Question

ActivityLuna I made the 3.8 x 10^5 kilometers trip to the moon in 2.16 x 10^3 minutes. In this activity, you will determine the average speed of Luna I using division.Part AWrite an expression for the speed of Luna I in kilometers/minute. Represent this expression as a quotient of two numbers expressed in scientific notation.Part BRewrite the expression you found in part A as the product of two fractions. The first fraction should be the first factors and the other fraction should be the powers of 10.Part CEstimate the value of the first fraction in part B to the hundredths place. Simplify the second fraction using the properties of exponents.Part DIs the value you found in part C represented in scientific notation? If not, convert the value to represent it in scientific notation. Also, write the average speed of Luna I in the most appropriate unit of measurement.

Answer 1

Part A.

The speed is given by

$\text{speed}=\frac{dis\tan ce}{time}$

In our case, the distance is 3.8x10^5 kilometers and the time is 2.16x10^3 minutes. By substituting these values we get

$\text{speed}=\frac{3.8\times10^5\text{ km}}{2.16\times10^3\text{ min}}\text{ }$

Part B.

We can rewrite the speed as

$\text{speed}=(\frac{3.8\times10^5\operatorname{km}}{1})(\frac{1}{2.16\times10^3\min})$

which is the product of two fractions.

Part C.

The first fraction is

$\frac{3.8\times10^5\operatorname{km}}{1}=3.8\times10^5\operatorname{km}$

and the second fraction is

$\frac{1}{2.16\times10^3\min}$

This fraction can be written as

$\frac{1}{2160\min}=0.463\times10^{-3}(\frac{1}{\min})$

Therefore, the answers for part C are

$\begin{gathered} \text{first fraction: }3.8\times10^5\operatorname{km} \\ \text{second fraction: }0.463\times10^{-3}(\frac{1}{\min}) \end{gathered}$

Part D.

Yes, the values are written in scientific notation. Then, the speed is given by

$\text{speed}=(3.8\times10^5\operatorname{km})(\text{ }0.463\times10^{-3}\frac{1}{\min})$

which gives

$\text{speed}=(3.8\times0.463\times10^{5-3})(\frac{\operatorname{km}}{\min }$

then, the result is

$\text{speed}=1.76\times10^2\frac{\operatorname{km}}{\min }$