course Phy 201
I apologize that my work is out of order. I had done this work previously and saved it to my computer and I just got my internet fix yesterday.
Assign. 4, Version 7A 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?
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First we have to find the final distance. We do this by solving for x in the following equation provided in Chapter 2 Pg. 27 of the textbook.
X = 83 m + 4m/s(8s) + ½(.9 m/s^2)(8s)^2
= 83 m +32 m/s^2 +28.8 m/s^2
= 143.8 m
By solving for X we can now solve for the velocity at t=8 from another equation provided in chapter 2.
V^2 = (4m/s)^2 + 2(.9 m/s^2)(143.8 m – 83m)
= 16 m^2/s^2 + (1.8 m/s^2)(60.8 m)
= 16 m^2/s^2 + 109.44 m^2/s^2
Next take the square root of both sides and we get,
V = 11.2 m/s
- Therefore, the velocity at clock time t=8 seconds is 11.2 m/s.
- The average velocity is
vAve = `ds/`dt
= 11.2 m/s + 4 m/s / 2
= 22.4 m/s
- At clock time t=8 seconds the bee is 60.8 m from the observer; 143.8m-83m = 60.8m
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Excellent solution, but be sure you understand my additional comments:
The equations given in the text are consistent with the equations as I give them. The equations I use are for motion on an interval and are more specific about the difference between clock time t and a time interval `dt. As long as they are used correctly, the text equations are fine.
The text equations have advantages and disadvantages, as do the equations as I state them. You understand everything well enough to use either, or to use the two versions interchangeably.
It is still necessary to be able to reason out situations involving motion and represent them on graphs. The equations can be used to get the same results, but not with the same degree of understanding.
This entire problem can be reasoned out from the definitions of average velocity and average acceleration. You get change in velocity by multiplying acceleration by time interval, add this to the initial velocity to get the final velocity, average initial and final velocities to get average velocity and multiply by time interval to get displacement. Then you add the displacement to the intial position to get the final position.