Finding upper and lower estimates

[Mr.0]

I really need help with this question. I given my resulting answers but they don't seem to be right. I used recatngle area methods to try and solve them.

Roger runs a marathon. His friend Jeff rides behind him on a bicycle and clocks his speed every 15 minutes. Roger starts out strong, but after an hour and a half he is so exhausted that he has to stop. Jeff's data follows.
Time since start (min) 0 15 30 45 60 75 90
Speed (mph) 12 11 10 10 8 7 0

Assuming that Roger's speed is never increasing, give upper and lower estimates for the distance Roger ran during the first half hour.
upper = 15*(12+11)=345, lower= 15*(10+11)=315
Give upper and lower estimates for the distance Roger ran in total during the entire hour and a half.
upper= 15*(12 + 11 + 10 + 10 + 8 + 7 )=870, lower= 15*(11+10 +10 +8 +7 +0 )690
How often would Jeff have needed to measure Roger's speed in order to find lower and upper estimates within 0.1 mile of the actual distance he ran?
every abs(f(a)-f(b))delta(t)=.1=.1/12

 

[wjm11]

I given my resulting answers but they don't seem to be right. I used recatngle area methods to try and solve them.

Roger runs a marathon. His friend Jeff rides behind him on a bicycle and clocks his speed every 15 minutes. Roger starts out strong, but after an hour and a half he is so exhausted that he has to stop. Jeff's data follows.
Time since start (min) 0 15 30 45 60 75 90
Speed (mph) 12 11 10 10 8 7 0

Assuming that Roger's speed is never increasing, give upper and lower estimates for the distance Roger ran during the first half hour.
upper = 15*(12+11)=345, lower= 15*(10+11)=315

Your first problem is that you have ignored the units stated in the problem.
The time is given in minutes; the speed is given in miles/hour.

You need to adjust one or the other to make them compatible. I suggest changing your time into hours:

15 min = .25 hr

Now let’s look at the first two data points (the first 15 minute interval). If Jeff ran at 12 mph the entire time (and only dropped down to 11 mph at the last second), then his distance would be:

d = vt
d = (12 mi/hr)(.25 hr) = 3 mi

On the other hand, if Jeff dropped down to 11 mph just a second after his speed of 12mph was recorded, then his distance is only:

d = (11 mi/hr)(.25 hr) = 2.75 mi

These are the upper and lower bounds. Apply this to the rest of the data.