# Application of Integrals Check My Work

I would like someone to double-check my answers on this question:

River currents (miles per hour) at a certain location are given below. The current direction at this location was “from the north” during the time interval shown, and the current did not exhibit any severe fluctuations other than those shown in the chart.

Chart:
time of day: 6:00 6:10 6:20 6:30 6:40
speed (mph): 17 20 22 21 17

A. Using trapezoids, estimate the average river current speed from the north from 6:00 AM until 6:40 AM. (i got 0.333miles)

B. At approximately what time between 6:00 AM and 6:40 AM would you estimate that the river current had the average velocity? ( i got approx. 6:10)

C. A message in a bottle near this location is released at 6:00 AM. Assuming that the bottle travels along with the river’s current, approximately how far south will the bottle be at 6:40 AM? ( i got the distance traveled would be the 13.33 mi)

D. What is the average acceleration of this bottle from 6:00 AM to 6:40 AM? ( i got 0.025 mi/min^2)

Thank you for any help!

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1. a. how can the average be much less than the lowest river velocity?

b. Doesn't the data show at 6:10 the velocity was 20?

c. distance=avgvelocity*time
time= 2/3 hr So I don't know exactly what you used for avg velocity. See a) above.

d.avg acceleration=(vf-vi)/time
= (17-17)/time= zero

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2. I did
(0.5)(10)(0.283+0.333) + (0.5)(10)(0.333+0.367) + (0.5)(10)(0.367+0.350) + (0.5)(10)(0.350+0.283)
= 13.33 miles
Avg value: (13.33) / 40 = 0.333 mi/min

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posted by Pax
3. so 19.98 miles per hour (converting the mi/min to mph) would be my actual average i suppose whoops

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posted by Pax
4. Which would give B. as being 6:10, or maybe about 6:09 since the velocity is riiiight under 20

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posted by Pax
5. for c.) d=v*t where v=19.98 and t=2/3, 19.98*(2/3)= 13.32 (which is actually one decimal off of my answer oops)

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posted by Pax
6. I think the problem was I hadn't converted my units. Does this make sense now?

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posted by Pax

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