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 approx. 20 miles/hr)

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 and 6:33)

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? ( Same as the distance in A.: 40/3)

D. What is the average acceleration of this bottle from 6:00 AM to 6:40 AM? (I got 0 mi/hr^2)


Please check, and give feedback

Answer 1

looks good to me.

Answer 2

A
(17+20)/2 + (20+22)/2 + (22+21)/2 + (21+17)/2 ] /4
= 20 sure enough
B
agree


C
if average current is 20 miles/hr and you drift for 40 min which is 2/3 of an hour you go 20 * 2/3 = 13.33 miles

D. time intervals are 10 min which is 1/6 hr delta t
first one +3/(1/6)
second +2/(1/6)
third -1/(1/6)
fourth -3/(1/6
av = (1/(4*6) ) (3 + 2 -1-4)
= (1/24)(0)
0 miles per hour squared

Answer 3

For D, isn't it supposed to be av = (1/(4*6) ) (3 + 2 -1-3) instead of av = (1/(4*6) ) (3 + 2 -1-4) ? Correct me if I'm wrong, I'm just confused about why it's a 4 instead of a 3.