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### Distance, Time and Speed

```
Date: 8/1/96 at 23:16:42
From: Anonymous
Subject: Distance, Time and Speed

Kassim took 45 mins to drive to the airport. If he drove continuously
at an average speed of 76 km/h, what distance did he drive?
```

```
Date: 9/1/96 at 17:57:59
From: Doctor Jerry
Subject: Re: Distance, Time and Speed

The standard meaning of average speed for an automobile driving from
point u at time U to point v at time V is AvSp=(v-u)/(V-U).  The
problem statement above gives AvSp and V-U = 45/60 = 3/4, so you can
solve for v-u.

I find 76 = (v-u)/(3/4).  So, v-u = 76*(3/4) = 57km.

I hope this is clear.  I wanted to be sure both of us understand the
same thing by average speed.

1. First, when I calculated V-U = 45/60, I was substituting into the
formula AvSp = (v-u)/(V-U) for average speed.  I took point AvSp = 76,
the starting time U to be 0, and the ending time V to be 45min = 45/60
hour.

2.  If a person drives from point u to point v at a variable speed and
the clock changes from U to V during the drive, the average speed
formula gives the constant speed at which a person must drive from
time U to time V so that this person travels exactly the same distance
as the person who drives at a variable speed.

For example, if someone drives 30km during a 1/2 hour time and does it
by first driving very slowly and, towards the end of the time, very
fast, his or her average speed is (v-u)/(V-U) = 30/(1/2) = 60 km/hr.
If a second person drives for 1/2 hour at the constant speed of 60
km/hr, he or she will have driven the same distance, namely,
60*(1/2) = 30 km.

-Doctor Jerry,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
Middle School Algebra

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