The following graph shows the distance Thomas is from his home (in meters) for a period of 2 minutes. Determine the average rate
of change for the first 50 seconds, explain what this means in context of the problem. Then find the average rate of change for the 70 remaining seconds
The answer: <span>the average rate of change for the first 50 seconds: it is v= 100 m/ 50 s = 2 m/s </span><span>Thomas speed is 2 m in each second for the first 50 seconds.
</span><span>The average rate of change for the 70 remaining seconds: the time is between [70; 120] so the distance is d= 200m - 40m= 160m the average rate is v= 160m / 70 s = 2 . 28m /s
