Problem: At the rate of 8 meters/second per second, by how much will an the velocity of an object initially traveling at 8.5 meters/second change in 13 seconds? How fast will the object the moving at the end of the 13 seconds? What will be the average velocity of the object? How far will the object have moved in 13 seconds?
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Solution: If every second velocity increases by 8 meters per second, then in 13 seconds it will increase by 786.5 meters per second. Starting at 8.5 meters per second, this increase will result in a velocity of 112.5 meters per second. The average velocity will be ( 8.5 + 112.5)/2 meters per second = 60.5 m/s. At this average velocity, the distance in 13 seconds will be ( 60.5)( 13) meters.
Generalized Response: If the rate at which velocity changes is denoted a , then in time interval `dt the velocity changes by a `dt. If the velocity starts at v0 and increases by a `dt, the velocity attained will be vf = v0 + a `dt. Since the rate at which velocity changes is constant, the average velocity will be the average of the initial and final velocities: vAve = ( v0 + vf ) / 2, where vf = v0 + a `dt (as above). The displacement will therefore be `ds = vAve * `dt, using the vAve just obtained. This sequence of calculations will get us to `ds. We could obtain a single expression for `ds: substituting vf = v0 + a `dt into vAve = (v0 + vf) / 2 we get vAve = (v0 + [v0 + a `dt] ) / 2. Simplifying this expression we have vAve = (2 v0 + a `dt) / 2 = v0 + a `dt / 2. Using this expression in `ds = vAve * `dt we obtain `ds = (v0 + a `dt / 2) * `dt = v0 `dt + a `dt^2 / 2.
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Figure description:
The 'blue' relationship in the first figure below shows what we know, that the displacement can be obtained from average velocity and time interval:
`ds = vAve * `dt.
. The information given does not include average velocity. However average velocity can be obtained for a uniform acceleration situation if initial and final velocities are known, as they are here. The 'black' relationship shows that average velocity is obtained from v0 and vf. The relationships among various variables, including the 'payoff' relationship `ds = (v0 + vf) / 2 * `dt, are shown in red.
We characterize uniform acceleration by the five parameters `ds, `dt, v0, vf and a. The present sequence of problems and associated figures has demonstrated two fundamental relationships among `ds, `dt, v0, vf and a:
`ds = (vf + v0) / 2 * `dt
and
a = (vf - v0) / `dt.
The five parameters are all we need to completely characterize any uniform acceleration situation. In fact we only need three of these quantites; it we know the values of any three we can find the values of the other two using the two equations listed above. Two more equations,
vf^2 = v0^2 + `a `ds
and
`ds = v0 `dt + .5 a `dt^2, can be derived from the first two and are also very useful. The second figure below outlines the derivation of these equations.
The remaining two figures show how we might proceed when we have values for v0, vf and `dt, and when we know v0, a and `ds.
If we know v0, vf and `dt, we can proceed in one of two directions. Starting with v0 and vf, we can find either vAve (the 'black' path upward from vf and v0) or `dv (the 'blue' path downward from vf and v0). When we find vAve, we can put it together with `dt to find `ds (the 'red' path). When we find `dv, we put it together with `dt to get a.
If we know v0, a and `ds, there is no way to directly reason out vAve or `dv, or anything else. We simply have to either set up a system of simultaneous equations, or we use E3. It turns out that E3 is much easier. We put all our information into E3 and we solve the resulting equation to get vf. This is indicated by the three 'blue' lines connecting v0, a and `ds to vf. Having obtained vf, we can put it together with v0 to get vAve. Then, known that vAve, `ds and `dt for a 'triangle' we use vAve and `ds to find `dt.
These relationships will be explored more fully in your text. They are included here for your information.
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vf_and_dt_from_v0_a_and_ds.gif
