SOLUTION: On a recent trip to Florida, we averaged 50 miles per hour. On the return trip we averaged 60 miles per hour. What do you think the average speed of the trip to Florida and back
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Question 68427This question is from textbook Intermediate Algebra
: On a recent trip to Florida, we averaged 50 miles per hour. On the return trip we averaged 60 miles per hour. What do you think the average speed of the trip to Florida and back was? Defend your position. Algebraically, determine the average speed of the trip to Florida and back. This question is from textbook Intermediate Algebra
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Distance(d)=rate(r) times time(t) or d=rt and t=d/r
Time down=d/50
Time back=d/60
Average Speed equals total distance divided by total time
Total distance=2d
Total time=d/50+d/60 so
Average Speed=2d/(d/50+d/60)
DENOMINATOR: LCM is 300: (d/50+d/60)=(6d+5d)/300=11d/300 now we have
Average Speed=2d/(d/50+d/60) =2d/(11d/300)
Average Speed=2d(300/11d)=600/11=54.545 mph---------------------answer
Defense: average speed is defined as total distance divided by total time and that's what we have calculated
