PHYS101: Introduction to Mechanics

Example 1.1 Unit Conversions: A Short Drive Home

Suppose that you drive the 10.0 km from your university to home in 20.0 min. Calculate your average speed (a) in kilometers per hour (km/h) and (b) in meters per second (m/s). (Note: Average speed is distance traveled divided by time of travel).

Strategy

First we calculate the average speed using the given units. Then we can get the average speed into the desired units by picking the correct conversion factor and multiplying by it. The correct conversion factor is the one that cancels the unwanted unit and leaves the desired unit in its place.

Solution for (a)

(1) Calculate average speed. Average speed is distance traveled divided by time of travel. (Take this definition as a given for now - average speed and other motion concepts will be covered in a later module). In equation form,

$\text{average speed =} \, \dfrac{\text { distance }}{\text { time }}.$

(2) Substitute the given values for distance and time.

$\text{average speed =} \, \dfrac{10.0 \mathrm{~km}}{20.0 \mathrm{~min}}=0.500 \dfrac{\mathrm{km}}{\mathrm{min}}.$

(3) Convert $\mathrm{km} / \mathrm{min}$ to $\mathrm{km} / \mathrm{h}$ : multiply by the conversion factor that will cancel minutes and leave hours. That conversion factor is $60 \mathrm{~min} / \mathrm{hr}$. Thus,

$\text{average speed =} \, 0.500 \dfrac{\mathrm{km}}{\min } \times \dfrac{60 \mathrm{~min}}{1 \mathrm{~h}}=30.0 \dfrac{\mathrm{km}}{\mathrm{h}}.$

Discussion for (a)

To check your answer, consider the following:

(1) Be sure that you have properly cancelled the units in the unit conversion. If you have written the unit conversion factor upside down, the units will not cancel properly in the equation. If you accidentally get the ratio upside down, then the units will not cancel; rather, they will give you the wrong units as follows:

$\dfrac{\mathrm{km}}{\min } \times \dfrac{1 \mathrm{hr}}{60 \mathrm{~min}}=\dfrac{1}{60} \dfrac{\mathrm{km} \cdot \mathrm{hr}}{\min ^{2}},$

which are obviously not the desired units of km/h.

(2) Check that the units of the final answer are the desired units. The problem asked us to solve for average speed in units of km/h and we have indeed obtained these units.

(3) Check the significant figures. Because each of the values given in the problem has three significant figures, the answer should also have three significant figures. The answer 30.0 km/hr does indeed have three significant figures, so this is appropriate. Note that the significant figures in the conversion factor are not relevant because an hour is defined to be 60 minutes, so the precision of the conversion factor is perfect.

(4) Next, check whether the answer is reasonable. Let us consider some information from the problem - if you travel 10 km in a third of an hour (20 min), you would travel three times that far in an hour. The answer does seem reasonable.

Solution for (b)

There are several ways to convert the average speed into meters per second.

(1) Start with the answer to (a) and convert km/h to m/s. Two conversion factors are needed - one to convert hours to seconds, and another to convert kilometers to meters.

(2) Multiplying by these yields

$\text { Average speed }=30.0 \dfrac{\mathrm{km}}{\mathrm{h}} \times \dfrac{1 \mathrm{~h}}{3,600 \mathrm{~s}} \times \dfrac{1,000 \mathrm{~m}}{1 \mathrm{~km}},$

$\text { Average speed }=8.33 \dfrac{\mathrm{m}}{\mathrm{s}} \text {. }$

Discussion for (b)

If we had started with 0.500 km/min, we would have needed different conversion factors, but the answer would have been the same: 8.33 m/s.

You may have noted that the answers in the worked example just covered were given to three digits. Why? When do you need to be concerned about the number of digits in something you calculate? Why not write down all the digits your calculator produces? The module Accuracy, Precision, and Significant Figures will help you answer these questions.