1.) Your velocity is given by v(t) = t^2 + 6 in m/sec, with t in seconds.
Estimate the distance, S, traveled between t= 0 and t = 5.
Use an overestimate with data every one second.
The distance is approximately ___________ m.
Please show work and explain. I'm so confused!
For the next question, I just need someone to check my answers because I did a problem just like this, and the computer said it was wrong and this is my last chance to get it right so if I did it wrong, explain how to do it right.
2.) 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 follow:
Time since start (min) 0 15 30 45 60 75 90
Speed (mph) 14 13 12 12 10 9 0
In case the table messed up, the points are (0, 14) (15, 13) (30, 12) (45, 12) (60, 10) (75, 9) (90, 0)
a.) Assuming that Roger's speed is never increasing, give upper and lower estimates for the distance Roger ran during the first half hour.
Lower estimate = 375 miles
Upper estimate = 405 miles
b) Give upper and lower estimates for the distance Roger ran in total during the entire hour and a half.
Lower estimate = 840 miles
Upper estimate = 1050 miles
Thanks so much!
bbb
1.) To estimate the distance traveled between t = 0 and t = 5, you can use an overestimate with data taken every one second.
To do this, you will need to find the average velocity over small time intervals and then sum them up to get an estimate of the total distance.
First, divide the time interval from t = 0 to t = 5 into smaller intervals of 1 second each. So, you will have 5 small intervals: [0, 1], [1, 2], [2, 3], [3, 4], and [4, 5].
Next, calculate the average velocity for each interval by finding the value of the velocity function at the midpoint of each interval. The midpoint of each interval is the average of the two endpoints.
For example, for the interval [0, 1], the midpoint is (0 + 1)/2 = 0.5 seconds. Substituting this value into the velocity function v(t) = t^2 + 6, we get v(0.5) = (0.5)^2 + 6 = 6.25 m/s. Repeat this process for the other intervals.
The estimated distance traveled in each interval is the product of the average velocity and the width of the interval (1 second). So, for the interval [0, 1], the estimated distance is 6.25 m/s * 1 s = 6.25 m.
Finally, sum up all the estimated distances to get the total estimated distance traveled between t = 0 and t = 5.
Estimated distance = 6.25 m + (sum of estimated distances for the other intervals)
2.) For the second question, let's address each part separately.
a) To estimate the distance Roger ran during the first half hour, you can use the upper and lower estimates. The upper estimate corresponds to assuming Roger's speed decreases continuously at the highest rate observed during the given data. The lower estimate corresponds to assuming Roger's speed remains constant at the lowest value observed during the given data.
Looking at the given data, the lowest speed observed during the first half hour is 12 mph. So, the lower estimate for the distance Roger ran during the first half hour is:
Lower estimate = 12 mph * (30 min / 60 min) = 6 miles
Similarly, the upper estimate is the initial speed of 14 mph until the first speed measurement at 15 minutes. So, the upper estimate for the distance Roger ran during the first half hour is:
Upper estimate = 14 mph * (15 min / 60 min) = 3.5 miles
Therefore, the estimated distance Roger ran during the first half hour is between 3.5 miles and 6 miles.
b) To estimate the distance Roger ran during the entire hour and a half, you need to sum up the estimated distances for each time interval.
The interval [0, 15] has an estimated distance of 14 mph * (15 min / 60 min) = 3.5 miles.
The interval [15, 30] has an estimated distance of 13 mph * (15 min / 60 min) = 3.25 miles.
Similarly, calculate the estimated distances for the remaining intervals [30, 45], [45, 60], [60, 75], [75, 90] using the given data and the same method used above.
Finally, sum up all the estimated distances from the intervals to get the total estimated distance Roger ran during the entire hour and a half.
Lower estimate = sum of lower estimates for each interval
Upper estimate = sum of upper estimates for each interval
Based on your calculations, the lower estimate for the total distance is 840 miles and the upper estimate is 1050 miles.