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by smallsorrow » Sat Mar 14, 2009 4:35 am
A driver completed the first 20 miles of a forty miles trip at an avg. of 50 mph. At what avg. speed must the driver complete the last 20 miles to achieve an avg. speed of 60mph for the entire 40 miles trip?

Is there a particular formula?

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by mals24 » Sat Mar 14, 2009 5:05 am
Ok i think you've posted in the wrong section. This is the DS section :)

Average Speed = Total Distance/Total Time.
Total time = Total distance/average speed.

To find the average speed of the last 20 miles, we need the total time it took to cover the last 20 miles.

Time taken to cover last 20 miles = time taken to cover 40 miles - time taken to cover first 20 miles.

= 40/60-20/50
=4/15 hours.

Hence average speed of the last 20 miles = 20/(4/15) = 75mph.
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by cramya » Sat Mar 14, 2009 5:18 am
Yes there is a formula but what Mals did works just fine also.


If 2 equal distances are covered at different speeds then

average speed = 2 * speed1 * speed2 /(speed1 + speed2)

Given three equal distances are covered by different speeds lets say x,y,z

then average speed = 3xyz / xy + yz + zx


Using formula 1 above

2 * 50 * speed2 / (50+ speed2) = 60

100speed2 = 3000 + 60speed2

speed2 = 75


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by mals24 » Sat Mar 14, 2009 5:24 am
cramya wrote: average speed = 2 * speed1 * speed2 /(speed1 + speed2)

Given three equal distances are covered by different speeds lets say x,y,z

then average speed = 3xyz / xy + yz + zx
Hey cramya thanks for sharing the formula :) Im looking at it for the first time.
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by cramya » Sat Mar 14, 2009 5:25 am
No probs Mals. Hope u r doing well...

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by logitech » Sat Mar 14, 2009 9:13 am
Here comes the logic behind the formula
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by logitech » Sat Mar 14, 2009 9:14 am
So using the formula above:

2/Vav = 1/v1 + 1/v2

2/60 = 1/50 + 1/X

1/x = 1/30-1/50

x=75
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