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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.

I can add some comments:  

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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