Set 1 Problem number 2

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

Over a period of 2.079 seconds, the average rate at which the position of an object changes is 2.904 meters per second.   How far will the object move during   this time?

Solution

The average rate at which a quantity changes in time is equal to the change in that quantity divided by the time required for the change.  

• It follows that the change in the quantity is equal to the product of the rate and the time interval. 
• Thus the change in position is equal to the product of the 2.904 m/s rate and the 2.079 s time interval.
• Change in position is therefore

• position change = 2.904 m/s * 2.079 s = 6.037416 meters.

Generalized Solution

The average  time rate of change of position over a time interval is

• average rate = change in position / time interval.  

If we let s stand for position, v for rate and `dt for time interval we can write

• vAve = `ds / `dt. 

Thus

• `ds = vAve * `dt. 

We call the rate at which position changes the velocity.