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--- 183_notes:examples:averagevelcompare [2014/07/10 20:11] – [Setup] caballero
+++ 183_notes:examples:averagevelcompare [2014/07/10 20:23] (current) – [Solution] caballero
@@ Line 6: / Line 6: @@
 ==== 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 ====
@@ Line 11: / Line 13: @@
   * 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 66.8 $\dfrac{m}{s}$.
+  * For the next 2.5 hours (9000 s), you drive at 29.9 $\dfrac{m}{s}$.
 === Lacking ===
@@ Line 29: / Line 31: @@
 ==== Solution ====
-==== 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).