Question 283355: A dirt biker must circle a 5-mile track twice. His mean speed must be 60 mph. On his first lap, he averaged 30 mph. How fast must he travel on his second lap in order to qualify?
a)30 m.p.h. b) 60 m.p.h. c) 90 m.p.h. d) 120 m.p.h. e) none of these
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! Your answer is selection E (none of the above).
He needs to do 10 laps at an average speed of 60 miles per hour.
Since Rate * Time = Distance, this means that 60 * T = 10 because:
Average rate needs to be 60 miles per hour.
2 laps at 5 miles per lap = total of 10 miles.
Solve for T and you get T = 10/60 = 5/30.
On the first lap, he traveled at 30 miles per hour to cover 5 miles.
Since R*T = D, this means that:
30*T = 5 which means that T = 5/30.
Since he used up all of his available time on the first lap, he will never be able to average 60 miles per hour on the second lap because he has 0 time left in which to do it in.
If you want to use the Rate * Time = Distance Formulas a different way, it would come out like this:
30*T1 = 5 where T1 is the time it takes for him to do the first lap.
60*(T1 + T2) = 10 here T1 is the time it takes for him to do the first lap and T2 is the time it takes for him to do the second lap.
Since 30*T1 = 5, then T1 = 5/30 so we can susbtitute for T1 in the second equation to get:
60*(5/30 + T2) = 10
We divide both sides of this equation by 60 to get:
5/30 + T2 = 10/60
We subtract 5/30 from both sides of this equation to get:
T2 = 10/60 - 5/30
We multiply 5/30 by 2/2 to get 10/60
Out equation becomes:
T2 = 10/60 - 10/60 = 0
In order for him to average 60 miles per hour overall, T2 has to be 0 which is impossible.
If x = rate he requires on the second lap, then formula for second lap becomes:
x*T2 = 5
If we divide both sides of this equation by T2, then we get x = 5/T2.
Since T2 = 0, then this equation is invalid and there is no value of x that will satisfy it.
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