course PHY 231
7/9 00:34
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?v_1 = 4m/s
s_1 = -83m
t_1 = 3sec
t_2 = 8sec
a_avg = 0.9m/sec^2
'dt = 5sec
v_2 = ?
s_2 = ?
v_avg = ?
a_avg * 'dt = 'dv
'dv = (0.9m/sec^2) * 5sec = 4.5m/sec
'dv = v_2 - v_1
v_2 = 'dv + v_1 = 4.5m/sec + 4m/sec = 8.5m/sec
v_avg = 0.5(v_2+v_1) = 0.5(4m/sec+8.5m/sec) = 6.25m/sec
v_avg = 'ds/'dt
'ds = v_avg*'dt = (6.25m/sec)*8sec = 50m
'ds = s_2 - s_1
s_2 = 'ds + s_1 = 50m - 83m = -33m
v_2 = 8.5m/sec
v_avg = 6.25m/sec
s_2 = -33m"
Very good.
At the end you should interpret your results to specifically answer the question. THe bee is moving at 8.5 m/s and is 33 m from the observer.
&#Good work. See my notes and let me know if you have questions. &#
#$&*