Problem: If velocity increases linearly from 8 meters per second to 98 meters per second in 6 seconds, what is the average velocity, and what is the distance moved in the 6 seconds?
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Solution: Since the rate of velocity increase during the acceleration phase is constant, average velocity will be halfway between 8 and 98 meters per second, or 11.5 meters per second (this halfway velocity can be found as the average of the 8 and the 98 meters per second, or ( 8 + 98) / 2 = 11.5 meters per second). At 11.5 meters per second, in 6 seconds an object would move 69 meters.
Generalized Response: We know that displacement ds is equal to the product of average velocity vAve and time interval dt. We know dt, but we have two velocities, which we call v0 (for starting, or zero-time, velocity) and vf (for final velocity). We don't have vAve. However, knowing that velocity increases at a constant rate we can make the approximation that the average velocity is the average of the initial velocity v0 and the final velocity vf. The average of two numbers is obtained by adding them and dividing by 2. In this case the average velocity is vAve = (v0 + vf) / 2. We can now obtain the displacement: ds = vAve * dt = [ (v0 + vf) / 2 ] * dt.
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Figure description:
The figure below is slightly more complicated than the previous figures. This figure actually shows two relationships, one with black lines and one with blue. For the moment ignore the figure with blue lines and look at the relationship between v0, vf and vAve. This part of the figure shows that the quantities vf and v0 determine vAve; the equation below vAve shows the specific symbolic relationship.
Note that the title of the figure specifies that the acceleration is uniform. This means that velocity is changing at a constant rate, as opposed for example to faster and faster, slower and slower, or sometimes faster and sometimes slower. The relationship vAve = (v0 + vf) / 2 is precisely accurate whenever acceleration is uniform. When acceleration is not uniform the relationship is usually not completely accurate; however in some cases of nonuniform acceleration this expression for average velocity provides a reasonable approximation.
Now look at the 'blue' relationship. You should recognize this as the relationship you saw before between vAve, `dt and `ds. The only difference here is that we had to obtain vAve from other information, namely from v0 and vf.
So the figure shows that to obtain `ds, we use vAve and `dt in the usual manner; but first we must obtain vAve from v0 and vf.
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