\(\displaystyle Angular\hspace{1mm}speed= \frac{16\pi }{1 s}=16\pi\frac{rad}{s}\)

Since the plane is traveling NE, we know that it is traveling at an angle of 45 degrees north of east. Therefore, we can represent this in a diagram:

We have all of the information we need to solve for the velocity of the plane. We have an angle and a side adjacent to that angle. Therefore, we can use the cosine function to solve for the velocity.

\(\displaystyle cos(45^{\circ}) = \frac{300}{v}\)

Rearranging for v, we get:

\(\displaystyle v = \frac{300}{cos(45^{\circ})} = 424 \text{ mph}\)

If the car is traveling NW, it means that the angle \(\displaystyle \theta\) is \(\displaystyle 45\) degrees above the x-axis. Using this angle, and the fact that the magnitude of the velocity is \(\displaystyle 60\) mph, we can find the x and y components, or the west and north components using trigonometry.

\(\displaystyle \text{West: } \cos(45)=\frac{x}{60}\)

\(\displaystyle x=42.43\)

\(\displaystyle \text{North: } \sin(45)=\frac{y}{60}\)

\(\displaystyle y=42.43\)

Write the formula for the linear velocity.

\(\displaystyle v=\omega r\)

The diameter is twice the radius.

\(\displaystyle d=2r\)

If the diameter was decreased by one half, the radius will also decrease by one half.

Therefore:

\(\displaystyle v=\omega \cdot \frac{1}{2} r\)

\(\displaystyle v=\frac{1}{2}( \omega r)\)

The linear velocity will decrease by one half.

Write the linear velocity formula.

\(\displaystyle v=\frac{x}{t}\)

Since the answer is required in meters per second, two conversions will need to be made.

Convert 30 kilometers to meters.

\(\displaystyle 30 \: \textup{kilometers}\left(\frac{1000\:\textup{meters}}{1 \textup{ kilometer}}\right)=30000 \:\textup{meters}\)

Convert 2 hours to seconds.

\(\displaystyle 2\:\textup{hours}\left(\frac{60\textup{ minutes}}{1\textup{ hour}}\right)\left(\frac{60 \textup{ seconds}}{1 \textup{ minute}}\right)=7200\:\textup{seconds}\)

Apply the formula.

\(\displaystyle v=\frac{x}{t}=\frac{30000\textup{ meters}}{7200\textup{ seconds}}=4.1\bar{6}=4\)