course Phy 121
Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.
A bee is making a beeline for its hive. Its velocity is measured at a distance of 83 meters from the observer, when clock time is t = 3 sec and its velocity is 4 m/s, and again at clock time t = 8 sec. It it accelerates uniformly at .9 m/s/s during the time interval, then what is its velocity at t = 8 sec, and its average velocity during this time interval? How far is the bee from the observer at t = 8 sec?
slope = (y2 - y1)/(x2 - x1) is the same as acceleration, 0.9m/s^2 = (vf - 4m/s)/(8s - 3s)
So (0.9m/s^2)(5s) = vf -4m/s
(4.5m/s)change in velocity = vf - 4m/s
vf = 4.5m/s + 4m/s = 8.5 m/s
Now we can calculate average velocity: vAve = (vf + vo)/2
vAve = (8.5m/s + 4m/s)/2 = (12.5m/s)/2 = 6.25 m/s
Finally, we answer the question, ""How far is the bee from the observer at t = 8 sec?""
A distance of 83 meters is given to us in the first sentence, but this is the distance of the observer from the bee. It has nothing to do with the distance of the bee from its hive, so it is irrelevant information.
Since 8 seconds is the final time, we can say that the position at t = 8s is the same as the distance traveled over the interval. The distance traveled or change in position can be solved this way:
`ds = vAve x `dt = (6.25m/s)(8s - 3s) = (6.25m/s)(5s) = 31.25 m "
This looks very good. Let me know if you have questions.