7.1.3 Average Speed and Distance Traveled:
         Better Estimates

Table 2   Speed of car
between two
traffic lights

Time	Speed (mph)	Speed (miles/min)
`0`	`0`	`0`
`0.5`	`8.51`	`0.142`
`1`	`14.0`	`0.233`
`1.5`	`15.0`	`0.250`
`2`	`15.0`	`0.250`
`2.5`	`15.0`	`0.250`
`3`	`18.2`	`0.303`
`3.5`	`24.5`	`0.409`
`4`	`30.0`	`0.500`
`4.5`	`31.6`	`0.526`
`5`	`31.0`	`0.517`
`5.5`	`30.7`	`0.511`
`6`	`31.0`	`0.517`
`6.5`	`31.5`	`0.525`
`7`	`30.0`	`0.500`
`7.5`	`25.1`	`0.419`
`8`	`20.0`	`0.333`
`8.5`	`17.7`	`0.295`
`9`	`17.0`	`0.283`
`9.5`	`16.2`	`0.269`
`10`	`15.0`	`0.250`
`10.5`	`13.8`	`0.230`
`11`	`13.0`	`0.217`
`11.5`	`12.7`	`0.211`
`12`	`11.0`	`0.183`
`12.5`	`6.40`	`0.107`
`13`	`0`	`0`

We would obtain a better estimate of the total distance if we divided the `13` minutes into half-minute intervals (see Table 2). We could then estimate the distance traveled in each half-minute interval as we did before, taking the speed to be constant for that half-minute. The improvement in the estimate arises from the fact that our intervals are shorter than before, so the assumption of constant speed is closer to being true.

Now the estimated distance traveled in the first half-minute interval from `t=0` to `t=0.5` is `vtext[(]0 text[)] times 0.5.` In the second half-minute interval, the estimated distance is `vtext[(]0.5 text[)] times 0.5`. In the `k`th half-minute interval, the estimated distance is `vtext[((]k-1text[)] times 0.5text[)] times 0.5`. In summation notation, the sum we need to calculate is

When we carry out this calculation, we find an estimated distance of `4.11`, slightly different from the estimate of `4.09` that we calculated from one-minute intervals.

 

Activity 3

Since multiplying by `0.5` is the same as dividing by `2`, the sum just calculated can also be written as

1. Explain why this is the same as

   ${\displaystyle \frac{1}{2}\sum _{k=1}^{26}v\left(\frac{k-1}{2}\right)\mathrm{.}}$

2. Which expression involves fewer calculations in its evaluation?

Comment on Activity 3

We can continue dividing the total time into more and more subintervals, in expectation that the resulting estimates would be better and better approximations of the total distance traveled. Here is the procedure: We divide the `13`-minute time interval into `n` subintervals of length `Delta t = 13//n` minutes and name the endpoints of the subintervals `t_0`, `t_1`, `t_2`, and so on, up to `t_n`. For each `k` from `1` to `n`, we estimate the distance traveled in the `k`th subinterval, the one from `t_(k-1)` to `t_k`, by assuming the speed throughout that subinterval is constant. That is, we assume `vtext[(]t_(k-1)text[)]` is the speed until time `t_k.` Then our estimate for the total distance traveled is

In Table 3 we list the results from our calculations so far and from evaluation of this summation formula for several larger values of `n`. We conclude that, to an accuracy of three decimal places, the distance traveled is `4.12` miles.

Activity 4

1. What is the average speed for the trip?

2. Revisit Checkpoint 1, and print a copy of the page. Add appropriate labels to the tick marks on the `v`-axis. Draw a horizontal line on the figure at the level of `v` equal to the average speed.

3. What is the area of the rectangle under your horizontal line and above the `t`-axis?

4. How do you think this area compares to the area under the speed curve and above the `t`-axis?

Comment on Activity 4

By the way, "area" has to be interpreted carefully here because our horizontal and vertical scales are so different. If we had made the scales the same, the picture would have the appearance of Figure 4. This figure obscures the features of the velocity curve that we wanted to see, but the true area you calculated in Activity 4 is the one shown in Figure 4.