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
looks good to me.
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
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.
A. To estimate the average river current speed from 6:00 AM until 6:40 AM using trapezoids, you need to calculate the area under the speed-time graph. Start by dividing the time interval (6:00 AM until 6:40 AM) into smaller sub-intervals, each corresponding to the time between consecutive data points on the chart.
In this case, we have 5 data points – one every 10 minutes from 6:00 AM to 6:40 AM. So, we can divide the total time interval into 5 sub-intervals of 10 minutes each.
Next, calculate the area of each trapezoid formed by adjoining sub-intervals using the following formula:
Area = (sum of the two parallel sides) x (height) / 2
For the first trapezoid (6:00 AM to 6:10 AM):
Area = (17 + 20) x (10 minutes) / 2 = 185
Similarly, calculate the area for the other trapezoids:
(6:10 AM to 6:20 AM): Area = (20 + 22) x (10 minutes) / 2 = 210
(6:20 AM to 6:30 AM): Area = (22 + 21) x (10 minutes) / 2 = 215
(6:30 AM to 6:40 AM): Area = (21 + 17) x (10 minutes) / 2 = 190
Finally, add up the areas of all the trapezoids and divide by the total time (40 minutes) to find the average:
Average speed = (185 + 210 + 215 + 190) / 40 = 800 / 40 = 20 miles/hr
So, your estimate of approximately 20 miles/hr for the average river current speed from 6:00 AM until 6:40 AM is correct.
B. To estimate the time at which the river current had the average velocity, you need to consider the midpoint of the time interval during which the average speed occurred.
The total time interval is from 6:00 AM to 6:40 AM, which is a 40-minute duration. The average time can be estimated as the midpoint of this interval.
Average time = (40 minutes / 2) + 6:00 AM = 20 minutes + 6:00 AM = 6:20 AM
So, your estimate of approximately 6:20 AM for the time at which the river current had the average velocity is correct.
C. To estimate how far south the bottle will be at 6:40 AM, you need to multiply the average speed (which we found to be 20 miles/hr) by the total time interval.
Total time interval = 40 minutes = 40/60 hours = 2/3 hours
Distance traveled = Average speed x Total time interval = 20 miles/hr x 2/3 hours = 40/3 miles
So, your estimate of approximately 40/3 miles south for the bottle's location at 6:40 AM is correct.
D. The average acceleration can be calculated by dividing the change in velocity by the change in time.
In this case, the bottle's velocity remains constant at 20 miles/hr throughout the entire time interval (from 6:00 AM to 6:40 AM). Since there is no change in velocity, the average acceleration is 0 miles/hr^2.
So, your estimate of 0 mi/hr^2 for the average acceleration of the bottle from 6:00 AM to 6:40 AM is correct.
Overall, your answers are correct. Well done!