course Phy 121
A bee is making a beeline for its hive. Its velocity is measured at a distance of 75 meters from teh observer and again at distance 145 meters from the observer. The clock times at these two positions are t = 9 sec and t = 16 sec, and the measured velocities are 3 m/sec and 13.5 m/sec. What is its average velocity during this time? What is its average acceleration during this time? It is possible that the acceleration is uniform?First of all to find average velocity:
vAve = (Vf-Vo)/2 = (13.5-3)/2 = 5.25 m/sec
&#This represents the difference of the two velocities divided by 2. To average two quantities you don't subtract and divide by 2, you add and divide by 2.Had you added and divided by 2 you would have obtained the average of the initial and final velocities.However this is not necessarily the average velocity. Average velocity is the average rate of change of position with respect to clock time. &#
vAve = 'ds/'dt = (145m-75m)/(16s-9s) = 10 m/sec
This is correct.
I include both of these because I initially thought that the answer provided by the first formula was correct; however, (as best I can figure) these are not the initial and final velocities. They are just the velocities I am provided with. Therefore, I am going to conclude that 10 m/sec is the correct average velocity.
Finding average acceleration:
aAve = 'dV/'dt = (13.5-3)/(16-9) = 1.5 m/sec
good
Is the acceleration uniform?
I am going to say yes, based upon the fact that the time difference is 7 seconds. The average acceleration (1.5 m/sec) over the time period of 7 seconds (1.5*7) gives a result of 10.5, which is very close to the average velocity."
&#Average velocity is defined as average rate of change of position, vAve = `ds / `dt.Average velocity is equal to the average of initial and final velocities, (vf + v0) / 2, if acceleration is uniform. If average velocity is not equal to the average of initial and final velocities, then acceleration is not uniform. This information can be used to determine whether or not acceleration is uniform. What is your conclusion? &#
&#Please respond with a copy of this document, including my comments. Insert your revisions and/or questions and mark them with &#&#. &#
course Phy 121
A bee is making a beeline for its hive. Its velocity is measured at a distance of 75 meters from teh observer and again at distance 145 meters from the observer. The clock times at these two positions are t = 9 sec and t = 16 sec, and the measured velocities are 3 m/sec and 13.5 m/sec. What is its average velocity during this time? What is its average acceleration during this time? It is possible that the acceleration is uniform?First of all to find average velocity:
vAve = (Vf-Vo)/2 = (13.5-3)/2 = 5.25 m/sec
&#This represents the difference of the two velocities divided by 2. To average two quantities you don't subtract and divide by 2, you add and divide by 2.Had you added and divided by 2 you would have obtained the average of the initial and final velocities.However this is not necessarily the average velocity. Average velocity is the average rate of change of position with respect to clock time. &#
vAve = 'ds/'dt = (145m-75m)/(16s-9s) = 10 m/sec
This is correct.
I include both of these because I initially thought that the answer provided by the first formula was correct; however, (as best I can figure) these are not the initial and final velocities. They are just the velocities I am provided with. Therefore, I am going to conclude that 10 m/sec is the correct average velocity.
Finding average acceleration:
aAve = 'dV/'dt = (13.5-3)/(16-9) = 1.5 m/sec
good
Is the acceleration uniform?
I am going to say yes, based upon the fact that the time difference is 7 seconds. The average acceleration (1.5 m/sec) over the time period of 7 seconds (1.5*7) gives a result of 10.5, which is very close to the average velocity."
&#Average velocity is defined as average rate of change of position, vAve = `ds / `dt.Average velocity is equal to the average of initial and final velocities, (vf + v0) / 2, if acceleration is uniform. If average velocity is not equal to the average of initial and final velocities, then acceleration is not uniform. This information can be used to determine whether or not acceleration is uniform. What is your conclusion? &#
&#Please respond with a copy of this document, including my comments. Insert your revisions and/or questions and mark them with &#&#. &#