Set 1 Problem number 7

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

An object is observed for a total time of 50.3131 seconds. During that time it moves a distance of 9.007 meters. What therefore is its average velocity?

Solution

9.007 meters in 50.3131 seconds is 9.007/ 50.3131 meters in 1 second.

• The answer is therefore ( 9.007 meters) / ( 50.3131 seconds) = 5.586 meters per second.

Generalized Solution

Generalized Response: Average velocity is average displacment per unit of time.

• If displacement is `ds and the time interval is `dt, then `ds / `dt is the average displacement per time unit.
• Thus vAve = `ds / `dt.

We also recognize this as the average rate at which position changes.

In this problem we simply think in terms of ratios. 

• We 'feel' average velocity as the ratio of displacement to elapsed time. 
• Just as we know to divide the money we earn by the number of hours it takes to earn it, when we are familiar with the concept of velocity we don't have to quote equations--we simply know to divide displacement by elapsed time.

Explanation in terms of Figure(s), Extension

We can again use the triangular diagram to depict the relationship among `ds, `dt and vAve.

• In this case we can use vAve = `ds / `dt.

This in fact is usually taken as the definition of average velocity. It coincides with the concept of average velocity as average rate of change of position.

Figure(s)