Question

If you travel from Yakima to Ellensburg (Yakima to Ellensburg is 50 miles) with a speed of 60 miles/hour for half of the

Answer 1

Answer:

$\displaystyle \frac{480}{7}\approx 68.6\; \rm mph$.

Explanation:

The average speed of an object is equal to total distance over total time.

• Distance traveled: $\rm 50 \; mi$.

How much time is taken? This trip is divided into two halves, each of distance $\displaystyle \frac{50}{2} = 25\;\rm mi$.

Time spent on the first half of the trip:

$\displaystyle t_1 = \frac{s_1}{v_1} = \frac{25}{60} = \frac{5}{12}\; \rm hours$.

Similarly, time spent on the second half of the trip:

$\displaystyle t_2 = \frac{s_2}{v_2} = \frac{25}{80} = \frac{5}{16}\; \rm hours$.

In total:

$\displaystyle \frac{5}{12} + \frac{5}{16} = \frac{35}{48} \; \rm hours$.

Average speed:

$\begin{aligned} \text{Average speed} &= \frac{\text{Total Distance}}{\text{Total Time}}\\ &= 50 \left/\frac{35}{48}\right.\\ &= 50 \cdot \frac{48}{35} \\&= \frac{480}{7}\approx 68.6\; \rm mph \end{aligned}$.

This value turned out to be slightly different from the average of the speed during the two halves of the journey. The reason is that the object traveled at each speed for a different amount of time. It spent more time at the slower speed, which gives that speed a greater weight in the average. That explains why the average speed is closer to $\rm 60\; mph$ rather than $\rm 80\; mph$.