Chapter 5.4, Problem 33IP

To determine

To find: The order of group of rates from least to greatest.

Answer to Problem 33IP

The order of rates from least to greatestis $\text{7}\text{yd/min}<\text{500}\text{m/h}<\text{6}\text{in}\text{./s}$ .

Explanation of Solution

Given information: The given rates are $\text{500}\text{m/h}$ , $\text{7}\text{yd/min}$ and $\text{6}\text{in}\text{./s}$ .

Calculation:

For comparing different distance, convert the rates to the same basic unit.

To convert meter and yards into inches, use $\text{1}\text{m}=\text{39}\text{.37}\text{in}$ and $\text{1}\text{yd}=\text{36}\text{in}$ .

To convert hours and minutes to seconds,use $\text{1}\text{h}=\text{3600}\text{sec}$ and $\text{1}\text{min}=\text{60}\text{sec}$ .

The rate $\text{500}\text{m/h}$ into inches per seconds can be converted as:

  $\begin{array}{c}\text{500}\text{m/h}=\frac{\text{500}\text{m}}{\text{1}\text{h}}\cdot \frac{\text{39}\text{.37}\text{in}}{\text{1}\text{m}}\cdot \frac{\text{1}\text{h}}{\text{3600}\text{sec}}\\ =\text{5}\text{.47}\text{in/sec}\end{array}$

The rate $\text{7}\text{yd/min}$ into inches per seconds can be converted as:

  $\begin{array}{c}\text{7}\text{yd/min}=\frac{\text{7}\text{yd}}{\text{1}\text{min}}\cdot \frac{\text{36}\text{in}}{\text{1}\text{yd}}\cdot \frac{\text{1}\text{min}}{\text{60}\text{sec}}\\ =\text{4}\text{.2}\text{m/sec}\end{array}$

Compare the calculated value with 26inches per seconds.

  $\text{7}\text{yd/min}<\text{500}\text{m/h}<\text{6}\text{in}\text{./s}$

Therefore, the order of rates from least to greatest as required is $\text{7}\text{yd/min}<\text{500}\text{m/h}<\text{6}\text{in}\text{./s}$ .