SOLUTION: 30 milSincePlease help me to solve the problem below. I have tried to solve it using the average speed formula S=D/T, but it did not work. The result I got make no sense (30=0).

Algebra ->  Average -> SOLUTION: 30 milSincePlease help me to solve the problem below. I have tried to solve it using the average speed formula S=D/T, but it did not work. The result I got make no sense (30=0).       Log On


   

Question 174775: 30 milSincePlease help me to solve the problem below. I have tried to solve it using the average speed formula S=D/T, but it did not work. The result I got make no sense (30=0).

A car went around a one mile track at a speed of 30 miles per hour. At what speed must the car travel on its second time around the track if the driver wants to average 60 miles per hour for the two mile trip?
Thank you very much in advance for your help.
Camelia
Found 2 solutions by stanbon, actuary:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A car went around a one mile track at a speed of 30 miles per hour. At what speed must the car travel on its second time around the track if the driver wants to average 60 miles per hour for the two mile trip?
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Time required to go one mile at 30mph:
t = d/r = 1mi/30 mph = (1/30)of an hour = 2 minutes
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Time required to go two miles at 60 mph:
t = d/r = 2mi/60 mph = (2/60) of an hour = 2 minutes
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So, for the car to average 60mph it would have to cover the 2 miles in 2 minutes.
But the car has already taken 2 minutes to go 1 mile at 30 mph.
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So there is no speed that the car can travel on the 2nd mile to average
60 mph for the 2-mile journey.
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Cheers,
Stan H.

Answer by actuary(112) About Me  (Show Source):
You can put this solution on YOUR website!
Since the car traveled the first lap (one mile) at 30 miles per hour, it took 2 minutes (T= D/S = 1 mile/30 mph = 1/30 hour = 2 minutes)to finish the first lap. Since the goal is is to cover the track twice (2 miles) at an average speed of 60 miles per hour, the car will take T = D/S
2 miles/60 mph = 1/30 hour = 2 minutes for both laps.
I'm getting the same nonsensical answer as you did. Are you sure that you transcribed the problem correctly?