Question

II As the earth rotates, what is the speed of (a) a physics student in Miami, Florida, at latitude ${26}^{°}$, and (b) a physics student in Fairbanks, Alaska, at latitude ${65}^{°}$? Ignore the revolution of the earth around the sun. The radius of the earth is $6400\text{km}$.

Short answer

Short answer

The speed of the physics student in Miami is $418\text{m}/\text{s}$.

The speed of the student in Fairbanks is$197\text{m}/\text{s}$.

Step-by-step solution

Part (a): Given information

The latitude of Miami, Florida is${26}^{°}$ and the radius of earth is $6400\text{km}$.

Part (a): Explanation

The speed of the physics student is,

$v=\frac{d}{t}\mathrm{...}\left(\text{l}\right)$

Here, v is the speed of the student, d is the total distance travelled by the student in 1 day, and t is the time of travel.

The total distance travelled by the student is,

$d=2\pi \left(r\cos \theta \right)$

Here, r is the radius of earth, $\theta$is the latitude of Miami, and $\theta$is the latitude of Miami.

Substitute $6400\text{km}$for $r$and ${26}^{°}$for $\theta$in the above equation to find d.

$d=2\pi \left((6400\text{km})\cos \left({26}^{°}\right)\right)=36142.7\text{km}$Conclusion:

Substitute $36142.7\text{km}$for $d$and 1 day for $t$in the equation (I) to find $v$.

$v=\frac{36142.7\text{km}}{\left(1\text{day}\times \frac{24\text{h}}{1\text{day}}\times \frac{60\min }{1\text{h}}\times \frac{60\text{s}}{1\text{min}}\right)}=\left(0.418\text{km}/\text{s}\times \frac{1000\text{m}}{1\text{km}}\right)=418\text{m}/\text{s}$Therefore, the

speed of the physics student in Miami is $418\text{m}/\text{s}$.

Part (b) Step 1. Given information

The latitude of Miami, Florida is ${65}^{°}$and the radius of earth is$6400\text{km}$.

Part (b) Step 2. Explanation

The speed of the physics student is,

$v=\frac{d}{t}$

Here, $v$is the speed of the student, $d$is the total distance travelled by the student in 1 day, $t$and is the time of travel.

The total distance travelled by the student is,

$d=2\pi \left(r\cos \theta \right)$

Here, $r$is the radius of earth, and $\theta$is the latitude of Miami.

Substitute $6400\text{km}$for $r$and ${65}^{°}$for $\theta$in the above equation to find $d$.

$\begin{array}{l}d=2\pi \left((6400\text{km})\cos \left({65}^{°}\right)\right)\\ =16994.48\text{km}\end{array}$

Substitute $16994.48\text{km}$for $d$and 1 day for $t$in the equation (II) tom find $v$.

$\begin{array}{l}v=\frac{16994.48\text{km}}{\left(1\text{day}\times \frac{24\text{h}}{1\text{dy}}\times \frac{60\min }{1\text{h}}\times \frac{60\text{s}}{1\text{min}}\right)}\\ =\left(0.197\text{km}/\text{s}\times \frac{1000\text{m}}{1\text{km}}\right)\\ =197\text{m}/\text{s}\end{array}$

Therefore, the speed of the student in Fairbanks is$197\text{m}/\text{s}$.