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course phy121
11:11 pm 11/25/13
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?Points for the graph (3 sec , 4 m/sec) and (8 sec, x m/sec)
The slope is also the same as acceleration so find the slope
(x m/sec - 4 m/sec) / (8 sec - 3 sec) = .9 m/sec/sec
= .9 m/sec/sec = (x m/sec - 4 m/sec) / 5 sec
5 sec * .9 m/sec/sec = (x m/sec - 4 m/sec)
4.5 m/sec = x m/sec - 4 m/sec
8.5 m/sec = x m/sec
So x = 8.5. m/sec is the velocity at t = 8 sec which is also vf
Now we want the average velocity so (v0 + vf)/2
= so `vf =( 8.5 m/sec and + 4 m/sec )/2
= 6.25 m/sec is the average velocity
To find how far the bee is from the observer at t =8 you multiply the average velocity by 8 seconds so 6.25 m/sec * 8 sec = 50 m and then you take into consideration the original 83 meter distance from the problem getting:
50+83 = 133 meters, if the bee is travelling away from the observer and 83-50 = 33 meters if the bee is travelling toward the observer.
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