SOLUTION: Donna drove half the distance of a trip at 40 mi/h. At what speed would she have to drive for the rest of the distance so that the average speed for the entire trip would be 45 mi/

Question 247408: Donna drove half the distance of a trip at 40 mi/h. At what speed would she have to drive for the rest of the distance so that the average speed for the entire trip would be 45 mi/h?
Found 2 solutions by oberobic, ankor@dixie-net.com:
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
We know that:
d = rt, where d = distance, r = rate, t = time.
.
We are told she drove 1/2d at r=40. We are not told how long it took.
.
We want to find how fast she needs to drive for her average speed to be 45.
.
The obvious answer is 50, since (40+50)/2 = 45.
.
In other words, she drove 1/2d * 40 + 1/2d * 50 = 20 + 25 = 45.
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The problem is that time has been ignored with this solution.
.
Assume the distance to be traveled is 80 miles.
At 40 mph, the first half of the trip will take 1 hr @ 40 mph.
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To average 45 mph for the whole 80 miles, the total time is:
80 = 45t
45t = 80
t = 80/45 = 16/9 = 1 + 7/9 hrs.
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Since the first hour is gone, the remaining distance must be traveled in 7/9 hr.
40 = r * t = r * 7/9
360 = 7r
7r = 360
r = 51.428
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Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
half the distance of a trip at 40 mi/h.
At what speed would she have to drive for the rest of the distance so that the average speed for the entire trip would be 45 mi/h?
:
Let s = speed required on the 2nd half to average 45 mph for the trip
Let d = total distance of the trip
:
Write time equation; time = dist/speed
+ =
multiply equation by 360s to get rid of the denominator:
9s(.5d) + 360(.5d) = 8s(d)
:
4.5sd + 180d = 8sd
:
Divide thru by d
4.5s + 180 = 8s
:
180 = 8s - 4.5s
:
180 = 3.5s
s =
s = 51.43 mph
:

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