## Question

###### 4. (0/1 Points] DETAILS PREVIOUS ANSWERS SBIOCALC1 5.1.007.MI.SA. MY NOTES PRACTICE ANOTHER This question has several...

4. (0/1 Points] DETAILS PREVIOUS ANSWERS SBIOCALC1 5.1.007.MI.SA. MY NOTES PRACTICE ANOTHER This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise The speed of a runner increased steadily during the first three seconds of a race. Her speed at half-second intervals is given in the table. Find lower and upper estimates for the distance that she traveled during these three seconds 10/0.5 1.0 1.5 2.0 2.5 3.0 v (ft/s) 0 6.2 10.8 14.9 18.1 19.4 20.2 Step 1 We will use either L6 or Rg for the upper and lower estimates Since the runner's speed is an increasing function, then L6 will give the lower estimate, and R6 will give the upper estimate X Submit Skir. you cannot come back

## Answers

Please refer below images for the solution.

2.0/2.5 3.0 Ito) 0.5 18.1 /19.4 (20.2 6.2/10 8/14.9 IV (ft/lo distance upper she To find lowes and estimates for teavelled during these 3 seconds Given to lowes estimate R6 upper estimate During each time interval of 0.5s the velocito ih approximately constant. case 1: For lowes estimate of distance we take the velocity at the beginning at the beginning of each time is terval (left end point) So lowel estinate for the distance travelled for each interval is Distance = 0.5x0=oft (form t-os to t = 0.55) Distance 0.5x6.2 = 3.1ft Distance 0 5X 10.8 = 5.4ft. Distance = 0.5x14.9 = 7.45ft Distance = 0.5X16.1 = 9.05ft Distance - 05x19.4 = 9.1ft so total distance = 0+3-1+5.4 +7.95 +9.05+9.7 (26) -34-1 ft Lowe estimate distance: 34.7ftcase 2: For uppel estonte distance we take the velocity at the end of each time interval (Right and point) So upper estimate for distance travelled will be Distance - 0.5x6.2 = 318t (from tos to t= 0.55) Distance = 0.5x/0.8 5.4ft Distance = 0.5X 14.9 - 1.45ft Distance = 0.5x = 9.05ft 18.1 = 9.7ft Distance - 0.5X 19.4 0.5 20.2 = lo.lft Distance - 34+ so total Distance = 5.4 + 7.45 +9.05 +9.7+ 101 44.8ft upper estimate distance - 44.8 ft, eft please upvote

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