Distance, Time and SpeedDate: 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/ |
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