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===== Example: Comparing two ways of calculating the average velocity =====

You have learned about [[:183_notes:displacement_and_velocity|two ways of computing the average velocity]]. The //arithmetic// average is an approximation and it can be a poor one. Consider the driving from East Lansing to Chicago (222 miles or 358 km). To get to Chicago, you drive at 55.0 mph (24.6 $\dfrac{m}{s}$) for 1 hour and 66.8 mph (29.9 $\dfrac{m}{s}$) for 2.5 hours. Compare the average velocity to the //arithmetic// average velocity.

==== Setup ====

You will compare the two ways of computing the average velocity using the information provided and any information that you can collect or assume.

=== Facts ====

  * The distance from East Lansing to Chicago is 3.58$\times10^5m$.
  * For the first hour (3600 s), you drive at 24.6 $\dfrac{m}{s}$.
  * For the next 2.5 hours (9000 s), you drive at 29.9 $\dfrac{m}{s}$.

=== Lacking ===

  * Information about stops for gas, breaks, etc. are not known.

=== Approximations & Assumptions ===

  * You drive straight through with no breaks.
  * You use cruise control and do not deviate from the above speeds.
  * The problem can be considered to be in "one dimension" (along the road to Chicago).

=== Representations ===

  * The average velocity is given by $v_{avg,x} = \dfrac{\Delta x}{\Delta t}$.
  * The //arithmetic// average velocity is an approximation to the average velocity and is given by $v_{avg,x} \approx \dfrac{v_i + v_f}{2}$.

==== Solution ====

For this situation, the average velocity can be computed,

$$v_{avg,x} = \dfrac{\Delta x}{\Delta t} = \dfrac{3.58\times10^5m}{3600 s + 9000 s} = 28.4 \dfrac{m}{s}$$

You can compare that to the //arithmetic// average velocity,

$$v_{avg,x} \approx \dfrac{v_i + v_f}{2} = \dfrac{24.6 \dfrac{m}{s} + 29.9 \dfrac{m}{s}}{2} = 27.3 \dfrac{m}{s}$$

You can see that the //arithmetic// average is (in this case) less than the average velocity. It also under-predicts how far you would have driven,

$$\Delta x = v_{avg,x} \Delta t =  27.3 \dfrac{m}{s} (3600s+9000s) = 3.43\times10^5 m = 343 km$$

which is leaves you at the "outskirts" of Chicago (about 15 $km$ away).