You have learned about 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.

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

• 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 $\dfrac{m}{s}$.

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

• 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).

• 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 $v_{avg,x} \approx \dfrac{v_i + v_f}{2}$.

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).