Q. 103 - Precalculus Enhanced with Graphing Utilities

Question

Linear Speed on Earth Earth rotates on an axis through its poles. The distance from the axis to a location on Earth 30° north latitude is about 3429.5 miles. Therefore, a location on Earth at 30° north latitude is spinning on a circle of radius of 3429.5 miles. Compute the linear speed on the surface of Earth at 30° north latitude.

Step-by-Step Solution

Answer

The linear speed on the surface of the earth at $30°$  the north latitude is  $898 \frac{\mathrm{miles}}{\mathrm{hour}}$

Step 1. Given information.

Consider the given information,

The distance from the axis to a location on Earth at 30° north latitude is about 3429.5 miles. Therefore, a location on Earth at 30° north latitude is spinning on a circle of radius of 3429.5 miles.

Step 2. Explanation to find the value.

Since-, the earth rotates 360 degrees in 24 hours. Thus,

$1°=\frac{24 \mathrm{hours}}{360°}$

Now, check for 30 degrees.

$\begin{array}{rcl}t& =& \frac{24 \mathrm{hours}}{360°}\times 30°\\ & =& \frac{{\displaystyle 24 \mathrm{hours}}}{{\displaystyle 12}}\\ & =& 2 \mathrm{hours}\end{array}$

Convert the angle 30 degrees into radian.

$\begin{array}{rcl}\theta & =& 30°\times \frac{\mathrm{\pi }}{180°}\\ & =& \frac{\mathrm{\pi }}{6}\end{array}$

Step 3. Find the linear speed.

Since angular speed is $\omega =\frac{\theta }{t}$ and linear speed is $v=r\omega$. Thus,

$v=\frac{\theta }{t}·r$

Substitute the angle $\frac{\mathrm{\pi }}{6}$, time is 2 hours and the radius is $3429.5$. Substitute the values in the formula,

$\begin{array}{rcl}v& =& \frac{\frac{\pi }{6}}{2}·3429.5\\ & =& \frac{\pi }{12}·3429.5\\ & =& \frac{\frac{22}{7}}{12}·3429.5\\ & \approx & 898 \mathrm{miles}/\mathrm{hours}\end{array}$